· abcdefghijklmnopqrstuvwxyz. Preface T his text book is an initiative by the BOS Mathematics, Savitribai Phule Pune University. The syllabus of Savitribai Phule Pune University

abcdefghijklmnopqrstuvwxyz.

Savitribai Phule Pune University

A Text Book for S.Y.B.Sc./S.Y.B.A. Mathematics (2013 Pattern)

PAPER I-MT 211

MULTIVARIABLE CALCULUS I

Panel of Authors

Dr.Andhare P.G. (Convenor)Prof. Ghanwat A.JProf. Kumkar T.K

Editors

Dr.P.M. Avhad Dr.S.A.Katre

Conceptualized by Board of Studies (BOS) in Mathematics, Savitribai Phule Pune University,Pune.

Contents

Chapter 1. Limit and Continuity of Multivariable Functions 11.1. Functions of Several Variables 11.2. Limit and Continuity 5

Chapter 2. Partial Derivatives 132.1. Definition and Examples 132.2. Second Order Partial Derivative 192.3. Higher Order Partial Derivative 22

Chapter 3. Differentiability 25

Chapter 4. Extreme Values 47

Chapter 5. Multiple Integrals 605.1. Double Integrals over Rectangles 60

i


Preface

This text book is an initiative by the BOS Mathematics, Savitribai Phule Pune University.The syllabus of Savitribai Phule Pune University is always considered commendable, with a listof reputed books in the subject recommended for reference. Many times teachers face difficultyin doing justice to every aspect covered collectively by all these books. So, while preparing newsyllabus for S.Y.B.Sc./S.Y.B.A. it was thought the University should prepare textbooks for Math-ematics with the following objectives:1. Uniformising notations, definitions and to focus on important aspects of revised syllabus whichneed to be stressed and understood.2. Collecting all relevant topics, problems, questions prescribed in syllabus.3. Providing a ready reference to teachers and students.The book is written in accordance with the new prescribed syllabus of S.Y.B.Sc., S.Y.B.A. Math-ematics (2013 Pattern), for the Paper I - MT 211: Multivariable Calculus I (First Term). Thesyllabus deals with important topics in Mathematics, Vector Calculus.This book consists of a detail introduction at the beginning of each chapter, several illustrativeexamples and problems for practice with hints and solutions given at the end of each section. Aproper understanding of all the illustrative examples would go a long way in making the subjectfully comprehensible.In case of queries/suggestions, send an email to: sumati.munot @ gmail.comWe hope our endeavour will benefit both students and teachers.

-Authors

ii


Acknowledgment

We sincerely thank the following University authorities for their constant motivation, guidanceand valuable help in the preparation of this book.

• Dr. W. N. Gade, Hon. Vice Chancellor, Savitribai Phule Pune University, Pune.• Dr. V. B. Gaikwad, Director BCUD, Savitribai Phule Pune University, Pune.• Dr. K. C. Mohite, Dean, Faculty of Science, Savitribai Phule Pune University, Pune.• Dr. B. N. Waphare, Professor, Department of Mathematics, Savitribai Phule Pune Uni-

versity, Pune.• Dr. M. M. Shikare, Professor, Department of Mathematics, Savitribai Phule Pune Uni-

versity, Pune.• Dr. V. S. Kharat, Professor, Department of Mathematics, Savitribai Phule Pune Univer-

sity, Pune.• Dr. V. V. Joshi, Professor, Department of Mathematics, Savitribai Phule Pune University,

Pune.• Mr. Dattatraya Kute, Senate Member, Savitribai Phule Pune University; Manager, Sav-

itribai Phule Pune University Press.• All the staff of Savitribai Phule Pune University press.

iii

Syllabus:- Paper I MT 211: Multivariable Calculus I

(1) Limit and Continuity of Multivariable functions: . . . [06]1.1 Functions of several variables, graphs and level curves of function of two variables .1.2 Limit and continuity in higher dimensions .

(2) Partial Derivatives: . . . [04]2.1 Definition and examples .2.2 Second order partial derivative, the mixex derivative theorem .2.3 Partial derivatives of higher order.

(3) Differentiability: . . . [12]3.1 Differentiability, the increment theorem for functions of two variables(without proof).3.2 Chain rules for composite function.3.3 Directional derivatives, gradient vectors.3.4 Tangent planes, normal lines and differentials.

(4) Extreme Values: . . . [10]4.1 Extreme values, First derivative test and Second derivative test for local extreme

values.4.2 Lagrange’s multipliers method for finding extreme values of constraint function (One

Constraint)4.3 Taylors Formula for two variables.

(5) Multiple Integrals: . . . [16]5.1 Double Integral over rectangles, Fubini’s theorem for calculating double integrals

(Without proof)5.2 Double integrals in polar form.5.3 Triple integral in cylindrical and spherical coordinates.5.4 Triple integral in cylindrical and spherical coordinates.5.5 Substitution in multiple integrals, Application to are and volumes.

Text book: Prepared by the BOS Mathematics, Savitribai Phule Pune University, Pune.Recommended Book: Thomas’ Calculus’ 11th Edition, G.B. Thomas. Revised by MauriceD. Weir, Joel Hass and Frank R. Giordano. Pearson Education 2012. Articles: 14.1 to 14.10,15.1,15.3,15.4,15.6,15.7Reference Books:

(1) Basic Multivariable Calculus, J.E. Marsden, A.J. Tromba, A. Weinstein. Springer Verlag(Indian Edition).

(2) Shanti Narayan, R.K. Mittal, A Text-book of Vector Calculus, S. Chand and Company.(3) D.V.Widder, Advanced Calculus 2 (2 nd Edition)(4) T.M. Apostol. Calculus Vol.2 (2 nd Edition), John Wiley, Newyork (1967).

iv

CHAPTER 1

Limit and Continuity of Multivariable Functions

1.1. Functions of Several Variables

Recall for the function of one variable, the underlying set was R, the set of real numbers. Fortwo variables, corresponding role is played by the set R2, the cartesian product of R with itself.R2 = {(x, y)/x, y ∈ R} where R2 is represented by points in the plane. Similarly, points in spaceare represented by R3 = {(x, y, z)/x, y, z ∈ R} and Rn = {(x1, x2, . . . , xn)/x1, x2, . . . , xn ∈ R}.

Definition 1.1. Real Valued Functions of Two Independent Variables:Let D = {(x, y)/x, y ∈ R} ⊂ R2. A real valued function f on D is a rule that assigns a uniquesingle real number w = f(x, y) to each element in D.

Definition 1.2. Real Valued Functions of Three Independent Variables:Let D = {(x, y, z)/x, y, z ∈ R} ⊂ R3. A real valued function f on D is a rule that assigns a uniquesingle real number w = f(x, y, z) to each element in D.

Definition 1.3. Real Valued Functions of n Independent Variables:Let D = {(x1, x2, · · ·xn)/x1, x2, · · ·xn ∈ R} ⊂ Rn.A real valued function f on D is a rule thatassigns a unique single real number w = f(x1, x2, · · ·xn) to each element in D.

The set D is the domain of function. The set f(D) = {w ∈ R/w = f(x1, x2, · · ·xn)} is rangeof function. w is dependent variable (output variable) and x1, x2, · · ·xn are independent variables(input variables).

Example 1.1. Sketch the domain of f(x, y) =√y − x2. What is the range of function?

Solution: The domain D is the set of all pairs (x, y) in the plane for which√y − x2 is real

∴ y − x2 ≥ 0. ∴ y ≥ x2.Therefore, domain D is the set of points that lie above and on the parabola y = x2.The range of function is the set of all non-negative numbers w ≥ 0.

Example 1.2. What are the domain and range of f(x, y) =1

xy.

Solution: The domain D is the set of all pairs in the plane for which xy 6= 0.The range of function is the set {w ∈ R/w ∈ (−∞, 0) ∪ (0,∞)} .

Example 1.3. What are the domain and range of f(x, y) =xy

(x2 − y2).

Solution: The domain of function consists of all points in the plane except the points for whichx2 − y2 = 0.i.e. all points in plane except points on the lines y = x and y = −x.The range of function is entire set of real numbers i.e. −∞ < w <∞.

Example 1.4. Find the domain and range of f(x, y) =1

(x2 + y2 + z2).

Solution: The domain D is the set of points in the space for whichx2 + y2 + z2 6= 0.i.e. the domain is set of points (x, y, z) 6= (0, 0, 0).The range of function is the set of all positive numbers i.e. 0 < w <∞. Interior and Boundarypoints

Definition 1.4. Interior point: A point (x0, y0) in a region R in the xy plane is an interiorpoint of R if every disk centered at (x0, y0) of positive radius lies entirely in R.

Definition 1.5. Boundary point: A point (x0, y0) is boundary point of R if every diskcentered at (x0, y0) of positive radius contains points that lie in R as well as points lie outside ofR.

1

Chapter 1. Limit and Continuity of Multivariable Functions § 1.1. Functions of Several Variables

Note: The boundary point itself need not belong to R.

Definition 1.6. Open Region: A region is open if it consists entirely of interior points.

Definition 1.7. Closed Region: A region is closed if it contains all its interior and boundarypoints.

Definition 1.8. Bounded Region: A region in the plane is bounded if it lies inside a disk offixed radius.

Definition 1.9. Unbounded Region: A Region in the plane is unbounded if it is not bounded.

For example:

(i) Bounded regions: Triangles, rectangles, circles,disks.(ii) Unbounded regions: lines, coordinate axes, quadrants, half planes.

In Fig.1.1,function is defined only where (y − x2) ≥ 0.i.e.domain is closed, unbounded region.

Example 1.5. If f(x, y) =√y − x then

(1) find the boundary of domain of function.(2) determine whether domain is an open region,closed region or neither(3) decide whether domain is bounded or unbounded.

Solution: Let f(x, y) =√y − x

The domain of f(x, y) is set of all pairs in the plane for which (y − x) ≥ 0i.e. y ≥ x and range of f(x, y) is set of all non negative numbers i.e. w ≥ 0

Boundary of domain of function is√

(y − x) = 0⇒ y = xi.e. The straight line passing through origin.Therefore, domain is closed region, which is unbounded.

1.1.1. Graphs and Level curves of function of two variables.

Definition 1.10. Level Curve: The set of all points in the plane where a function f(x, y)has a constant value ’c’ is called a level curve of f .i.e.{(x, y) ∈ D/f(x, y) = c}

Definition 1.11. Graph of Function: The set of all points {x, y, f(x, y)} in a space , for(x, y) in domain of f is called the graph of the function f .The graph of the function f is also called surface z = f(x, y).

Example 1.6. Plot the level curves for the function f(x, y) =(x+ y)

(x− y), x 6= y, if c = 0, 1.

Solution: For Level curve

f(x, y) = c, c ∈ R(x+ y)

(x− y)= c

y =

(c− 1

c+ 1

)x

i.e.The level curves are lines passing through origin with slope(c− 1)

(c+ 1)

(1) For c = 0, slope is -1∴ The level curve is the line y = −x.

(2) For c = 1, slope is 0∴ The level curve is y = 0 i.e. x axis.

Fig.1.2 Level curves for f(x, y) =(x+ y)

(x− y), x 6= y

Example 1.7. Plot the graph and level curves of the function f(x, y) = 100 − x2 − y2 forc = 0, 51, 75.

Page|2

§ 1.1. Functions of Several Variables Chapter 1. Limit and Continuity of Multivariable Functions

Solution: Let f(x, y) = 100− x2 − y2 The domain D of f is entire xy plane and range of f isthe set of real numbers less than or equal to 100.For Level curves:

f(x, y) = c, c ∈ R100− x2 − y2 = c

x2 + y2 = 100− c

Therefore, level curves are circles centered at origin with radius√

100− c.(1) For c = 0, x2 + y2 = 100 Therefore, level curve is circle with center origin and radius 10(2) For c = 51, x2 + y2 = 49 The level curve is circle with center origin and radius 7(3) For c = 75, x2 + y2 = 25 The level curve is circle with center origin and radius 5

Fig.1.3 Level curves for f(x, y) = 100− x2 − y2 For Graph of f(x, y):The graph is the paraboloid z = 100− x2 − y2. as shown in Fig.1.4also the level curve f(x, y) = 100 consists of the origin alone.

Example 1.8. Sketch the level curves of f(x, y) = −(x− 1)2 − y2 + 1for f(x, y) = 1, 0,−3,−8.

Solution: Let f(x, y) = −(x− 1)2 − y2 + 1For level curves

f(x, y) = c, c ∈ R(x− 1)2 + y2 = 1− c

i.e.level curves are circles centered at (1,0)and radius√

1− chere,c = 1, 0,−3,−8i.e.level curves are circles centered at (1,0) and respective radii are 0,1,2,3.

Example 1.9. Find an equation of the level curve of f(x, y) = 16−x2− y2 that passes throughthe point (2

√2,√

2).

Solution: Let level curves f(x, y) = 16− x2 − y2 that passes through the point (2√

2,√

2).

∴ f(x, y) = 16− (2√

2)2 − (√

22)

= 6

16− x2 − y2 = 6

x2 + y2 = 10

Level curve is a circle with center at origin and radius is√

10. EXERCISE 1.1 In example 1-12

Q1: Find domain and range of functions.Q2: Describe the level curves of functions.Q3: Find the boundary of domain of functions.Q4: Determine whether the domain is an open region, closed region or neither.Q5: Decide whether domain is bounded or unbounded.

(1) f(x, y) = y − x.

(2) f(x, y) =√

(y − x.(3) f(x, y) = 4x2 + 9y2.(4) f(x, y) = x2 − y2.(5) f(x, y) = xy.

(6) f(x, y) =y

x2.

(7) f(x, y) =1√

16− x2 − y2.

(8) f(x, y) =√

9− x2 − y2.(9) f(x, y) = ln(x2 + y2).

(10) f(x, y) = e−(x2+y2).(11) f(x, y) = sin−1(y − x).

Page|3

Chapter 1. Limit and Continuity of Multivariable Functions § 1.1. Functions of Several Variables

(12) f(x, y) = tan−1(y

x). In example 13-16. Find an equation for the level curve of the function

f(x, y) that passes through the given point.(13) f(x, y) = 16− x2 − y2 , (2

√2,√

2).

(14) f(x, y) =√x2 − 1 , (1, 0).

(15) f(x, y) =∫ yx

dt

1 + t2, (−√

2,√

2).

ANSWERS 1.1

(1) (i) Domain :- all points in xy Plane. Range :- all real numbers.(ii) level curves are straight lines y − x = c

i.e. parallel to y = x(iii) No boundary points.(iv) Both open and closed.(v) Unbounded.

(2) (a) Domain :-set of all (x, y) such that y ≥ xRange :- z ≥ 0.

(b) straight lines y − x = c, c ≥ 0(c) boundary is

√y − x = 0 i.e.straight lines y = x.

(d) closed.(e) Unbounded.

(3) (a) Domain :-all points in xy Plane.Range :- z ≥ 0.

(b) f(x, y) = 0 , the origin. f(x, y) = c > 0,ellipses with center(0,0), major and minoraxis along X andY axis respectively.

(c) No boundary points(d) Both open and closed.(e) Unbounded.

(4) (i) Domain :-all points in xy Plane.Range :- all real Numbers.

(ii) Level curves:For f(x, y) = 0 ,the union of lines y = ±x for f(x, y) = c 6= 0 ,hyperbolas centered at (0,0)with foci on the x axis if c > 0 and on the Y axis ifc < 0 .

(iii) No boundary points.(iv) Both open and closed.(v) Unbounded.

(5) (i) Domain :-all points in xy Plane.Range :- all real Numbers.

(ii) Level curves:For f(x, y) = 0 ,the X and Y axis for f(x, y) 6= 0 , hyperbolas with Xand Y axis as asymptotes .

(iii) No boundary points.(iv) Both open and closed.(v) Unbounded.

(6) (i) Domain :-all (x, y) 6= (0, y)Range :- all real Numbers.

(ii) Level curves:For f(x, y) = 0 ,the X axis without origin for f(x, y) 6= 0, parabolasminus origin.

(iii) boundary is line x = 0.(iv) open region.(v) Unbounded.

(7) (i) Domain :-all pairs (x,y) satisfying x2 + y2 < 16.

Range :- z ≥ 1

4.

(ii) Level curve :are circles centered at origin with radii r < 4.(iii) boundary is the circle x2 + y2 = 16.(iv) open region.(v) bounded.

(8) (i) Domain :-all pairs (x,y) satisfying x2 + y2 ≤ 9.Range :- 0 ≤ z ≤ 3.

(ii) Level curves are circles centered at origin with radii r ≤ 3

Page|4

§ 1.2. Limit and Continuity Chapter 1. Limit and Continuity of Multivariable Functions

(iii) boundary is the circle x2 + y2 = 9.(iv) closed region.(v) bounded.

(9) (i) Domain :-(x, y) 6= (0, 0).Range :-all real Numbers.

(ii) Level curves are circles centered at origin with radii r > 0.(iii) boundary is the single point (0,0).(iv) open region.(v) unbounded.

(10) (i) Domain :-all points in xy Plane.Range :-0 < z < 1.

(ii) Level curves are origin itself and the circles centered at origin with radii r > 0.(iii) No boundary points.(iv) both open and closed region.(v) unbounded.

(11) (i) Domain :-all (x,y) satisfying −1 ≤ (y − x) ≤ 1.

Range :-−π2≤ z ≤ π

2.

(ii) Level curves are straight lines of the form y − x = c , −1 ≤ c ≤ 1.(iii) boundary are two straight lines y = 1 + x and y = −1 + x.(iv) closed region.(v) unbounded.

(12) (i) Domain :-all (x, y) , x 6= 0.

Range :-−π2< z <

π

2(ii) Level curves are straight lines of the form y − x = c , −1 ≤ c ≤ 1.(iii) boundary are two straight lines y = 1 + x and y = −1 + x.(iv) closed region.(v) unbounded.

(13) x2 + y2 = 10(14) x = 1 or x = −1(15) tan−1y − tan−1x = 2tan−1

√2

1.2. Limit and Continuity

Limit of Function If the values of a function z = f(x, y) can be made as close as we like to afixed number L by taking the point (x, y) close to the point (x0, y0) but not equal to (x0, y0) , Wesay that L is the limit of f(x, y) as (x, y) approaches to (x0, y0).Since,

|x− x0| =√

(x− x0)2 ≤√

(x− x0)2 + (y − y0)2 and

|y − y0| =√

(y − y0)2 ≤√

(x− x0)2 + (y − y0)2

then√

(x− x0)2 + (y − y0)2 < δ

⇒|x− x0| < δ and |y − y0| < δ

conversely , if for some δ > 0 , both |x− x0| < δ and |y − y0| < δ

⇒√

(x− x0)2 + (y − y0)2 <√δ2 + δ2 =

√2δ thus , the open square |x − x0| < δ and |y −

y0| < δ lies inside the open disc√

(x− x0)2 + (y − y0)2 <√

2δ and contains the open disc√(x− x0)2 + (y − y0)2 < δ

Definition 1.12. Limit of Function of two variables: A function f(x, y) approaches thelimit L ,as (x, y) approaches (x0, yo) , if there exists a corresponding number δ > 0 such that forall (x, y) in the domain of f ,

|f(x, y)− L| < ε whenever 0 <√

(x− x0)2 + (y − y0)2 < δ

or 0 < |x− x0| < δ and 0 < |y − y0| < δ

In symbols , we can write as

lim(x,y)→(x0,y0)

f(x, y) = L

Page|5

Chapter 1. Limit and Continuity of Multivariable Functions § 1.2. Limit and Continuity

Theorem 1.1. Properties of Limits:If lim

(x,y)→(x0,y0)f(x, y) = L , lim

(x,y)→(x0,y0)g(x, y) = M , L,M ∈ R then

(i) lim(x,y)→(x0,y0)

[f(x, y)± g(x, y)] = L±M

(ii) lim(x,y)→(x0,y0)

[f(x, y).g(x, y)] = L.M

(iii) lim(x,y)→(x0,y0)

Kf(x, y) = KL ,K,L ∈ R and

(iv) lim(x,y)→(x0,y0)

f(x, y)

g(x, y)=

L

M,M 6= 0.

Note:-If a function f(x, y) has different limits along two different paths as (x, y) → (x0, y0)then lim

(x,y)→(x0,y0)f(x, y) does not exists. continuous Function:

Definition 1.13. continuous Function of two variables: A function f(x, y) is said to becontinuous at the point (x0, yo) , if

(i) f(x, y) is defined at (x0, yo)i.e. f(x0, yo) exists.

(ii) lim(x,y)→(x0,y0)

f(x, y) exists.

(iii) lim(x,y)→(x0,y0)

f(x, y) = f(x0, yo)

Definition 1.14. A function f(x, y) is said to be continuous in a region R if it is continuousat every point of its domain.

Theorem 1.2. Properties of continuous function:If f(x, y) and g(x, y) are both continuous at point (x0, y0) then

(i) f(x, y)± g(x, y) are continuous at point (x0, y0)(ii) Kf(x, y) is continuous at point (x0, y0)(iii) f(x, y).g(x, y) is continuous at point (x0, y0)

(iv)f(x, y)

g(x, y)is continuous at point (x0, y0) , g(x, y) 6= 0

(v) |f(x, y)| is continuous at point (x0, y0)

Theorem 1.3. Let f(x, y) be continuous function at point (x0, y0) then f(x, y0) is continuousfunction at x = x0 and f(x0, y) is continuous function at y = y0,where f(x, y0) and f(x0, y) beingfunction of one variable.

Proof. Since,f(x, y) is continuous function at point (x0, y0)lim

(x,y)→(x0,y0)f(x, y) = f(x0, yo)

For given ε > 0 , there exists δ > 0 such that

|x− x0| < δ and |y − y0| < δ ⇒ |f(x, y)− f(x0, y0)| < ε

|x− x0| < δ and |y0 − y0| < δ ⇒ |f(x, y0)− f(x0, y0)| < ε

|x− x0| < δ ⇒ |f(x, y0)− f(x0, y0)| < ε

∴ f(x, y0)is continuous function at x = x0

Similarly, |y − y0| < δ ⇒ |f(x0, y)− f(x0, y0)| < ε

∴ f(x0, y)is continuous function at y = y0

The converse of above theorem need not be true.

For example: Let f(x, y) =

2xy

x2 + y2, (x, y) 6= (0, 0);

0, (x, y) = (0, 0).

limx→0

f(x, 0) = limx→0

0 = 0

limy→0

f(0, y) = limx→0

0 = 0

Page|6


∴ f(x, y) is continuous function at (0, 0) when considered as a single variable x or yNow ,considered along path y = mx

lim(x,y)→(0,0)

f(x, y) = limx→0

f(x,mx)

= limx→0

2mx2

x2 +m2x2

=2m

1 +m2

∴ this limit depends on m.∴ limit does not exists at (0, 0)i.e. Function f(x, y) of two variable is not continuous at (0, 0)

Definition 1.15. continuity of composite Function: If f(x, y) is continuous at (x0, y0)and g is single variable function continuous at f(x0, y0) then the composite function h = gofdefined by h(x, y) = g(f(x, y)) is continuous at (x0, y0).

for example: e(x−y) , cos(xy

x2 + 1) , ln(1 + x2y2)

Example 1.10. Evaluate the limits , if they exists

(a) lim(x,y)→(0,0)

3x2 − y2 + 5

x2 + y2 + 2

(b) lim

(x,y)→(π

2,0)

cosy + 1

y − sinx

(c) lim(x,y,z)→(0,−2,0)

ln√x2 + y2 + z2

Solution:

(a) lim(x,y)→(0,0)

3x2 − y2 + 5

x2 + y2 + 2=

3(0)2 − (0)2 + 5

(0)2 + (0)2 + 2

=5

2

(b) lim

(x,y)→(π

2,0)

cosy + 1

y − sinx=

cos(0) + 1

(0)− sin(π

2)

=2

−1= −2

(c) lim(x,y,z)→(0,−2,0)

ln√x2 + y2 + z2 = ln

√(0)2 + (−2)2 + (0)2

= ln√

4

= ln2

Example 1.11. Find lim(x,y)→(2,0)

√2x− y − 2

2x− y − 4, 2x− y 6= 4

Solution:

lim(x,y)→(2,0)

√2x− y − 2

2x− y − 4= lim

(x,y)→(2,0)

√2x− y − 2

(√

2x− y)2 − (2)2

= lim(x,y)→(2,0)

√2x− y − 2

[√

2x− y + 2][√

2x− y − 2]

= lim(x,y)→(2,0)

1√2x− y + 2

=1√

4 + 2

=1

4

Page|7


Example 1.12. Let f(x, y) =x+ y

2 + cosxand ε = 0.02 Show that there exists δ > 0 such that for

all (x, y) ,√x2 + y2 < δ ⇒ |f(x, y)− f(0, 0)| < ε.

Solution:

f(x, y) =x+ y

2 + cosx, ε = 0.02

Since, −1 ≤ cosx ≤ 1

∴ (−1 + 2) ≤ (2 + cosx) ≤ (2 + 1)

∴ 1 ≤ (2 + cosx) ≤ 3

∴1

3≤ 1

2 + cosx≤ 1

∴|x+ y|

3≤ | x+ y

2 + cosx| ≤ |x+ y| ≤ |x|+ |y|

∴ | x+ y

2 + cosx| ≤ |x|+ |y|

consider, |f(x, y)− f(0, 0)| = | x+ y

2 + cosx− 0|

= | x+ y

2 + cosx| ≤ |x|+ |y|

∴ if |x| < δ, |y| < δ then |f(x, y)− f(0, 0)| ≤ |x|+ |y| < 2δ = ε

∴ there exists δ =ε

2=

0.02

2= 0.01 such that

|x| < δ, |y| < δ ⇒ |f(x, y)− f(0, 0)| < ε

Example 1.13. Let f(x, y, z) = tan2x+ tan2y + tan2z and ε = 0.03.Show that there exists a , δ > 0 such that for all (x, y, z),√x2 + y2 + z2 < δ ⇒ |f(x, y, z)− f(0, 0, 0)| < ε.

Solution: f(x, y, z) = tan2x+ tan2y + tan2z, ε = 0.03Consider,

|f(x, y, z)− f(0, 0, 0)| = |tan2x+ tan2y + tan2z|≤ |tan2x|+ |tan2y|+ |tan2z|= tan2x+ tan2y + tan2z

Now if |x| < δ, |y| < δ, |z| < δ

⇒ |f(x, y, z)− f(0, 0, 0)| ≤ tan2x+ tan2y + tan2z

≤ tan2δ + tan2δ + tan2δ

= 3tan2δ = ε

∴ there exists tan2δ =ε

3=

0.03

3= 0.01

∴ tanδ = 0.1

∴ there exists δ = tan−1(0.1)

such that |x| < δ, |y| < δ, |z| < δ ⇒ |f(x, y, z)− f(0, 0, 0)| < ε

Example 1.14. Show that f(x, y) =x2

x2 − yhas no limit as (x, y → (0, 0) by considering

different paths.

Solution: Along X-axis (i.e.y = 0):

lim(x,y)→(0,0)

f(x, y) = limx→0

f(x, 0) = limx→0

x2

x2

= 1

Page|8


Along Y -axis (i.e.x = 0):

lim(x,y)→(0,0)

f(x, y) = limy→0

f(0, y) = limy→0

0

0− y= 0

Along y = kx2, (k 6= 1):

lim(x,y)→(0,0)

f(x, y) = limx→0

f(x, kx2) = limx→0

x2

x2 − kx2

= limx→0

x2

x2(1− k)

=1

1− k∴ different limits for different paths.∴ limit does not exist.

Example 1.15. Show that lim(x,y)→(0,0)

tan−1(|x|+ |y|x2 + y2

) =π

2by using polar coordinates.

Solution: by polar coordinates,

x = rcosθ, y = r sin θ and r → 0 as (x, y)→ (0, 0)

lim(x,y)→(0,0)

tan−1(|x|+ |y|x2 + y2

) = limr→0

tan−1(|rcosθ|+ |rsinθ|

r2)

= limr→0

tan−1(|r|(|cosθ|+ |sinθ|)

r2)

consider,

limr→0+


r2) = lim

r→0+tan−1(

(|cosθ|+ |sinθ|)r

)

= limr→0+

tan−1(∞)

=π

2

consider,

limr→0−


r2) = lim

r→0−tan−1(

−(|cosθ|+ |sinθ|)r

)

=π

2

∴ lim(x,y)→(0,0)

tan−1(|x|+ |y|x2 + y2

) =π

2

Example 1.16. At what points the function f(x, y) = sin1

xyis continuous ?

Solution:

f(x, y = sin1

xyis continuous at all points(x, y) except x = 0 or y = 0.

Example 1.17. At what points(x,y,z) in a space the function f(x, y, z) =1

|xy|+ |z|is contin-

uous ?

Solution:

f(x, y, z) =1

|xy|+ |z|is continuous at all points(x, y, z) except (0, y, 0) or (x, 0, 0).

Example 1.18. Define f(0, 0) , so that f(x, y) = ln(3x2 − x2y2 + 3y2

x2 + y2) is continuous at origin.

Page|9


Solution: Since f(x, y) is continuous at origin.

lim(x,y)→(0,0)

f(x, y) = f(0, 0)

∴ f(0, 0) = lim(x,y)→(0,0)

f(x, y)

= lim(x,y)→(0,0)

ln(3x2 − x2y2 + 3y2

x2 + y2)

By using polar coordinates x = rcosθ, y = rsinθ and r → 0 as (x, y)→ (0, 0)

∴ f(0, 0) = limr→0

ln(3r2 cos2 θ − r4 cos2 θsin2θ + 3r2 sin2 θ

r2)

= limr→0

ln(3− r2 cos2 θ sin2 θ)

= ln 3

EXERCISE 1.2

(1) Evaluate the limits, if they exists

(i) lim(x,y)→(0,0)

3x2 − y2 + 5

x2 + y2 + 2

(ii) lim(x,y)→(0,4)

x√y

(iii) lim(x,y)→(3,4)

√x2 + y2 − 1

(iv) lim(x,y)→(2,−3)

(1

x+

1

y)2

(v) lim

(x,y)→(0,π

4)

secxtany.

(vi) lim(x,y)→(0,0)

cos(x2 + y3

x+ y + 1)

(vii) lim(x,y)→(0,ln2)

ex−y

(viii) lim(x,y)→(1,1)

ln|1 + x2y2|

(ix) lim(x,y)→(0,0)

eysinx

x

(x) lim(x,y)→(1,1)

cos 3√|xy| − 1

(xi) lim(x,y)→(1,0)

xsiny

x2 + 1

(xii) lim

(x,y)→(π

2,0)

cosy + 1

y − sinx

(xiii) lim(x,y,z)→(1,3,4)

(1

x+

1

y+

1

z)

(xiv) lim(x,y,z)→(1,−1,−1)

2xy + yz

x2 + z2

(xv) lim(x,y,z)→(3,3,0)

(sin2x+ cos2y + sec2z)

(xvi) lim

(x,y,z)→(−1

4,π

2,2)

tan−1(xyz)

(xvii) lim(x,y,z)→(π,0,3)

ze−2ycos2x

(xviii) lim(x,y,z)→(0,−2,0)

ln√x2 + y2 + z2

(2) Find the limits of following functions

(i) lim(x,y)→(1,1)

x2 − 2xy + y2

x− y, x 6= y

(ii) lim(x,y)→(1,1)

x2 − y2

x− y,x 6= y

Page|10


(iii) lim(x,y)→(1,1)

xy − y − 2x+ 2

x− 1,x 6= 1

(iv) lim(x,y)→(2,−4)

y + 4

x2y − xy + 4x2 − 4x

(v) lim(x,y)→(0,0)

x− y + 2√x− 2

√y

√x−√y

,x 6= y.

(vi) lim(x,y)→(2,2)

x+ y − 4√x+ y − 2

(vii) lim(x,y)→(2,0)

√2x− y − 2

2x− y − 4

(viii) lim(x,y)→(4,3)

√x−√y + 1

x− y − 1(3) Each of following example gives a function f(x, y) and a positive number ε, show that

there exists a δ > 0 such that for all (x, y),√x2 + y2 < δ ⇒ |f(x, y)− f(0, 0)| < ε(i) f(x, y) = x2 + y2, ε = 0.01

(ii) f(x, y) =y

x2 + 1, ε = 0.05

(iii) f(x, y) =x+ y

x2 + 1, ε = 0.01

(iv) f(x, y) =x+ y

2 + cosx, ε = 0.02

(4) Each of following example gives a function f(x, y, z) and a positive number ε, show thatthere exists a δ > 0 such that for all (x, y, z),√x2 + y2 + z2 < δ ⇒ |f(x, y, z)− f(0, 0, 0)| < ε(i) f(x, y, z) = x2 + y2 + z2, ε = 0.015(ii) f(x, y, z) = xyz, ε = 0.008

(iii) f(x, y, z) =x+ y + z

x2 + y2 + z2 + 1, ε = 0.015

(iv) f(x, y) = tan2x+ tan2y + tan2z, ε = 0.03(5) Show that following functions have no limit as(x, y) → (0, 0), by considering different

paths.

(i) f(x, y) = − x√x2 + y2

(ii) f(x, y) =x4

x4 + y2

(iii) f(x, y) =x4 − y2

x4 + y2

(iv) f(x, y) =xy

|xy|(v) f(x, y) =

x− yx+ y

(vi) f(x, y) =x+ y

x− y

(vii) f(x, y) =x2 + y

y

(viii) f(x, y) =x2

x2 − y(6) By using Polar coordinates,find the limit of following functions as (x, y)→ (0, 0) if exists.

(i) f(x, y) =x3 − xy2

x2 + y2

(ii) f(x, y) = cos(x3 − y3

x2 + y2)

(iii) f(x, y) =y2

x2 + y2

(iv) f(x, y) =2x

x2 + x+ y2

(v) f(x, y) = tan−1(|x|+ |y|x2 + y2

)

Page|11


(vi) f(x, y) =x2 − y2

x2 + y2

(7) At what points the following functions are continuous?(i) f(x, y) = sin(x+ y)(ii) f(x, y) = ln(x2 + y2)

(iii) f(x, y) =x+ y

x− y(iv) f(x, y) =

y

x2 + 1

(v) f(x, y) = sin1

xy

(vi) f(x, y) =x+ y

2 + cosx

(vii) f(x, y) =x2 + y2

x2 − 3x+ 2

(viii) f(x, y) =1

x2 − y(ix) f(x, y, z) = x2 + y2 − 2z2

(x) f(x, y, z) =√x2 + y2 − 1

(xi) f(x, y, z) = lnxyz(xii) f(x, y, z) = ex+ycosz

(xiii) f(x, y, z) = xysin1

z

(xiv) f(x, y, z) =1

x2 + z2 − 1

(xv) f(x, y, z) =1

|y|+ |z|(xvi) f(x, y, z) =

1

|xy|+ |z|(8) Define f(0, 0), so that function f(x, y) be continuous at the origin ?

(i) f(x, y) = ln(3x2 − x2y2 + 3y2

x2 + y2)

(ii) f(x, y) =3x2y

x2 + y2

(9) Show that function f(x, y, z) = x+ y − z is continuous at every point (x0, y0, z0).(10) Show that function f(x, y, z) = x2 + y2 + z2 is continuous at origin.

ANSWERS 1.2

(1)(i)5

2(ii) 0 (iii)2

√6 (iv)

1

36(v) 1 (vi) 1 (vii)

1

2(viii) ln2 (ix) 1 (x) 1 (xi) 0 (xii) −2 (xiii)

19

12(xiv)

−1

2(xv) 2 (xvi) tan−1(

−π4

) (xvii) 3 (xviii) ln2

(2)(i)]0 (ii) 2 (iii) −1 (iv)1

2(v) 2 (vi) 4 (vii)

1

4(viii)

1

4(3) (i)δ = 0.1 (ii) δ = 0.05 (iii) δ = 0.005 (iv) δ = 0.01(4) (i)

√0.015 (ii) δ = 0.2 (iii) δ = 0.005 (iv) δ = tan−1(0.1)

(6) (i)0 (ii) 1 (iii) does not exist (iv) does not exist (v)π

2(vi) does not exists

(7) (i)All (x, y) (ii)All (x, y) except (0, 0) (iii)All (x, y) so that x 6= y (iv) All (x, y) (v) All (x, y)except x = 0 or y = 0 (vi)All (x, y) (vii)All (x, y) so that x 6= 2 and x 6= 1 (viii)All (x, y) so thaty 6= x2 (ix)All (x, y, z) (x)All (x, y, z) except interior of cylinder x2 +y2 = 1 (xi)All (x, y, z) so thatxyz > 0 (xii)All (x, y, z) (xiii)All (x, y, z) with z 6= 0 (xiv)All (x, y, z) with x2 + y2 6= 1 (xv)All(x, y, z) except (x, 0, 0) (xvi)All (x, y, z) except (0, y, 0) or (x, 0, 0)(8) (i)ln3 (ii)0

Page|12

CHAPTER 2

Partial Derivatives

2.1. Definition and Examples

If z = f(x, y) is a function of two variable x and y.It is possible to consider the rate of change ofz w.r.t. x ,keeping y as a constant and also rate of change of z w.r.t. y ,keeping x as a constant,Weget a partial derivative.To distinguish partial derivative from ordinary derivative we use the symbol ’∂’rather than d whichis previously used.

Definition 2.1. Partial derivative of function w.r.t.x: The partial derivative of functionf(x, y) w.r.t.x at point(x0, y0) is

∂f

∂x|(x0,y0) = lim

h→0

f(x0 + h, y0)− f(x0, y0)

h, if limit exists.

It is denoted by fx(x0, y0)

Note:

(1) An equivalent expression for the partial derivative isd

dxf(x, y0) at x = x0

(2) The slope of the curvez = f(x, y0) at the point P (x0, y0, f(x0, y0)) in the plane y = y0 isthe value of the partial derivative of f w.r.t. x at (x0, y0)The tangent line to the curve at point P is the line in the plane y = y0 that passes throughP with this slope.

Definition 2.2. Partial derivative of function w.r.t.y: The partial derivative of functionf(x, y) w.r.t.y at point(x0, y0) is

∂f

∂y|(x0,y0) = lim

k→0

f(x0, y0 + k)− f(x0, y0)

k, if limit exists.

It is denoted by fy(x0, y0) ord

dyf(x0, y)|y=y0

Note:The slope of the curvez = f(x0, y) at the point P (x0, y0, f(x0, y0)) in the plane x = x0 is thevalue of the partial derivative of f w.r.t. y at (x0, y0)The tangent line to the curve at point P is the line in the plane x = x0 that passes through P withthis slope.

Example 2.1. Find partial derivative∂f

∂xand

∂f

∂yof following functions

(a) f(x, y) = 5xy − 7x2 − y2 + 3x− 6y + 2 at (2,−3)(b) f(x, y) = sin2(x− 3y)

Solution:

(a) Let f(x, y) = 5xy − 7x2 − y2 + 3x− 6y + 2

∂f

∂x= 5y − 14x+ 3

∂f

∂x|(2,−3) = 5(−3)− 14(2) + 3

= −40

∂f

∂y= 5x− 2y − 6

∂f

∂y|(2,−3) = 5(2)− 2(−3)− 6

= 10

13

Chapter 2. Partial Derivatives § 2.1. Definition and Examples

(b) f(x, y) = sin2(x− 3y)

∂f

∂x= 2sin(x− 3y)

∂[sin(x− 3y)]

∂x

= 2sin(x− 3y)cos(x− 3y)∂(x− 3y)

∂x= 2sin(x− 3y)cos(x− 3y)(1)

= sin[2(x− 3y)]

Similarly,

∂f

∂y= 2sin(x− 3y)

∂[sin(x− 3y)]

∂y

∂f

∂y= 2sin(x− 3y)cos(x− 3y)

∂(x− 3y)

∂y

= 2sin(x− 3y)cos(x− 3y)(−3)

= −3sin[2(x− 3y)]

Example 2.2. Find fx, fy and fz of following functions

(a) f(x, y, z) = x−√y2 + z2

(b) f(x, y, z) = sin−1(xyz)

Solution:

(a) Let f(x, y, z) = x−√y2 + z2

∂f

∂x=∂(x−

√y2 + z2)

∂x= 1

also,

∂f

∂y=∂(x−

√y2 + z2)

∂y

= − 1

2√y2 + z2

(2y)

= − y√y2 + z2

Similarly,

∂f

∂z=∂(x−

√y2 + z2)

∂z

= − 1

2√y2 + z2

∂(y2 + z2)

∂z

= − 1√y2 + z2

(2z)

= − z√y2 + z2

(b) f(x, y, z) = sin−1(xyz)

∂f

∂x=

1√1− x2y2z2

∂(xyz)

∂x

=yz√

1− x2y2z2

Similarly ,

∂f

∂y=

xz√1− x2y2z2

∂f

∂z=

xy√1− x2y2z2

Page|14

§ 2.1. Definition and Examples Chapter 2. Partial Derivatives

Example 2.3. By using limit definition of partial derivative,

compute the partial derivative∂f

∂xand

∂f

∂yof f(x, y) = 4 + 2x− 3y − xy2 at (−2, 1)

Solution:f(x, y) = 4 + 2x− 3y − xy2

∂f

∂x|(x0,y0) = lim

h→0

f(x0 + h, y0)− f(x0, y0)

h∂f

∂x|(−2,1) = lim

h→0

f(−2 + h, 1)− f(−2, 1)

h

= limh→0

[4 + 2(−2 + h)− 3− (−2 + h)]− [4− 4− 3 + 2]

h

= limh→0

[4− 4 + 2h− 3 + 2− h]− [−1]

h

= limh→0

h− 1 + 1

h= 1

Similarly,

∂f

∂y|(x0,y0) = lim

k→0

f(x0, y0 + k)− f(x0, y0)

k

∂f

∂y|(−2,1) = lim

k→0

f(−2, 1 + k)− f(−2, 1)

k

= limk→0

[4 + 2(−2)− 3(1 + k)− (−2)(1 + k)2]− [−1]

k

= limk→0

[4− 4− 3− 3k + 2 + 4k + 2k2 + 1]

k

= limk→0

2k2 + k

k= lim

k→02k + 1

= 1

Example 2.4. Write the definition of partial derivative∂f

∂zat (x0, y0, z0)

and use this to find∂f

∂zat (1, 2, 3) for f(x, y, z) = x2yz2

Solution: Let f(x, y, z) = x2yz2,

∂f

∂z|(x0,y0,z0) = lim

h→0

f(x0, y0, z0 + h)− f(x0, y0, z0)

h∂f

∂z|(1,2,3) = lim

h→0

f(1, 2, 3 + h)− f(1, 2, 3)

h

∂f

∂z|((1,2,3)) = lim

h→0

2(3 + h)2 − 2(9)

h

= limh→0

18 + 12h+ 2h2 − 18

h= lim

h→0(12 + 2h)

= 12

Example 2.5. Find∂x

∂zat (1,−1,−3), if the equation xz + ylnx− x2 + 4 = 0. defines x as a

function of y and z and partial derivative exists.

Page|15


Solution:Let xz + ylnx− x2 + 4 = 0

differentiating partially w.r.t.’z’

x+ z∂x

∂z+y

x

∂x

∂z− 2x

∂x

∂z= 0

∴ x+ (z +y

x− 2x)

∂x

∂z= 0

∴∂x

∂z=

−x2

zx+ y − 2x2

at point (1,-1,-3)

∂x

∂z=

−(1)2

−3− 1− 2

=1

6

Example 2.6. Find the slope of the tangent to the parabola at (1, 2, 5), if the plane x = 1intersects the paraboloid z = x2 + y2 in a parabola.

Solution: Equation of paraboloid is given by z = x2 + y2.Plane x = 1 intersects this paraboloid into parabola as shown in fig.2.1. The Slope is partial

derivative∂z

∂yat point (1, 2)

∴∂z

∂y=∂(x2 + y2)

∂y

= 2y

∂z

∂y|(1,2) = 4

Example 2.7. If resistors of R1, R2,and R3 ohms are connected in parallel to make an R-ohms

resistors , the value of R as1

R=

1

R1+

1

R2+

1

R3.

Find∂R

∂R2,when R1 = 30, R2 = 45,and R3 = 90 ohms.

Solution: Since,

1

R=

1

R1+

1

R2+

1

R3

∴∂

∂R2(

1

R) =

∂

∂R2(

1

R1+

1

R2+

1

R3)

∴ − 1

R2

∂R

∂R2= 0− 1

(R2)2+ 0

∴1

R2

∂R

∂R2=

1

(R2)2

∴∂R

∂R2= (

R

R2)2 · · · · · · · · · · · · (a)

Page|16

§ 2.1. Definition and Examples Chapter 2. Partial Derivatives

If R1 = 30, R2 = 45, R3 = 90 then

1

R=

1

R1+

1

R2+

1

R3

=1

30+

1

45+

1

90

=1

15∴ R = 15

∴ equation (a) becomes

∂R

∂R2= (

15

45)2

=1

9

EXERCISE 2.1

(1) Find∂f

∂xand

∂f

∂yof following functions.

(i) f(x, y) = x2 + 3xy + y − 1(ii) f(x, y) = ysin(xy)

(iii) f(x, y) =2y

y + cosx(iv) f(x, y) = 2x2 − 3y − 4(v) f(x, y) = x2 − xy + y2

(vi) f(x, y) =x2 − 1

y + 2(vii) f(x, y) = 5xy − 7x2 − y2 + 3x− 6y + 2(viii) f(x, y) = (xy − 1)2

(ix) f(x, y) = (2x− 3y)3

(x) f(x, y) =√x2 + y2

(xi) f(x, y) = [x3 +y

2]

2

3

(xii) f(x, y) =1

x+ y

(xiii) f(x, y) =x

x2 + y2

(xiv) f(x, y) =x+ y

xy − 1

(xv) f(x, y) = tan−1(y

x)

(xvi) f(x, y) = ex+y+1

(xvii) f(x, y) = e−x.sin(x+ y)(xviii) f(x, y) = ln(x+ y)(xix) f(x, y) = exylny(xx) f(x, y) = sin2(x− 3y)(xxi) f(x, y) = cos2(3x− y2)(xxii) f(x, y) = xy

(xxiii) f(x, y) = logyx

(xxiv) f(x, y) =y∫xg(t)dt , (g is continuous for all t)

(2) Find fx, fy and fz of following functions.(i) f(x, y, z) = 1 + xy2 − 2z2

(ii) f(x, y, z) = xy + yz + xz

(iii) f(x, y, z) = x−√y2 + z2

(iv) f(x, y, z) = (x2 + y2 + z2)−

1

2

(v) f(x, y, z) = sin−1(xyz)(vi) f(x, y, z) = sec−1(x+ yz)

(vii) f(x, y, z) = ln(x+ 2y + 3z)

Page|17


(viii) f(x, y, z) = yz ln(xy)

(ix) f(x, y, z) = e−(x2+y2+z2)

(x) f(x, y, z) = e−(xyz)

(xi) f(x, y, z) = tanh(x+ 2y + 3z)(xii) f(x, y, z) = sinh(xy − z2)(xiii) f(x, y, z) = x sin(y + 3z)

(3) Find the first order partial derivative of the following functions.(i) f(t, α) = cos(2πt− α)

(ii) g(u, v) = v2e

2u

v

(iii) h(ρ, φ, θ) = ρ sinφ cos θ(iv) g(r, θ, z) = r(1− cos θ)− z

(4) By using limit definition of partial derivative, Compute the partial derivative∂f

∂xand

∂f

∂yoff(x, y) = 1− x+ y − 3x2y at (1, 2)

(5) Write the definition of partial derivative∂f

∂yat (x0, y0, z0) use this to find

∂f

∂yat (−1, 0, 3)

for f(x, y, z) = −2xy2 + yz2.

(6) Find∂z

∂xat (1, 1, 1), if the equation xy + z3x− 2yz = 0 defines z as a function of x and y

and partial derivative exists.

(7) Find∂z

∂x, if the equation yz − ln z = x+ y defines z as a function of x and y and partial

derivative exists.

ANSWERS 2.1 (1) (i)∂f

∂x= 2x+ 3y,

∂f

∂y= 3x+ 1 (ii)

∂f

∂x= y2 cosxy,

∂f

∂y= xy cos(xy) + sin(xy)

(iii)∂f

∂x=

2y sinx

(y + cosx)2,∂f

∂y=

2 cosx

(y + cosx)2(iv)

∂f

∂x= 4x,

∂f

∂y= −3 (v)

∂f

∂x= 2x − y, ∂f

∂y= −x + 2y

(vi)∂f

∂x= 2x(y + 2),

∂f

∂y= x2 − 1 (vii)

∂f

∂x= 5y − 14x + 3,

∂f

∂y= 5x − 2y − 6 (viii)

∂f

∂x=

2y(xy−1),∂f

∂y= 2x(xy−1) (ix)

∂f

∂x= 6(2x−3y)2,

∂f

∂y= −9(2x−3y)2 (x)

∂f

∂x=

x√x2 + y2

,∂f

∂y=

y√x2 + y2

(xi)∂f

∂x=

2x2

3

√x3 +

y

2

,∂f

∂y=

1

3 3

√x3 +

y

2

(xii)∂f

∂x=

−1

(x+ y)2,∂f

∂y=

−1

(x+ y)2(xiii)

∂f

∂x=

y2 − x2

(x2 + y2)2,∂f

∂y=

−2xy

(x2 + y2)2(xiv)

∂f

∂x=−y2 − 1

(xy − 1)2,∂f

∂y=−x2 − 1

(xy − 1)2(xv)

∂f

∂x=

−yx2 + y2

,∂f

∂y=

x

x2 + y2(xvi)

∂f

∂x= ex+y+1,

∂f

∂y= ex+y+1 (xvii)

∂f

∂x= −e−x sin(x + y) + e−x cos(x + y),

∂f

∂y=

e−x cos(x + y) (xviii)∂f

∂x=

1

(x+ y),∂f

∂y=

1

(x+ y)(xix)

∂f

∂x= yexy ln y,

∂f

∂y= xexy ln y +

exy

y

(xx)∂f

∂x= 2 sin(x − 3y) cos(x − 3y),

∂f

∂y= −6 sin(x − 3y) cos(x − 3y) (xxi)

∂f

∂x= −6 cos(3x −

y2) sin(3x − y2),∂f

∂y= 4y cos(3x − y2) sin(3x − y2) (xxii)

∂f

∂x= yxy−1,

∂f

∂y= xy. lnx (xxiii)

∂f

∂x=

1

x ln y,∂f

∂y= − lnx

y(ln y)2(xxiv)

∂f

∂x= −g(x),

∂f

∂y= g(y) (xxv)

∂f

∂x=

y

(1− xy)2,∂f

∂y=

x

(1− xy)2

(2)(i)fx = 1 + y2, fy = 2xy, fz = −4z (ii)fx = y + z, fy = x + z, fz = y + x (iii)fx = 1, fy =

−y√y2 + z2

, fz =−z√y2 + z2

(iv)fx = −x(x2+y2+z2)

−3

2 , fy = −y(x2+y2+z2)

−3

2 , fz = −z(x2+y2+

z2)

−3

2 (v)fx =yz√

1− x2y2z2, fy =

xz√1− x2y2z2

, fz =xy√

1− x2y2z2(vi)fx =

1

|x+ yz|√

(x+ yz)2 − 1, fy =

z

|x+ yz|√

(x+ yz)2 − 1, fz =

y

|x+ yz|√

(x+ yz)2 − 1(vii)fx =

1

x+ 2y + 3z, fy =

2

x+ 2y + 3z, fz =

Page|18

§ 2.2. Second Order Partial Derivative Chapter 2. Partial Derivatives

3

x+ 2y + 3z(viii)fx =

yz

x, fy = z ln(xy) + z, fz = y ln(xy) (ix)fx = −2xe−(x2+y2+z2), fy =

−2ye−(x2+y2+z2), fz = −2ze−(x2+y2+z2) (x)fx = −yze−(xyz), fy = −xze−(xyz), fz = −xye−(xyz)

(xi)fx = sech2(x+2y+3z), fy = 2sech2(x+2y+3z), fz = 3sech2(x+2y+3z) (xii)fx = y cosh(xy−z2), fy = x cosh(xy − z2), fz = −2z cosh(xy − z2) (xiii)fx = sin(y + 3z), fy = x cos(y + 3z), fz =

3x cos(y + 3z) (3) (i)∂f

∂t= −2π sin(2πt− α),

∂f

∂α= sin(2πt− α) (ii)

∂g

∂u= 2ve

(2u

v) ,∂g

∂v= 2ve

(2u

v)−

2ue(2u

v)

(iii)∂h

∂ρ= sinφ cos θ,

∂h

∂φ= ρ cosφ cos θ,

∂h

∂θ= −ρ sinφ sin θ (iv)

∂g

∂r= 1 − cos θ,

∂g

∂θ=

r sin θ,∂g

∂z= −1 (4)

∂f

∂x= −13,

∂f

∂y= −2 (5)

∂f

∂y= 9 (6)

∂z

∂x= −2 (7)

∂z

∂x=

z

yz − 1

2.2. Second Order Partial Derivative

If we partially differentiate f(x, y) twice ,we get second order partial derivative.It is denotedby

∂

∂x(∂f

∂x) =

∂2f

∂x2= (fx)x = fxx = fx2

∂

∂x(∂f

∂y) =

∂2f

∂x∂y= (fy)x = fyx

∂

∂y(∂f

∂x) =

∂2f

∂y∂x= (fx)y = fxy

∂

∂y(∂f

∂y) =

∂2f

∂y2= (fy)y = fyy = fy2

The second order partial derivative at point (x0, y0) are defined as

fxx(x0, y0) = limh→0

fx(x0 + h, y0)− fx(x0, y0)

h

fxy(x0, y0) = limk→0

fx(x0, y0 + k)− fx(x0, y0)

k

fyx(x0, y0) = limh→0

fy(x0 + h, y0)− fy(x0, y0)

h

fyy(x0, y0) = limk→0

fy(x0, y0 + k)− fy(x0, y0)

k

Example 2.8. Find all the second order partial derivatives of function f(x, y) = tan−1(y

x).

Page|19

Chapter 2. Partial Derivatives § 2.2. Second Order Partial Derivative

Solution: Let f(x, y) = tan−1(y

x)

∂f

∂x=

1

1 + (y

x)2

∂

∂x(y

x)

=x2

x2 + y2(−yx2

)

=−y

x2 + y2

also,∂f

∂y=

1

1 + (y

x)2

∂

∂y(y

x)

=x2

x2 + y2(1

x)

=x

x2 + y2

∂2f

∂x2=

∂

∂x(∂f

∂x)

=∂

∂x(−y

x2 + y2)

= −y ∂∂x

(1

x2 + y2)

= −y[−1

(x2 + y2)2]∂

∂x(x2 + y2)

=y

(x2 + y2)2(2x)

=2xy

(x2 + y2)2

∂2f

∂y2=

∂

∂y(∂f

∂y)

=∂

∂y(

x

x2 + y2)

= x.∂

∂y(

1

x2 + y2)

= x[−1

(x2 + y2)2]∂

∂y(x2 + y2)

=−2xy

(x2 + y2)2

∂2f

∂x∂y=

∂

∂x(∂f

∂y)

=∂

∂x(

x

x2 + y2)

=(x2 + y2)− x(2x)

(x2 + y2)2

=y2 − x2

(x2 + y2)2

Page|20

§ 2.2. Second Order Partial Derivative Chapter 2. Partial Derivatives

∂2f

∂y∂x=

∂

∂y(∂f

∂x)

=∂

∂y(−y

x2 + y2)

=(x2 + y2)(−1)− (−y)2y

(x2 + y2)2

=y2 − x2

(x2 + y2)2

Theorem 2.1. The Mixed Derivative Theorem (Clairaut’s Theorem):Statement:If f(x, y) and its partial derivatives fx, fy, fxy, fyx are defined throughout an open re-gion containing a point (a, b) and are all continuous at (a, b) thenfxy(a, b) = fyx(a, b)

Proof. Given:Functions f, fx, fy, fxy, fyx are all defined on an open region containing point(a, b).Also all are continuous at point (a, b).Claim:fxy(a, b) = fyx(a, b).Since,f, fx, fy, fxy, fyx are all defined in the interior of a rectangle R in the xy plane containingpoint (a, b).Let h and k be the numbers such that the point (a+ h, b+ k) also lies in R. Consider,

(1) ∆ = F (a+ h)− F (a)

(2) Where, F (x) = f(x, b+ k)− f(x, b)

apply the Mean Value theorem to F on (a, a+ h),which is continuous because it is differentiable.∴ Equation (1) becomes,

(3) ∆ = hF ′(c1), c1 ∈ (a, a+ h)

From Equation (2)

F ′(x) = fx(x, b+ k)− fx(x, b)

∴ Equation (3) becomes,

(4) ∆ = h[fx(c1, b+ k)− fx(c1, b)]

apply Mean Value theorem to function g(y) = fx(c1, y)

∴ g(b+ k)− g(b) = kg′(d1), d1 ∈ (b, b+ k)

∴ fx(c1, b+ k)− fx(c1, b) = kfxy(c1, d1)

Equation (4) becomes ,

(5) ∆ = hkfxy(c1, d1), for some point(c1, d1) ∈ R′

Now,By using equation (2),equation (1)becomes

∆ = f(a+ h, b+ k)− f(a+ h, b)− f(a, b+ k) + f(a, b)

∆ = [f(a+ h, b+ k)− f(a, b+ k)]− [f(a+ h, b)− f(a, b)]

(6) ∆ = φ(b+ k)− φ(b)

(7) where, φ(y) = f(a+ h, y)− f(a, y)

apply Mean Value theorem to equation (6) we get,

(8) ∆ = kφ′(d2), d2 ∈ (b, b+ k)

from equation (7)

(9) φ′(y) = fy(a+ h, y)− fy(a, y)

∴ Equation (8) becomes ,

∆ = k[fy(a+ h, d2)− fy(a, d2)]

apply Mean Value theorem to function fy(x, d2),we get

fy(a+ h, d2)− fy(a, d2) = hfyx(c2, d2), c2 ∈ (a, a+ h)

Page|21

Chapter 2. Partial Derivatives § 2.3. Higher Order Partial Derivative

(10) ∴ ∆ = khfyx(c2, d2)

from equation (5) and (10)

(11) fxy(c1, d1) = fyx(c2, d2)

where,(c1, d1)and(c2, d2) both lie in R′. Since, fxy and fyx are both continuous at point (a, b)

∴ fxy(c1, d1) = fxy(a, b) + ε1

andfyx(c2, d2) = fyx(a, b) + ε2

(ε1, ε2)→ (0, 0)as(h, k)→ (0, 0)

∴ as(h, k)→ (0, 0),

fxy(a, b) = fyx(a, b)

Note: Clairauts theorem says that,to calculate mixed second order derivative , we may differ-entiable in either order ,provided the continuity conditions are satisfied.

Example 2.9. Verify that Wxy = Wyx for W = ex + x ln y + y lnx

Solution: Let W = ex + x ln y + y lnx

∴∂W

∂x= Wx = ex + ln y +

y

x

∴∂W

∂y= Wy =

x

y+ lnx

∂2W

∂x∂y= Wyx =

∂

∂x(x

y+ lnx)

=1

y+

1

x

∂2W

∂y∂x= Wxy =

∂

∂y(∂W

∂x) =

1

y+

1

x

∴Wxy = Wyx

2.3. Higher Order Partial Derivative

There is no theoretical limit to how many times we can differentiate as long as the derivativeexists.We get higher order derivative denoted by symbols,

∂3f

∂x∂y2= fyyx = fy2x

∂4f

∂x3∂y= fyxxx = fyx3

and so on.

Example 2.10. Find fyxyz if f(x, y, z) = 1− 2xy2z + x2y

textbfSolution: First we differentiate f(x, y, z) with respect to y,then with respect to x,then yand then z.

∴ fy = −4xyz + x2

fyx = −4yz + 2x

fyxy = −4z

fyxyz = −4

Example 2.11. Show that f(x, y, z) = 2z3 − 3(x2 + y2)z satisfies a Laplace equation∂2f

∂x2+

∂2f

∂y2+∂2f

∂z2= 0

Page|22

§ 2.3. Higher Order Partial Derivative Chapter 2. Partial Derivatives

Solution: Let f(x, y, z) = 2z3 − 3(x2 + y2)z

∂f

∂x= −6xz

∂2f

∂x2= −6z

also,∂f

∂y= −6yz

∂2f

∂y2= −6z

also,∂f

∂z= 6z2 − 3(x2 + y2)

∂2f

∂z2= 12z

∴∂2f

∂x2+∂2f

∂y2+∂2f

∂z2= −6z − 6z + 12z

∴∂2f

∂x2+∂2f

∂y2+∂2f

∂z2= 0

∴ f(x, y, z) satisfies Laplace equation. EXERCISE 2.2

(1) Find all second order partial derivatives of the following functions.(i) f(x, y) = x+ y + xy(ii) f(x, y) = sinxy(iii) g(x, y) = x2y + cos y + y sinx(iv) h(x, y) = xey + y + 1(v) r(x, y) = ln(x+ y)

(vi) s(x, y) = tan−1(y

x)

(2) Verify that Wxy = Wyx for the following functions.(i) W = ln(2x+ 3y)(ii) W = ex + x ln y + y lnx(iii) W = xy2 + x2y3 + x3y4

(iv) W = x sin y + y sinx+ xy(v) W = log(x− 2y)

(vi) W = tan−1(y

x)

(vii) W = ax2 + 2hxy + by2

(viii) W = sin−1(x

y)

(ix) W = log(x sin y + y sinx)

(x) W = log tan(y

x)

(xi) W = x2y2 cos(1

x)

(xii) W = x3y + exy2

(3) Show that each of following function satisfies a Laplace equation.(i) f(x, y, z) = x2 + y2 − 2z2

(ii) f(x, y, z) = 2z3 − 3(x2 + y2)z(iii) f(x, y) = e−2y cos(2x)

(iv) f(x, y) = ln√x2 + y2

(v) f(x, y, z) = (x2 + y2 + z2)

−1

2

(vi) f(x, y, z) = e3x+4y cos(5z)(vii) f(x, y) = log(x2 + y2)

(viii) f(x, y) = a log(x2 + y2) + b tan−1(y

x)

(ix) f(x, y) = 2(ax+ by)2 − (x2 + y2) , a2 + b2 = 1

(4) Show that each of following function are solutions of the wave equation∂2W

∂t2= c2∂

2W

∂x2.

(i) W = sin(x+ ct)

Page|23

Chapter 2. Partial Derivatives § 2.3. Higher Order Partial Derivative

(ii) W = cos(2x+ 2ct)(iii) W = sin(x+ ct) + cos(2x+ 2ct)(iv) W = ln(2x+ 2ct)(v) W = tan(2x− 2ct)(vi) W = 5 cos(3x+ 3ct) + ex+ct

(vii) W = f(u) , f is differentiable function of u and u = a(x+ ct) , a is constant.

(5) If f(x, y) = x2tan−1(y

x)− y2tan−1(

x

y) then prove that

∂2f

∂y∂x=x2 − y2

x2 + y2

(6) If f(x, y) = log(x3 + y3 − x2y − xy2)then show that∂f

∂x+∂f

∂y=

2

x+ y(7) If f(x, y) = log(tanx+ tan y + tan z)then show that

sin 2x∂u

∂x+ sin 2y

∂u

∂y+ sin 2z

∂u

∂z= 2

(8) If f(x, y) =x2 + y2

x+ ythen show that

[∂u

∂x− ∂u

∂y]2 = 4[1− ∂u

∂x− ∂u

∂y] = 4

(x− y)2

(x+ y)2

(9) If u = (x2 + y2 + z2)−

1

2 then show that

(∂u

∂x)2 + (

∂u

∂y)2 + (

∂u

∂z)2 = u4

(10) If z = f(x+ ay) + φ(x− ay) then prove that∂2z

∂y2= a2 ∂

2z

∂x2

(11) Evaluate1

a2

∂2z

∂x2+

1

b2∂2z

∂y2if,a2x2 + b2y2 − c2z2 = 0

(12) If f(x, y, z) = log(x2 + y2 + z2) Find∂2f

∂y∂z

(13) If f(x, y, z) = exyz then show that∂3f

∂x∂y∂z= (1 + 3xyz + x2y2z2)exyz

(14) If u = (x2 + y2 + z2)

1

2 then prove that∂2u

∂x2+∂2u

∂y2+∂2u

∂z2=

2

u

ANSWERS 2.2 (1) (i)fxx = 0, fyy = 0, fxy = 1, fyx = 1 (ii)fxx = −y2 sinxy, fyy = −x2 sinxy, fxy =fyx = cosxy − xy sinxy (iii)gxx = 2y − y sinx, gyy = − cos y, gxy = gyx = 2x + cosx (iv)hxx =

0, hyy = xey, hxy = hyx = ey (v)rxx =−1

(x+ y)2, ryy =

−1

(x+ y)2, rxy = ryx =

−1

(x+ y)2(vi)sxx =

2xy

(x2 + y2)2, syy =

−2xy

(x2 + y2)2, sxy = syx =

y2 − x2

(x2 + y2)2(11)

1

[c√a2x2 + b2y2]

(12)]−4yz

(x2 + y2 + z2)2

Page|24

CHAPTER 3

Differentiability

Introduction:In case of function of one variable, we know that if y = f(x) is a function of one variable x thenwe say that the function f is differentiable at x = x0 if the increment or change in f from x tox0 + ∆x,

∆y = f(x0 + ∆x)− f(x0) is expressed as∆y = f(x0 + ∆x)− f(x0)∆y = f ′(x0)∆x+ ε1∆x; where as ∆x→ 0, ε1 → 0.

Here, f ′(x0) is called the differential (total) of function f. It is denoted by df.Thus, df = differential of f = f ′(x0)hNow we shall extend this concept for the function of two variables.Suppose f(x, y) is a function of two variables x and y. Let (x0, y0) be a point in the domain R2 off(x, y) and (x0 +∆x, y0 +∆y) be any point in a neighbourhood of point (x0, y0) and in the domainof f.The increment (or change) in the function f is the diffrencef(x0 + ∆x, y0 + ∆y)− f(x0, y0) from point (x0, y0) to (x0 + ∆x, y0 + ∆y).This is denoted by ∆f(x0, y0) or ∆f.Thus, ∆f(x0, y0) = f(x0 + ∆x, y0 + ∆y)− f(x0, y0)e.q. : If

f(x, y) = x2y

∆f(x0, y0) = f(x0 + ∆x, y0 + ∆y)− f(x0, y0)

= (x0 + ∆x)2(y0 + ∆y)− x20 y0

∆f(x0, y0) = 2x0y0∆x+ x20 ∆y + y0(∆x)2 + 2x0∆x∆y + (∆x)2∆y · · · (i)

Now, if we put A = 2x0y0, B = x20, ε1 = y0∆x+ x0∆y and

ε2 = x0∆x+ (∆x)2 then expression (i) can be written as∆f(x0, y0) = A∆x+B∆y + ε1∆x+ ε2∆y; · · · (ii)where A and B are independent of ∆x and ∆y, and

lim(∆x,∆y)→(0,0)

ε1 = 0, lim(∆x,∆y)→(0,0)

ε2 = 0.

Here, the function f(x, y) is said to have a differential at point (x0, y0). It is denoted by df.Thus, df = A∆x+B∆y.Note that when ∆x and ∆y are sufficientely small df gives a goodapproximation of ∆f(x0, y0).

3.1: Definition (Differentiability)A function f(x, y) is said to be differentiable at a point (x0, y0) if there exists a neighbourhood(x0 + ∆x, y0 + ∆y) of (x0, y0) in which the increment ∆f(x0, y0) can be expressed in the form

∆f(x0, y0) = f(x0 + ∆x, y0 + ∆y)− f(x0, y0)

= A∆x+B∆y + ε1∆x+ ε2∆y;

where A and B are independent of ∆x and ∆y andlim

(∆x,∆y)→(0,0)ε1 = 0, lim

(∆x,∆y)→(0,0)ε2 = 0.

Theorem 1: (Necessary conditions for differentiability :-)suppose f(x, y) is a real valued function defined on a neighbourhood of (x0, y0). If f(x, y) is differ-entiable at (x0, y0) then(i) fx(x0, y0) and fy(x0, y0) both exists(ii) f(x, y) is continuous at (x0, y0).

Proof : Assume that f(x, y) is differentiable at point (x0, y0).

25

Chapter 3. Differentiability

(i) ∴ By the definition of differentiability at (x0, y0)

∆f(x0, y0) = f(x0 + ∆x, y0 + ∆y)− f(x0, y0)

= A∆x+B∆y + ε1∆x+ ε2∆y · · · (1),

where A and B are constants independent of ∆x,∆y and

lim(∆x,∆y)→(0,0)

ε1 = 0, lim(∆x,∆y)→(0,0)

ε2 = 0.

Equation (1) is true for small values of ∆x and ∆y.Put ∆y = 0 in equation (1), we get

f(x0 + ∆x, y0)− f(x0, y0) = A∆x+ ε1∆x

∴ (A+ ε1)∆x = f(x0 + ∆x, y0)− f(x0, y0)

∴ A+ ε1 =f(x0 + ∆x, y0)− f(x0, y0)

∆x

∴ lim∆x→0

[A+ ε1] = lim∆x→0

[f(x0 + ∆x, y0)− f(x0, y0)

∆x

]∴ A = lim

∆x→0

(f(x0 + ∆x, y0)− f(x0, y0)

∆x

)= = fx(x0, y0).

i.e. A = fx(x0, y0) exist.Similarly by putting ∆x = 0 in equation (1) we get B = fy(x0, y0).This proves condition (i).(ii) Taking limit as (∆x,∆y)→ (0, 0) of Equation (1) we get

lim(∆x,∆y)→(0,0)

[f(x0 + ∆x, y0 + ∆y)− f(x0, y0)] = 0

... the limit of each term on R.H.S. is 0.

∴ lim(∆x,∆y)→(0,0)

f(x0 + ∆x, y0 + ∆y) = f(x0, y0)

This shows that f(x, y) is continuous at (x0, y0).

Remark 1: A function f(x, y) is differentiable at (x0, y0) iff the partial derivatives fx(x0, y0) andfy(x0, y0) exists and

∆f = f(x0 + ∆x, y0 + ∆y)− f(x0, y0)

= fx(x0, y0)∆x+ fy(x0, y0)∆y + ε1∆x+ ε2∆y;

where ε1 → 0, ε2 → 0 as (∆x,∆y)→ (0, 0).

Remark 2: The converse of the above theorem is not true i.e. above conditions are not sufficient.

Theorem 2: (Increment theorem for function of two variables :-)Suppose that the first partial derivatives of function z = f(x, y) are defined through-out openregion R containing the point (x0, y0) and fx(x0, y0), fy(x0, y0) are continuous. Then the change (orincrement) ∆Z = f(x0+∆x, y0+∆y)−f(x0, y0) in the value of f from (x0, y0) to (x0+∆x, y0+∆y)in R satisfies ∆Z = fx(x0, y0)∆x+fy(x0, y0)∆y+ε1∆x+ε2∆y; where ε1 → 0, ε2 → 0 as ∆x,∆y →0.

example 1: Show that the function f(x, y) =√|xy| has first partial derivatives at the origin but

it is not differentiable at the origin.

Solution : Given that f(x, y) =√|xy|, (x0, y0) = (0, 0).

First let us find the first partial derivatives of f(x, y) at the origin.

∴ fx(0, 0) = lim∆x→0

(f(0 + ∆x, 0)− f(0, 0)

∆x

)fx(0, 0) = lim

∆x→0

(√|∆x, 0| ·

√|0|

∆x

)

= lim∆x→0

(0

∆x

)= 0

∴ fx(0, 0) = 0 · · · (i)Page|26


Similarly, fy(0, 0) = 0 · · · (ii)From (i) and (ii) both the first partial derivatives of f(x, y) exists at (0, 0).Now, suppose that f is differentiable at (0, 0) then by the definition of differentiability

f(∆x,∆y)− f(0, 0) = fx(0, 0)∆x+ fy(0, 0)∆y + ε1∆x+ ε2∆y;

∴√|∆x ·∆y| −

√|0| = 0 ·∆x+ 0 ·∆y + ε1∆x+ ε2∆y · · · (iii)

ε1, ε2 → 0 as (∆x,∆y)→ (0, 0).Since (iii) holds for all small values of ∆x and ∆y, put ∆y = ∆x in (iii), we get√

|(∆x)2| = ε1∆x+ ε2∆x

∴ |∆x| = ∆x(ε1 + ε2)

∴|∆x|∆x

= ε1 + ε2

Taking limit as ∆x→ 0 of both sides.

∴ lim∆x→0

|∆x|∆x

= lim∆x→0

(ε1 + ε2)

∴ ±1 = 0;

which is absurd. Hence f is not differentiable at (0, 0).Moreover,For continuity of f(x, y) at (0, 0). Consider

|f(x, y)− f(0, 0)| = |√|xy||

=√x · √y

≤ x2 + y2 < ε...√x ≤

√x2 + y2

√y ≤

√x2 + y2

⇒√x2 + y2 <

√ε(= δ) Thus, |f(x, y)− f(0, 0)| < ε, whenever

√x2 + y2 < δ

⇒ f(x, y) is continuous at (0, 0).

example 2: Show that the function f(x, y) = |x|(1 + y) is not differentiable at (0, 0) but iscontinuous at (0, 0).

Solution : Given that

f(x, y) = |x|(1 + y).

(x0, y0) = (0, 0)

∴ f(x0, y0) = f(0, 0)

= 0

Consider

lim∆x→0

f(x0 + ∆x, y0)− f(x0, y0)

∆x= lim

∆x→0

f(∆x, 0)− f(0, 0)

∆x

= lim∆x→0

|∆x|(1 + 0)− 0

∆x

= lim∆x→0

|∆x|∆x

Now

lim∆x→0+

|∆x|∆x

= lim∆x→0+

(∆x

∆x

)= 1 · · · (i)

lim∆x→0−

|∆x|∆x

= lim∆x→0−

(−∆x

∆x

)= −1 · · · (ii)

∴ lim∆x→0

(|∆x|∆x

)does not exist : (

... by (i) and (ii))

i.e. lim∆x→0

f(∆x, 0)− f(0, 0)

∆xdoes not exist, which means that fx(0, 0) does not exist.

Since existence of fx(0, 0) and fy(0, 0) is a necessary condition for differentiability, therefore f isnot differentiable at (0, 0).

Page|27


To show that f(x, y) is continuous at (0, 0) we will use ε− δ definition.Let ε > 0. Consider

|f(x, y)− f(0, 0)| = |f(x, y)− 0| = ||x|(1 + y)|= |x||1 + y|≤ 2|x|, if |y| < 1

∴ |f(x, y)− f(0, 0)| ≤ 2|x| < ε

∴ |f(x, y)− f(0, 0)| < ε, if |x| < ε

2= δ

Take δ = min{ε

2, 1}

then |f(x, y)− f(0, 0)| < ε when |x| < δ, |y| < δ

∴ lim∆x→0

f(x, y) = 0 = f(0, 0)⇒ f(x, y) is continuous at (0, 0).

example 3: Let

f(x, y) =2xy

x2 + y2, if (x, y) 6= (0, 0)

= 0, if (x, y) = (0, 0)

Show that f(x, y) is not differentiable at (0, 0) even though fx(0, 0) and fy(0, 0) exists.

Solution : First let us show that fx(0, 0) & fy(0, 0) exist

fx(0, 0) = lim∆x→0

f(∆x, 0)− f(0, 0)

∆x

= lim∆x→0

0− 0

∆x= 0

Similarly fy(0, 0) = 0 i.e. both fx(0, 0) & fy(0, 0) exist.Now, we will find the limit of f(x, y) along a path y = mx,m 6= 0.

∴ lim(x,y)→(0,0)

f(x, y) = lim(x,mx)→(0,0)

f(x,mx)

= limx→0

(2x ·mxx2 +m2x2

)=

2m

1 +m2

which depends upon the path. i.e. lim(x,y)→(0,0)

f(x, y) does not exist.

Hence, f is not continuous at (0, 0).Therefore f is not differentiable at (0, 0).

example 4: Let

f(x, y) = 2xyx2 − y2

x2 + y2, (x, y) 6= (0, 0)

= 0, (x, y) = (0, 0)

Show that f(x, y) is differentiable at (0, 0).

Solution :

fx(0, 0) = lim∆x→0

f(∆x, 0)− f(0, 0)

∆x

= lim∆x→0

0− 0

∆x= 0

Similarly fy(0, 0) = 0.∴ fx(0, 0) and fy(0, 0) both exist.Now,

∆f = f(x0 + ∆x, y0 + ∆y)− f(x0, y0)

∆f = f(∆x,∆y)− f(0, 0)

∴ f(∆x,∆y)− f(0, 0) = 0 ·∆x+ 0 ·∆y + ε1∆x+ ε2∆y; where

Page|28


ε1 =2(∆x)2∆y

(∆x)2 + (∆y)2, if (∆x,∆y) 6= (0, 0)

= 0, if (∆x,∆y) = (0, 0)

and

ε2 =−2(∆x)(∆y)2

(∆x)2 + (∆y)2, if (∆x,∆y) 6= (0, 0)

= 0, if (∆x,∆y) = (0, 0)

Here as (∆x,∆y) → (0, 0), ε1 → 0, ε2 → 0. ∴ f(∆x,∆y) − f(0, 0) = fx(0, 0)∆x + fy(0, 0)∆y +ε1∆x+ ε2∆y; ε1 → 0, ε2 → 0 as (∆x,∆y)→ (0, 0)Hence by the definition, f(x, y) is differentiable at (0, 0).

Theorem 3: (Sufficient Conditions for Differentiability :-)If f(x, y) is a function of two variables x and y such that

(i) fx(a, b) and fy(a, b) exist

(ii) One of the first partial derivatives fx, fy is continuous at (a, b).Then f(x, y) is differentiable at (a, b).

Proof : Suppose fy is continuous at (a, b) ⇒ fy exist in the neighbourhood of (a, b), (say squareδ- neighbourhood of (a, b)) i.e ∃δ > 0 so that the point (a + h, b + k) lies in the δ-neighbourhoodof (a, b) where |h| < δ, |k| < δ.Now

∆f = f(a+ h), b+ k)− f(a, b)

= f(a+ h, b+ k)− f(a+ h, b) + f(a+ h, b)− f(a, b) · · · ?

Dfine the function g(y) as g(y) = f(a+ h, y)Here g is derivable in (b, b+ k) and we have g′(y) = fy(a+ h, y).Also g is continuous in [b, b+ k].Hence by LMV T (IInd form)g(b+ k)− g(b) = kg′(b+ kθ); 0 < θ < 1.i.e. f(a+ h, b+ k)− f(a+ h, b) = kfy(a+ h, b+ kθ) · · · (1)Since fy is continuous at (a, b)

lim(h,k)→(0,0)

fy(a+ h, b+ kθ) = fy(a, b)

lim(h,k)→(0,0)

[fy(a+ h, b+ kθ)− fy(a, b)] = 0

If we put fy(a+ h, b+ kθ)− fy(a, b) = ψ(h, k) then lim(h,k)→(0,0)

ψ(h, k) = 0.

With this equation (1) becomes

f(a+ h, b+ k)− f(a+ h, b) = k(fy(a, b) + ψ(h, k))

f(a+ h, b+ k)− f(a+ h, b) = kfy(a, b) + kψ(h, k) · · · (2)

Now we hare

fx(a, b) = limh→0

f(a+ h, b)− f(a, b)

h

∴ limh→0

[f(a+ h, b)− f(a, b)

h− fx(a, b)

]= 0.

Put φ(h) =f(a+ h, b)− f(a, b)

h− fx(a, b) then lim

h→0φ(h) = 0 i.e. φ(h)→ 0 as (h, k)→ (0, 0).

∴ f(a+ h, b)− f(a, b) = hfx(a, b) + hφ(h, k) · · · (3)Putting (2), (3) and (1) in ? we get

∆f = f(a+ h, b+ k)− f(a, b) = hfx(a, b) + kfy(a, b) + hφ(h, k) + kψ(h, k); where φ(h, k)→ 0 andψ(h, k)→ 0 as (h, k)→ (0, 0).Hence, by the definition of differentiability, f(x, y) is differentiable at (a, b).

Differentials : Let z = f(x, y) be a differentiable function of two variables x and y. The differential

or total differential of z; denoted by dz; is defined as dz =∂z

∂xdx +

∂z

∂ydy; where dx and dy (are

called the differentials of x and y) are two new independent variables.

Page|29


Suppose z = f(x, y) is differentiable at (x0, y0). Then

∆z = f(x0 + ∆x, y0 + ∆y)− f(x0, y0)

∆z =∂z

∂x∆x+

∂z

∂y∆y + ε1∆x+ ε2∆y;

ε1, ε2 → 0 as (∆x,∆y)→ (0, 0).For small values of ∆x & ∆y∆z = dz + ε1∆x+ ε2∆y; where∆x,∆y are increments in x and y respectively.Hence, the increment ∆z is approximately equal to the differential dz.i.e. we can compute the approximate value of the given function by using differential.

Formula is

f(x0 + ∆x, y0 + ∆y) ≈ f(x0, y0) + df ; where

df =∂f

∂x(x0, y0)∆x+

∂f

∂y(x0, y0)∆y

Working Rule : Given any function f(x, y)

(i) Decide x0, y0 and ∆x,∆y.(ii) Find f(x0, y0).

(iii)

(∂f

∂x

)(x0,y0)

,

(∂f

∂y

)(x0,y0)

obtain these values.

(iv) Use the formula.

example 1: Using differentials find the approximate value of(2 · 01)(3 · 02)2.

Solution : Let f(x, y) = xy2

∴ f(x0 + ∆x, y0 + ∆y) = (2 · 01)(3 · 02)2.Here, x0 = 2, y0 = 3 and ∆x = 0 · 01, ∆y = 0 · 02.

∴ f(x0, y0) = f(2, 3) = 2(3)2 = 18

fx(x0, y0) =

(∂f

∂x

)(x0,y0)

= y20

∴ fx(2, 3) = (3)2 = 9

fy(x0, y0) =

(∂f

∂y

)(x0,y0)

= 2x0y0

∴ fy(2, 3) = 2(2)(3) = 12.

∴ df = fx(x0, y0)∆x+ fy(x0, y0)∆y∴ df = y2

0∆x+ 2x0y0∆y= 9(0 · 01) + 12(0 · 02)

∴ df = 0 · 33.

Hence,

f(x0 + ∆x, y0 + ∆y) ≈ f(x0, y0) + df

∴ (2 · 01)(3 · 02)2 ≈ 18 + 0 · 33

= 18 · 33

example 2: Find approximate value of

√4 · 1

25 · 01by using differentials.

Solution : Let f(x, y) =

√x

y.

Here, x0 = 4, y0 = 25 and ∆x = 0 · 1, ∆ = 0 · 01.

∴ f(x0, y0) = f(4 · 25) =

√4

25=

2

5.

fx(x0, y0) =1

2√x0 y0

∴ fx(4 · 25) =1

2√

4 · 25=

1

20

fy(x0, y0) =−1

2

√x0

y30

∴ fy(4 · 25) =−1

2

√4

(25)3=−1

2· 2

125.

Page|30


∴ df = fx(x0, y0)∆x+ fy(x0, y0)∆y

∴ df =1

20(0 · 1)− 1

125× (0 · 01)

= 0 · 005− 0 · 00008∴ df = 0 · 00492.

Hence,

f(x0 + ∆x, y0 + ∆y) ≈ f(x0, y0) + df

∴

√4 · 1

25 · 01≈ 2

3+ 0 · 00492

= 0 · 4 + 0 · 00492= 0 · 40492.

Exercise (A)

(1) Show that the function

f(x, y) =xy√x2 + y2

, if x2 + y2 6= 0

= 0, if x = y = 0.

is continuous at the origin but is not differentiable at the origin.(2) If

f(x, y) = x sin

(1

x

)+ y sin

(1

y

), y 6= 0

= 0, x, y = 0.

(3) If f(x, y) =x4 + y4

x2 + y2, x2 + y2 6= 0 and f(0, 0) = 0. Show that f(x, y) is differentiable at

(0, 0).

(4) Show that f(x, y) =x3 − y3

x2 + y2, x2 + y2 6= 0 is continuous at (0, 0) but not differentiable at

(0, 0).(5) Using differentials, find the approximate value of

(i) [3 · 8)2 + 2(2 · 1)3]1/5 (ii) f(5 · 12, 6 · 85)

(iii)√

(5 · 98)2 + (8 · 01)2 (iv) f(5 · 04× 9 · 98).(6) If f(x, y) = x3 − xy + 3y2 then find f(4 · 98, 4 · 1).

(7) If f(x, y, z) = (x2 + y2 + z2)1/2, find f(0 · 1, 4 · 01, 3 · 1).(8) Let z = x2 − xy + 3y2 then find dz if x = 5, y = 4,∆x = −2,∆y = 1.

Answers

(5) (i) 2 · 01 (ii) 5 · 10 (iii) 9 · 996 (iv) 50 · 3 (6) 140 · 7(7) 5 · 068 (8) −123.

3.2 : Composite Function : Chain Rule.For function of one variable, y = f(x) and x = φ(t) then y = f(φ(t)) is called a composite functionof t.

Its derivative w.r.t. ′t′ is given bydy

dt=dy

dx· dxdt

which is known as Chain Rule.

For function of two variables also we will have composite function and Chain Rule.

1. Suppose u = f(x, y) is a function of two independent variables x, y and x, y are themselvesfunction of single variable′t′ i.e. x = φ(t), y = ψ(t) thenu = f [φ(t), ψ(t)] = F (t) is called a composite function of single variable′t′.e.q.:(1) u = f(x, y) = x+ y and x = at, y = bt2

then u = f(at, bt2) = at+ bt2 is a composite function of single variable ′t′.

(2) u = sin(x+ y2) and x = cost, y = t2 + 1 thenu = sin[cost + (t2 + 1)2], a composite function of ′t′.

2. Suppose W = f(u, v) is a function of two variables u, v and u, v are functions of twovariables x and yi.e. u = φ(x, y), v = ψ(x, y) thenW = f [φ(x, y), ψ(x, y)] = F (x, y) is called a composite function of two variables x and y.

Page|31


e.q.: W = f(u, v) and u = x+ y, v = x− y thenW = f(x+ y, x− y) is a composite function of x and y.

3. Suppose Z = f(x) is a function of one variable x and x is itself a function of two variablesu and v i.e. x = φ(u, v) thenZ = f(φ(u, v)) is a composite function two variables u and v.

e.q.: Z = f(u); u = ax+ by then Z = f(ax+ by) is a composite function of x and y.

Theorem 4: Chain Rule (I):If u = f(x, y) is a differentiable function of x and y, x = φ(t), y = ψ(t) are themselves functions ofsingle variable ′t′. Then the composite function u = f [φ(t), ψ(t)] is differentiable function of single

variable ′t′ and its total derivative is given bydu

dt=∂u

∂x· dxdt

+∂u

∂y· dydt.

Proof : We have given that u = f(x, y) and x = φ(t), y = ψ(t). Let ∆x = φ(t + ∆t) − φ(t) and∆y = ψ(t + ∆t) − ψ(t) be the increments in x and y respectively corresponding to an increment∆t in t.

Since u = f(x, y) is differentiable, by increment theorem ∆u =∂u

∂x∆x+

∂u

∂y∆y+ε1∆x+ε2∆y · · · (1)

where ε1 → 0, ε2 → 0 as (∆x,∆y)→ (0, 0).

∴ ∆u =

(∂u

∂x+ ε1

)∆x+

(∂u

∂y+ ε2

)∆y

∴∆u

∆t=

(∂u

∂x+ ε1

)∆x

∆t+

(∂u

∂y+ ε2

)∆y

∆t· · · (2)

As x = φ(t), y = ψ(t) are differentiable functions at t∴ they are continuous at t and hence ∆x, ∆y → 0 as ∆t→ 0.∴ ε1 → 0, ε2 → 0 as ∆t→ 0.

Also lim∆t→0

∆x

∆t=dx

dtand lim

∆t→0

∆y

∆t=dy

dt.

Taking limit as ∆t→ 0 of equation (2)

lim∆t→0

∆u

∆t=

[lim

∆t→0

(∂u

∂x+ ε1

)][lim

∆t→0

∆x

∆t

]+

[lim

∆t→0

(∂u

∂y+ ε2

)][lim

∆t→0

∆y

∆t

]⇒ du

dt=∂u

∂x· dxdt

+∂u

∂y· dydt

which proves Chain Rule (I).

Theorem 5: Chain Rule (II):If W = f(u, v) is a differentiable function of two variables u and v.u = φ(x, y), v = ψ(x, y) are differentiable functions of x and y then the composite functionW = f [φ(x, y), ψ(x, y)] = F (x, y) is alsodifferentiable and

∂w

∂x=

∂w

∂u· ∂u∂x

+∂w

∂v· ∂v∂x,

∂w

∂y=

∂w

∂u· ∂u∂y

+∂w

∂v· ∂v∂y

Proof : Since u, v and w are differentiable functions, by Chain Rule (I)

∆u =∂u

∂x∆x+

∂u

∂y∆y + ε1∆x+ ε2∆y · · · (1)

∆v =∂v

∂x∆x+

∂v

∂y∆y + ε3∆x+ ε4∆y · · · (2)

∆w =∂w

∂u∆u+

∂w

∂v∆v + ε5∆u+ ε6∆v · · · (3);

where ε1, ε2, ε3, ε4 → 0 as (∆x,∆y)→ (0, 0) andε5, ε6 → 0 as (∆u,∆v)⇒ (0, 0).

Page|32


Now ∆w =

(∂w

∂u+ ε5

)∆u+

(∂w

∂v+ ε6

)∆v ∴ by (3), Vsing (1) and (2) in above we get

∆w =

(∂w

∂u+ ε5

)(∂u

∂x∆x+

∂u

∂y∆y + ε1∆x+ ε2∆y

)+

(∂w

∂v+ ε6

)(∂v

∂x∆x+

∂v

∂y∆y + ε3∆x+ ε4∆y

)∴ ∆w =

(∂w

∂u· ∂u∂x

+∂w

∂v· ∂v∂x

)∆x+

(∂w

∂u· ∂u∂y

+∂w

∂v· ∂v∂y

)∆y

+α1∆x+ α2∆y · · · (4);

where α1 and α2 are sum of terms each containing the factors ε1, ε2, · · · ε6.∴ α1 → 0, α2 → 0 as (∆x,∆y)→ (0, 0).From (4) w = F (x, y) is differentiable at (x, y).

Now put ∆y = 0 and divide by ∆x; Equation (4) becomes∆w

∆x=∂w

∂u· ∂u∂x

+∂w

∂v· ∂v∂x

+ α1 Taking

limit as ∆x → 0.∂w

∂x=∂w

∂u· ∂u∂x

+∂w

∂v· ∂v∂x

Similarly put ∆x = 0 and divide by ∆y to equation

(4); taking limit as ∆y → 0 we get∂w

∂y=∂w

∂u· ∂u∂y

+∂w

∂v· ∂v∂y

which proves the theorem.

# The Chain Rule for functions of three variables:-If W = f(x, y, z) is a differentiable function of three variables x, y, z and x, y, z are differentiablefunctions of single variable ′t′ then the composite function w = f(t) is also differentiable function

of ′t′ and its derivative isdw

dt=∂w

∂x· dxdt

+∂w

∂y· dydt

+∂w

∂z· dzdt. # Chain Rule for functions on

surfaces:-IfW = f(x, y, z) is a differentiable function of three variables x, y, z and x = g(r, s), y = h(r, s), z =k(r, s) are differentiable functions of two variables r and s. Then the composite function

w = f [g(r, s), h(r, s), k(r, s)] = F (r, s) is differentiable function of r and s; and∂w

∂r=∂w

∂x· ∂x∂r

+

∂w

∂y· ∂y∂r

+∂w

∂z· ∂z∂r,∂w

∂s=∂w

∂x· ∂x∂s

+∂w

∂y· ∂y∂s

+∂w

∂z· ∂z∂s, # Chain Rule for functions of many

variables:-If W = f(x1, x2, x3, · · · , xn) is a differentiable function of a finite set of varibles x1, x2, x3, · · · , xnand each x1, x2, x3, · · ·xn is a differentiable function of a finite set of variables p1, p2, p3, · · · , pr.Then W = f [p1, p2, p3, · · · , pr] is differentiable function of finite set of variables p1, p2, p3, · · · , prand we have

∂w

∂p1=

∂w

∂x1· ∂x1

∂p1+∂w

∂x2· ∂x2

∂p1+∂w

∂x3· ∂x3

∂p1+ · · ·+ ∂w

∂xn· ∂xn∂p1

,

∂w

∂p2=

∂w

∂x1· ∂x1

∂p2+∂w

∂x2· ∂x2

∂p2+∂w

∂x3· ∂x3

∂p2+ · · ·+ ∂w

∂xn· ∂xn∂p2

,

so on∂w

∂pr=

∂w

∂x1· ∂x1

∂pr+∂w

∂x2· ∂x2

∂pr+∂w

∂x3· ∂x3

∂pr+ · · ·+ ∂w

∂xn· ∂xn∂pr

example 1: If w = f(ax+ by) then show that b∂w

∂x− a∂w

∂y= 0.

Solution : We have given that w = f(ax+ by) put u = ax+ by then w = f(u).∴ By Chain Rule

∂w

∂x=

dw

du· ∂u∂x

= a · dwdu

∴ b∂w

∂x= ab

dw

du· · · (1)

∂w

∂y=

dw

du· ∂u∂y

= b · dwdu

∴ a∂w

∂y= ab

dw

du· · · (2)

Page|33


From (1) and (2) b∂w

∂x− a∂w

∂y= 0.

example 2: If z = f(y + ax) + g(y − ax) prove that zxx = a2zyy, assuming that second orderpartial derivative of f, g exist and a is constant.

Solution : Put u = y + ax, v = y − ax. Hence, z = f(u) + g(v); where

u = φ(y, x) = y + ax

v = ψ(y, x) = y − ax

∴ By Chain Rule

zx =∂z

∂x=

∂z

∂u· ∂u∂x

+∂z

∂v· ∂v∂x

= f ′(u)a+ g′(v) · (−a)

zx =∂z

∂x= a(f ′(u)− g′(v)) · · · (1)

Again differentiate w.r.t. x.

zxx =∂2z

∂x2=

∂

∂u[a(f ′(u)− g′(v))] · ∂u

∂x+

∂

∂v[a(f ′(u)− g′(v))] · ∂v

∂x

zxx = a2f ′′(u) + a2g′′(v) · · · (2)

Now zy =∂z

∂y=

∂z

∂u· ∂u∂y

+∂z

∂v· ∂v∂y

= f ′(u) + g′(v) · · · (3) Differentiate w.r.t. y again, we get

zyy =∂2z

∂y2= f ′′(u) + g′′(v) · · · (4) from (2) and (4) zxx = a2zyy proved.

example 3: If u = xy2 log(yx

)then find du.

Solution : We know that du =∂u

∂xdx+

∂u

∂ydy · · · (1) Now

∂u

∂x= y2 log

(yx

)+ xy2 · 1

y

x

(−1

x2

)· y

= y2 log(yx

)− y2 · · · (2)

∂u

∂y= 2xy log

(yx

)+ xy2 · 1

y

x

· 1

x

= 2xy log(yx

)+ xy · · · (3)

From (2) and (3) equation (1) becomes du =[y2 log

(yx

)− y2

]dx +

[2xy log

(yx

)+ xy

]dy ex-

ample 4: If u = x3 + y3; x = a cos t, y = b sin t then finddu

dt. Also verify the result.

Solution : We have given that u = x3 + y3; x = a cos t, y = b sin t.∴ u is a composite function of single variable t.

du

dt=

∂u

∂x· dxdt

+∂u

∂y· dydt

du

dt= 3x2(−a sin t) + 3y2(b cos t)

Putting values of x and y we getdu

dt= −3a3 cos2 t sin t + 3b3 sin2 t cos t · · · (1) Verification :

We have u = x3 + y3 putting values of x and y u = a3 cos3 t + b3 sin3 t. Differentiate w.r.t. t.

∴du

dt= −3a3 cos2 t · sin t+ 3b3 sin2 t · cos t which is equation (1).

example 5: If u = u

(y − xxy

,z − xxz

), show that x2∂u

∂x+ y2∂u

∂y+ z2∂u

∂z= 0 Solution : We have

Page|34


given that u = u

(y − xxy

,z − xxz

), put

r =y − xxy

=1

x− 1

yand

s =z − xxz

=1

x− 1

z

∴ u = u(r, s) i.e. u is composite function of x and y.

∴ Chain Rule∂u

∂x=∂u

∂r· ∂r∂x

+∂u

∂s· ∂s∂x· · · (1) Here

∂r

∂x= − 1

x2,∂r

∂y=

1

y2,∂r

∂z= 0. and

∂s

∂x= − 1

x2,∂s

∂y= 0,

∂s

∂z=

1

z2.

∴ Equation (1) becomes x2∂u

∂x= −∂u

∂r− ∂u

∂s· · · (2) Again

∂u

∂y=

∂u

∂r· ∂r∂y

+∂u

∂s· ∂s∂y

∴ y2∂u

∂y=

∂u

∂r· · · (3)

Also

∂u

∂z=

∂u

∂r· ∂r∂z

+∂u

∂s· ∂s∂z

∴ z2∂u

∂z=

∂u

∂s· · · (4)

Adding (2), (3) and (4) we get x2∂u

∂x+ y2∂u

∂y+ z2∂u

∂z= 0. example 6: If ϕ(x − az, cy − bz) = 0

show that ap+ bq = c; where p =∂z

∂x, q =

∂z

∂y.

Solution : We have ϕ(cx− az, cy − bz) = 0. Put r = cx− az, s = cy − bz so that ϕ(r, s) = 0.Differentiate r and s partially w.r.t. x, y.

∴∂r

∂x= c− a∂z

∂x,∂r

∂y= −a∂z

∂yand

∂s

∂x= −b∂z

∂x,∂s

∂y= c− b∂z

∂y.

Now by chain rule to ϕ we have∂ϕ

∂x=∂ϕ

∂r· ∂r∂x

+∂ϕ

∂s· ∂s∂x

∴ 0 =∂ϕ

∂r

(c− a∂z

∂x

)+∂ϕ

∂s

(−b∂z

∂x

)0 = c

∂ϕ

∂r+∂z

∂x

(−a∂ϕ

∂r− b∂ϕ

∂s

)

⇒ a∂z

∂x=

ac∂ϕ

∂r

a∂ϕ

∂r+ b

∂ϕ

∂s

· · · (1)

Also

∂ϕ

∂y=

∂ϕ

∂r· ∂r∂y

+∂ϕ

∂s· ∂s∂y

∴ 0 =∂ϕ

∂r

(−a∂z

∂y

)+∂ϕ

∂s

(c− b∂z

∂y

)0 = c

∂ϕ

∂s− ∂z

∂y

(a∂ϕ

∂r+ b

∂ϕ

∂s

)

⇒ b∂z

∂y=

bc∂ϕ

∂s

a∂ϕ

∂r+ b

∂ϕ

∂s

· · · (2)

Page|35


Adding (1) and (2) a∂z

∂x+ b

∂z

∂y=

ac∂ϕ

∂r+ bc

∂ϕ

∂s

a∂ϕ

∂r+ b

∂ϕ

∂s

= c = R.H.S. example 7: If w = xy and

x = cos t, y = sin t finddw

dtat t =

π

2.

Solution : We have w = xy

M-(I) : w = cos t · sin t =1

2sin 2t

∴dw

dt=

d

dt

(1

2sin 2t

)=

1

2· 2 cos 2t = cos 2t

∴

(dw

dt

)t=π

2

= cos 2 · π2

= −1.

M-(II) : By using Chain Rule

dw

dt=

∂w

∂x· dxdt

+∂w

∂y· dydt

∴dw

dt= y(− sin t) + x(cos t)

dw

dt= − sin2 t+ cos2 t = cos 2t

∴

(dw

dt

)t=π

2

= cos 2 · π2

= −1.

example 8: If z = f(x, y); x = uv, y =u+ v

u− v.

Then prove that 2x∂z

∂x= u

∂z

∂u+ v

∂z

∂v.

Solution : z = f(x, y); x = uv, y =u+ v

u− v.

∴ z = F [u, v] ∴ By Chain Rule.

∂z

∂u=

∂z

∂x· ∂x∂u

+∂z

∂y· ∂y∂u

=∂z

∂x(v) +

∂z

∂y

(u− v − (u+ v)

(u− v)2

)∂z

∂u= v

∂z

∂x− 2v

(u− v)2

∂z

∂y

⇒ u∂z

∂u= uv

∂z

∂x− 2uv

(u− v)2

∂z

∂y· · · (1)

Again

∂z

∂v=

∂z

∂x· ∂x∂v

+∂z

∂y· ∂y∂v

=∂z

∂x(u) +

∂z

∂y

(u− v − (u+ v)(−1)

(u− v)2

)∂z

∂v= u

∂z

∂x− 2u

(u− v)2

∂z

∂y

⇒ v∂z

∂v= uv

∂z

∂x− 2uv

(u− v)2

∂z

∂y· · · (2)

Adding (1) and (2)

u∂z

∂x+ v

∂z

∂y= 2uv

∂z

∂x

u∂z

∂x+ v

∂z

∂y= 2x

∂z

∂x

Hence the result is proved.

Page|36


example 9: If x = r cos θ, y = r sin θ then show that∂2θ

∂x∂y=

∂2

∂x2(log r) = − ∂2

∂y2(log r) =

−1

r2cos 2θ Solution : We have given that x = r cos θ, y = r sin θ. ∴ x2 + y2 = r2 and tan θ =

y

x⇒

θ = tan−1(y/x) · · · (1)

Differentiate (1) w.r.t. y partially ∴∂θ

∂y=

1

1 + (y/x)2· ∂∂y

(y/x) =x2

x2 + y2· 1

x=

x

x2 + y2· · · (2)

Now differentiate w.r.t. x partially ∴∂2θ

∂x∂y=

(x2 + y2)− x(2x)

(x2 + y2)2=

y2 − x2

(x2 + y2)2· · · (a) Again

r2 = x2 + y2 ⇒ r = (x2 + y2)1/2. Taking logaritham we get log r =1

2log(x2 + y2).

Differentiate w.r.t. x∂

∂x(log r) =

1

2· 2x

x2 + y2=

x

x2 + y2Again differentiate w.r.t. x

∂2

∂x2(log r) =

y2 − x2

(x2 + y2)2· · · (b) Similarly

∂2

∂y2(log r) =

x2 − y2

(x2 + y2)2=−(y2 − x2)

(x2 + y2)2· · · (c) Now, y2−x2 = r2 sin2 θ−

r2 cos2 θ = 0r2 cos 2θ · · · (d) ∴ from (a), (b), (c) and (d) we get∂2θ

∂x∂y=

∂2

∂x2(log r) = − ∂2

∂y2(log r) =

− 1

r2cos 2θ Hence proved

example 10: If u = f(r) and x = r cos θ, y = r sin θ then prove that∂2u

∂x2+∂2u

∂y2= f ′′(r) +

1

rf ′(r).

Solution : We have given that u = f(r) and x = r cos θ, y = r sin θ so that r2 = x2 + y2 and∂r

∂x=x

r.

Now∂u

∂x=df

dr· ∂r∂x,∂u

∂y=df

dr· ∂r∂y

∴∂u

∂x=df

dr· xr

∴∂2u

∂x2=

(d2f

dr2· ∂r∂x

)· xr

+df

dr

r − x · ∂r∂xr2

∴∂2u

∂x2=

(d2f

dr2· xr

)· xr

+df

dr

r − x · xrr2

=

d2f

dr2· x

2

r2+df

dr· r

2 − x2

r3

∂2u

∂x2=

d2f

dr2· x

2

r2+df

dr· y

2

r3· · · (1)

Similarly∂2u

∂y2=d2f

dr2· y

2

r2+df

dr· x

2

r3· · · (2) Adding (1) and (2) we get

∂2u

∂x2+∂2u

∂y2=

d2f

dr2· x

2 + y2

r2+df

dr· x

2 + y2

r3

=d2f

dr2· 1 +

df

dr· r

2

r3

=d2f

dr2+

1

r· dfdr

∂2u

∂x2+∂2u

∂y2= f ′′(r) +

1

r· f ′(r)

L.H.S. = R.H.S.

Hence proved.

example 11: If u = f(x, y) and x = r cos θ, y = r sin θ then Show that

(∂u

∂x

)2

+

(∂u

∂y

)2

=

Page|37

Chapter 3. Differentiability(∂u

∂r

)2

+1

r2

(∂u

∂θ

)2

Solution : We have

u = f(x, y)

u = f(r cos θ, r sin θ) = F (r, θ)

∴ By Chain Rule

∂u

∂r=

∂u

∂x· ∂x∂r

+∂u

∂y· ∂y∂r

∴∂u

∂r=

∂u

∂x· cos θ +

∂u

∂ysin θ · · · (1)

and

∂u

∂θ=

∂u

∂x· ∂x∂θ

+∂u

∂y· ∂y∂θ

∴∂u

∂θ=

∂u

∂x(−r sin θ) +

∂u

∂y(r cos θ)

∴1

r· ∂u∂θ

= −∂u∂x

sin θ +∂u

∂ycos θ · · · (2)

Squaring and adding (1) and (2) we get(∂u

∂r

)2

+1

r2

(∂u

∂θ

)2

=

(∂u

∂x

)2

(cos2 θ + sin2 θ) +

(∂u

∂y

)2

(sin2 θ + cos2 θ)(∂u

∂r

)2

+1

r2

(∂u

∂θ

)2

=

(∂u

∂x

)2

+

(∂u

∂y

)2

.

example 12: Show that u = f(x2y); where f is differentiable function of x and y satisfies x∂u

∂x=

2y∂u

∂y.

Solution : Put x2y = v then u = f(v) so that∂u

∂x=du

dv· ∂v∂x

= f ′(v) · 2xy ∴ x∂u

∂x= f ′(v) ·

2x2y · · · (1) and∂u

∂y=du

dv· ∂v∂y

= f ′(v) · x2 ∴ 2y∂u

∂y= f ′(v) · 2x2y · · · (2) From (1) and (2) we

observe that x∂u

∂x= 2y

∂u

∂y.

Exercise (B)

1. If z is a function of x and y if x = eu + e−v, y = e−u − e+v then prove that∂z

∂u− ∂z

∂v=

x∂z

∂x− y ∂z

∂y.

2. If z = f(x, y); x = eu cos v, y = eu sin v then show that

y∂z

∂u+ x

∂z

∂v= e2u ∂z

∂y.

3. If z = f(u, v); u = x+ y, v = x− y then show that∂z

∂x+∂z

∂y= 2

∂z

∂u.

4. If u = x3 + y3; x = a cos t, y = b sin t. Find the value of∂u

∂t.

5. If z = tan−1

(x

y

); x = u+ v, y = u− v then show that

∂z

∂u+∂z

∂v=

u− vu2 + v2

.

6. If w = f(x, y); x = u+ v, y = uv then show that

(i) (u− v)∂w

∂x= u

∂w

∂u− v∂w

∂v.

(ii) (u− v)∂w

∂y=∂w

∂u− ∂w

∂v.

7. If z = f(x, y); x = uv, y =u+ v

u− vthen show that 2x

∂z

∂x= u

∂z

∂u+ v

∂z

∂v.

Page|38


8. If u = sin−1

(x

y

)+ tan−1

(yx

)then find x

∂u

∂x+ y

∂u

∂y.

9. If z = 3(lx+my)2 − (x2 + y2); where l2 +m2 = 1 then show that∂2z

∂x2+∂2z

∂y2= 2.

10. If u = (1− 2xy + y2)−1/2, show that x∂u

∂x− y∂u

∂y= y2u3.

11. If z = f(x, y); x =1

2(u2 − v2), y = uv then show that

∂2z

∂x2+∂2z

∂y2=

1

u2 + v2

[∂2z

∂u2+∂2z

∂v2

].

3.3. Directional Derivatives :-

If f(x, y) is differentiable function and x = φ(t), y = ψ(t) thendf

dt=∂f

∂x· dxdt

+∂f

∂y· dydt. gives the

rate of change of f with respect to t. This also depends on the direction of motion along the curve.If the curve is straight line and the parameter t is the arc length measured from point p0(x0, y0)

in the direction of a given unit vector u thendf

dtis the rate of change of f with respect to distance

in the direction of u. These values ofdf

dtthrough p0 are called ’directional derivatives’.

Definition : (D.D. in the planes)Suppose the function f(x, y) is defined on a region R in the xy-plane. p0(x0, y0) is a point in Rand u = u1i+ u2j is a unit vector. x = x0 + su1, y = y0 + su2 are the parametric equations of aline through p0 parallel to u; where ’s’ is the arc length measured from p0 in the direction of u.

The derivative of f at p0(x0, y0) in the direction of u is

(df

ds

)u,p0

= lims→0

(f(x0 + su1, y0 + su2)− f(x0, y0)

s

)· · · (1)

if R.H.S. exist is called the ’directional derivative’ of f at p0. It is also denoted by (Duf)po .

Note that :

If u = i then (Duf)p0 gives∂f

∂xat p0 = (x0, y0) and if u = j then (Duf)p0 gives

∂f

∂yat p0 = (x0, y0).

e.q.1. Find the directional derivative of f(x, y) = x2 + xy at point (1, 2) in the direction of unit

vector u =1√2i+

1√2j.

Solution : f(x, y) = x2 + xy, p0(x0, y0) = (1, 2).

u = u1i+ u2j =1√2i+

1√2j.

(df

ds

)u,p0

= lims→0

(f(x0 + su1, y0 + su2)− f(x0, y0)

s

)

= lims→0

f(1 +s√2, 2 +

s√2

)− f(1, 2)

s

= lim

s→0

((1 +s√2

)2 + (1 +s√2

)(2 +s√2

)− (12 + 1 · 2)

s

= lims→0

5s√

2+ s2

s

= lim

s→0

(5√2

+ s

)=

5√2.

Page|39


∴

(df

ds

)u,p0

= (Duf)p0 =5√2.

i.e. the rate of change of f(x, y) = x2+xy at point (1, 2) in the direction of u =1√2i+

1√2j

is5√2.

e.q.2. Find the directional derivative of f(x, y, z) = x2+2y2+3z2 at point (1, 1, 0) in the directionof u = i− j + 2k.

Solution : We have given that f(x, y, z) = x2 + 2y2 + 3z2

p0(x0, y0, z0) = (1, 1, 0), u = i− j + 2k

∴ u =1√6

(i− j + 2k)

(Note that u is not unit vector ∴ make it unit vector)(df

ds

)u,p0

= lims→0

[f(x0 + su1, y0 + su2, z0 + su3)− f(x0, y0, z0)

s

]

= lims→0

f(1 +s√6, 1− s√

6,

2s√6

)− f(1, 1, 0)

s

(df

ds

)u,p0

= lims→0

((1 +s√6

)2 + 2(1− s√6

)2 + 3(2s√

6)2)− 3

s

= lims→0

(−2s√

6+

15s2

6

)s

= lim

s→0

(−2√

6+

15

6s

)=−2√

6.

∴

(df

ds

)u,p0

= (Duf)p0 =−2√

6.

The Gradient Vectors :

The definition of

(df

ds

)u,p0

can also be written by using chain rule as

(df

ds

)u,p0

=

(∂f

∂x

)p0

· dxds

+

(∂f

∂y

)p0

· dyds

=

(∂f

∂x

)p0

· u1 +

(∂f

∂y

)p0

· u2(df

ds

)u,p0

=

((∂f

∂x

)p0

i+

(∂f

∂y

)p0

j

)︸︷︷︸

gradient of f at p0

· (u1i+ u2j)︸︷︷︸direction of u

· · · ?

Definition : The gradient vector of f(x, y) at a point p0(x0, y0) is the vector Of =∂f

∂xi+

∂f

∂yj.

Note that : from ? the derivative of f in the direction of u at p0 is the dot product of u with thegradient of f at p0.

Theorem : If the partial derivatives of f(x, y) are defined at p0(x0, y0) then

(df

ds

)u,p0

= (Of)p0 · u

= scalar product of the gradient of f at p0 and u.

example 1. Find the derivative of f(x, y) = xey + cos(xy) at the point (2, 0) in the direction of

Page|40


3i− 4j.

Solution : We have given that f(x, y) = xey + cos(xy),p0 = (x0, y0) = (2, 0), suppose u = 3i− 4j.(Note that u is not unit vector ∴ make it unit)

So that u =3

5i− 4

5j.

Now

fx = ey − sin(xy) · yand fy = xey − sin(xy) · x

∴ fx(2, 0) = 1,

fy(2, 0) = 2.

∴ The gradient of f at (2, 0) = (Of)(2,0)

= fx(2, 0)i+ fy(2, 0)j= 1 · i+ 2j.The directional derivative of f at (2, 0) in the direction of 3i− 4j is(

df

ds

)u,p0

= (Duf)p0 = (Of)p0 · u

= (i+ 2j) ·(

3

5i− 4

5j

)= −1.

example 2: Find the derivative of f(x, y) = 2xy − 2y2 at the point p0(5, 5) in the direction ofA = 4i+ 3j.

Solution : f(x, y) = 2xy − 3y2; p0(x0, y0) = (5, 5), A = 4i+ 3j

∴ A =4

5i+

3

5j.

fx = 2y. ∴ (fx)(5,5) = 10; fy = 2x− 6y ∴ (fy)(5,5) = −20.

∴ The gradient of f at (5, 5) = (Of)(5,5) = 10i− 20j.

∴

(df

ds

)A,p0

= (Duf)p0 = (Of)p0 · A

= (10i− 20j) ·(

4

5i+

3

5j

)= −4.

example 3: Find the derivative of f(x, y, z) = x2+2y2−3z2 at the point p0(1, 1, 1) in the directionof i+ j + k.

Solution : We have given that f(x, y, z) = x2 + 2y2 − 3z2,

p0(x0, y0, z0) = (1, 1, 1); suppose u = i+ j + k so that u =1√3

(i+ j + k).

Now fx = 2x ∴ (fx)(1,1,1) = 2, fy = 4y ∴ (fy)(1,1,1) = 4, fz = −6z∴ (fz)(1,1,1) = −6.∴ The gradient of f at (1, 1, 1) = (Of)(1,1,1)

= (fx)(1,1,1)i+ (fy)(1,1,1)j + (fz)(1,1,1)k

= 2i+ 4j − 6k.

The derivative of f at point p0 is(df

ds

)u,p0

= (Duf)p0 = (Of)p0 · u

= (2i+ 4j − 6k) · 1

3(i+ j + k)

= 0.

example 4: Find the derivative of f(x, y, z) = cos(xy) + eyz + log(zx) at the point p0(1, 0, 1/2) inthe direction of i+ 2j + 2k.

Solution : We have f(x, y, z) = cos(xy) + eyz + log(zx),

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p0(x0, y0, z0) =

(1, 0,

1

2

), suppose u = i+ 2j + 2k. so that u =

1

3(i+ 2j + 2k).

Now

fx = − sin(xy) · y +1

zx· z ∴ (fx)(1,0,1/2) = 0 + 1 = 1

fy = − sin(xy) · x+ eyz · z ∴ (fy)(1,0,1/2) = 0 +1

2=

1

2

fz = eyz · y +1

zx· x ∴ (fy)1,0,1/2) = 0 + 2 = 2.

∴ The gradient of f at (1, 0, 1/2) = (Of)(1.0.1/2)

= (fx)(1,0,1/2)i+ (fy)(1,0,1/2)j + (fz)(1,0,1/2)k

= i+1

2j + 2k.

Hence, the derivative of f at (1, 0,1

2) in the direction of

u =

(df

ds

)u,p0

= (Duf)p0 = (Of)p0 · u

=

(i+

1

2j + 2k

)· 1

3(i+ 2j + 2k)

= 2.

# Properties of Directional Derivatives :-The directional derivative definition revels that

Duf = Of · u = |Of ||u| cos θ

= |Of | cos θ... u is unit vector.

It has following properties

1. The function f increase most rapidly when cos θ = 1 or when u is in the direction of Of.i.e. Duf = |Of | cos(0) = |Of |.

2. The function f decrease most rapidly when cos θ = −1 or when u is in the direction of−Of.i.e. Duf = |Of | cos(π) = −|Of |.

3. Any direction u orthogonal to the gradient is a direction of zero change in f when θ =π

2.

i.e. Duf = |Of | cos(π

2

)= |Of | · 0 = 0.

example 1: Find the directions in which f(x, y) =x2

2+y2

2.

(a) increase most rapidly at point (1, 1).

(b) decrease most rapidly at point (1, 1).

(c) What are the directions of zero change in f at (1, 1)?

Solution : We have f(x, y) =x2

2+y2

2.

(a) So that

(Of)(1,1) = (fx)(1,1)i+ (fy)(1,1)j

= i+ j

∴ Its direction is

|(Of)(1.1)| =1√2i+

1√2j

= u.

(b) f decrease most rapidly in the direction of −(Of)(1,1)

∴ −u = − 1√2i− 1√

2j.

(c) The directions of zero change at (1, 1) are the directions orthogonal to Of.

∴ n = − 1√2i+

1√2j and −n =

1√2i− 1√

2j.

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example 2:(a) Find the derivative of f(x, y, z) = x3 − xy2 − z at point p0(1, 1, 0) in the directionof 2i− 3j + 6k.

(b) In what direction does f change most rapidly at p0 and what are rates of change in thesedirections?

Solution : (a) Suppose A = 2i− 3j + 6k so that A =2

7i− 3

7j +

6

7k.

(fx)(1,1,0) = 2, (fy)(1,1,0) = −2, (fz)(1,1,0) = −1.

∴ (Of)(1,1,0) = 2i− 2j − k.Hence, the derivative of

f = (Duf)p0 = (Of)(1,1,0) · A

= (2i− 2j − k) ·(

2

7i− 3

7j +

6

7k

)=

4

7.

(b) The function f increase most rapidly in the direction ofOf = 2i− 2j − k and decrease most rapidly in the direction of −Of.The rate of change in the directions are |Of | = 3 and −|Of | = −3.

3.4: Tangent Planes, Normal Lines and Differentials :

If r = g(t)i+ h(t)j + k(t)k is a smooth curve on the level Surface f(x, y, z) = c of a differentiable

function f. Then f(g(t), h(t), k(t)) = c Differentiating both sides∂f

∂x· dgdt

+∂f

∂y· dhdt

+∂f

∂z· dkdt

=

0 ∴

(∂f

∂xi+

∂f

∂yj +

∂f

∂zk

)︸︷︷︸

(Of)p0

·(dg

dti+

dh

dtj +

dk

dtk

)︸︷︷︸

dr

dt

= 0 At every point p0 along the curve, Of is

orthogonal to the velocity vector of the curve.

Definition : The tangent plane at the point p0(x0, y0, z0) on the surface f(x, y, z) = c is the planethrough p0 normal to (Of)p0 .∴ The equation of the tangent plane is (fx)p0(x−x0)+(fy)p0(y−y0)+(fz)p0(z−z0) = 0. Definition: The normal line of the surface at p0 is the line through p0 parallel to (Of)p0 .∴ The equations of normal line are x = x0 + (fx)p0t, y = y0 + (fy)p0t, z = z0 + (fz)p0t example1: Find the tangent plane and normal line of the surface

f(x, y, z) = x2 + y2 + z − 9 = 0 at the point p0(1, 2, 4).

Solution : We have f(x, y, z) = x2 + y2 + z−9 = 0, A circular parabolaid p0(x0, y0, z0) = (1, 2, 4).∴ The gradient of f at

p0 = (Of)(1,2,4)

= (2xi+ 2yj + k(1,2,4)

= 2i+ 4j + k.

∴ By using the formula, equation of tangent plane is2(x− 1) + 4(y − 2) + 1 · (z − 4) = 0 i.e. 2x+ 4y + z = 14.and equation of normal line to the surface at point p0 is x = 1+2t, y = 2+4t, z = 4+ t. example2: Find the equation of tangent plane and normal line of the surface

(a) g(x, y, z) = x2 + 2xy − y2 + z2 − 7 = 0 at point (1,−1, 3).

(b) h(x, y, z) = x2 + y2 − 2xy − x+ 3y − z = 4 at point (2,−3, 18).

Solution : (a) We have g(x, y, z) = x2 + 2xy − y2 + z2 − 7 = 0.p0(x0, y0, z0) = (1,−1, 3)

The gradient of g at

p0 = (Og)(1,−1,3)

= 4j + 6k

∴ equation of tangent plane is0(x− 1) + 4(y + 1) + 6(z − 3) = 0 i.e. 4y + 6z − 14 = 0.

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and the equation of normal line at p0 is x = 1, y = −1 + 4t, z = 3 + 6t. (b) We have h(x, y, z) =x2 + y2 − 2xy − x+ 3y − z = 4,p0(x0, y0, z0) = (2,−3, 18).The gradient of h at

p0 = (Oh)(2,−3,18)

= 9i− 7j − k.

∴ Equation of tangent plane at p0 is9(x− 2)− 7(y + 3)− 1(z − 18) = 0 i.e. 9x− 7y − z − 21 = 0.and equation of normal line to the surface at point p0 is x = 2 + 9t, y = −3 − 7t, z = 18 − t.example 3: The surfaces f(x, y, z) = x2 +y2−2, a cylinder and g(x, y, z) = x+ z−4 = 0, a planemeet in an ellipse.Find parametric equations for the line tangent to ellipse at the point p0(1, 1, 3).

Solution : The tangent line is orthogonal to both Of and Og at p0.∴ It is parallel to v = Of × Og.Hence, the components of v and p0 gives equations of the tangent line.Here, (Of)(1,1,3) = 2i+ 2j and (Og)(1,1,3) = i+ k

∴ v = (Of)(1,1,3) × (Og)(1,1,3)

= (2i+ 2j)× (i+ k)

v =

∣∣∣∣∣∣i j k2 2 01 0 1

∣∣∣∣∣∣∴ v = 2i− 2j − 2k

p0(x0, y0, z0) = (1, 1, 3).∴ Equations of tangent line are x = 1+2t, y = 1−2t, z = 3−2t. example 4: Find the parametricequations for the tangent line to the curve of intersection of the surfaces f(x, y) = x2 + y2 − 4 = 0and g(x, y, z) = x2 + y2 − z = 0 at the point p0(

√2,√

2, 4).

Solution : The tangent line is orthogonal to both Of and Og at p0

∴ It is parallel to v = Of × Og.Therefore the components of v and p0 gives equations of tangent line.Here,

(Of)(√

2,√

2,4) = 2√

2i+ 2√

2j and

(Og)(√

2,√

2,4) =√

2i+ 2√

2j − k

Now,

v = (Of)(√

2,√

2,4) × (Og)(√

2,√

2,4)

v =

∣∣∣∣∣∣i j k

2√

2 2√

2 0

2√

2 2√

2 −1

∣∣∣∣∣∣v = 2

√2i+ 2

√2j

∴ equations of tangent line are x =√

2 − 2√

2t, y =√

2 + 2√

2t, z = 4. # Planes Tangent toSurface z = f(x, y) :To find an equation for the plane tangent to a surface z = f(x, y) at a point p0(x0, y0, z0); z0 =f(x0, y0).The equation z = f(x, y) can be written as f(x, y)− z = 0.∴ The surface z = f(x, y) is zero level surface of the functionF (x, y, z) = f(x, y)− z.

Differentiating this Fx =∂

∂x(f(x, y)− z) = fx − 0 = fx Similarly Fy = fy, and Fz = −1.

∴ Equation of the tangent plane to the level surfaces z = f(x, y) at point p0(x0, y0, z0) is (fx)(x0,y0)(x−x0) + (fy)(x0,y0)− (z− z0) = 0. example 1: Find the plane tangent to the surface z = x cos y−yexat point p0(0, 0, 0).

Page|44


Solution :

(fx)(0,0) = (cos y − yex)(0,0)

= 1, (fy)(0,0)

= (−x sin y − ex)(0,0)

= −1.

∴ Equation of tangent plane is1(x− 0)− 1(y − 0)− (z − 0) = 0 i.e. x− y − z = 0.

example 2: Find the equation of tangent plane to the surfacez = 4x2 + y2 at the point (1, 1, 5).

Solution : z = f(x, y) = 4x2 + y2

∴ (fx)(1,1,5) = (8x)(1,1,5) = 8

(fy)(1,1,5) = (2y)(1,1,5) = 2

∴ Equation of tangent plane is8(x− 1) + 2(y − 1)− (z − 5) = 0 i.e. 8x+ 2y − z − 5 = 0.

# Increment / Change and Distance :To estimate the change in the value of function f when we move a small distance ds from a pointp0 in a perticular direction of u. df = (Of)p0 · u︸︷︷︸

directional derivative

· ds︸︷︷︸distanceincrement

example 1. Estimate how

much the value of f(x, y, z) = xey + yz will change if the point p(x, y, z) moves 0.1 unit fromp0(2, 0, 0) straight forward to p1(4, 1,−2).

Solution : We have p0(2, 0, 0) and p1(4, 1,−2)∴ The direction vector p0 p1 = 2i+ j − 2k ≡ uso that u =

2

3i+

1

3j − 2

3k.

Now,

(Of)(2,0,0) = (ey i+ (xey + z)j + yk)(2,0,0) = i+ 2j

∴ (Of)p0 · u = (i+ 2j) ·(

2

3i+

1

3j − 2

3k

)=

4

3

Hence, the change in the value of

f = df = ((Of)p0 · u) · ds

=

(4

3

)· (0 · 1)

≈ 0 · 13.

Exercise (C)

1. Find the directional derivative of functions at the given point in the given direction.

(i) f(x, y) = 2x2 + y2 at the point (−1, 1) in the direction 3i− 4j.(ii) h(x, y) = tan−1(y/x) +

√3 sin−1(xy/2), at the point (1, 1) in the direction 3i− 2j.

(iii) g(x, y) = x− (y2/x) +√

3 sec−1(2xy), at the point (1, 1) in the direction 12i+ 5j.(iv) f(x, y, z) = xy + yz + zx, at the point (1,−1, 2) in the direction 3i+ 6j − 2k.(v) f(x, y, z) = x2 + 2y2 − 3z2, at the point (1, 1, 1) in the direction i+ j + k.(vi) g(x, y, z) = 3ex cos(yz), at the point (0, 0, 0) in the direction 2i+ j − 2k.

2. Find the directions in which the function (a) increase most rapidly (b) decrease mostrapidly at a given point.

(i) f(x, y) = x2 + xy + y2 at the point (−1, 1).(ii) g(x, y) = x2y + exy sin y at the point (1, 0).

(iii) f(x, y, z) =x

y− yz at the point (4, 1, 1).

(iv) g(x, y, z) = xey + z2 at the point (1, log 2, 1/2).(v) h(x, y, z) = log(xy) + log(yz) + log(xz) at the point (1, 1, 1).

(vi) w(x, y, z) = log(x2 + y2 − 1) + y + 6z at the point (1, 1, 0).3. Find the equation of tangent plane and normal line of the surfaces

Page|45


(i) f(x, y, z) = x2 + y2 + z2 = 3 at the point (1, 1, 1).(ii) g(x, y, z) = x2 + y2 − z2 = 18 at the point (3, 5,−4).(iii) h(x, y, z) = 2z − x2 = 0 at the point (2, 0, 2).(iv) w(x, y, z) = x2 − xy − y2 − z = 0 at the point (1, 1,−1).(v) p(x, y, z) = x+ y + z = 1 at the point (0, 1, 0).

4. Find on equation for the tangent to the ellipsex2

4+ y2 = 2 at the point (−2, 1). (Ans:

x− 2y = −4)5. Find parametric equations for the tangent line to the curve of intersection of surfaces at

the given point(i) f(x, y, z) = x+ y2 + xz = 4, g(x, y, z) = x = 1 at the point (1, 1, 1).(ii) f(x, y, z) = xyz = 1, g(x, y, z) = x2 + 2y2 + 3z2 = 6 at the point (1, 1, 1).(iii) f(x, y, z) = x2 + 2y + 2z = 4, g(x, y, z) = y = 1 at the point (1, 1, 1/2).(iv) f(x, y, z) = x+ y2 + z = 2, g(x, y, z) = y = 1 at the point (1/2, 1, 1/2).(v) f(x, y, z) = x3 + 3x2y2 + y3 + 4xy − z2 = 0, g(x, y, z) = x2 + y2 + z2 = 11 at the

point (1, 1, 3).6. Find the plane tangent to the level surfaces at the given point

(i) z = log(x2 + y2) at the point (1, 0, 0).

(ii) z = e(x2+y2) at the point (0, 0, 1).(iii) z =

√y − x at the point (1, 2, 1).

7. By about how much will f(x, y, z) = log√x2 + y2 + z2 change if the point p(x, y, z) moves

from p0(3, 4, 12) a distance of ds = 0 · 1 units in the direction of 3i+ 6j − 2k?8. Estimate how much value of f(x, y, z) = ex cos(yz) will change if the point p(x, y, z) moves

from the origin a distance of ds = 0 · 1 units in the direction of 2i+ 2j − 2k?9. How much value of g(x, y, z) = x+x cos z− y sin z+ y change if the point p(x, y, z) moves

from p0(2,−1, 0) a distance of ds = 0 · 2 units towards the point p1(0, 1, 2)?10. Estimate how much the value of h(x, y, z) = cos(πxy) + xz2 will change if the point

p(x, y, z) moves from p0(−1,−1,−1) a distance of ds = 0 · 1 units towards the origin?11. Is there a direction A in which the rate of change of f(x, y) = x2 − 3xy + 4y2 at point

p(1, 2) equals 14? Give reason for your answer.12. In what two directions is the derivative of f(x, y) = x2− y2/x2 + y2 at point p(1, 1) equal

to zero?13. In what direction is the derivative of f(x, y) = xy + y2 at point p(3, 2) equal to zero?14. The derivative of f(x, y) at p0(1, 2) in the direction of i + j is 2

√2 and in the direction

of −2j is −3. What is the derivative of f in the direction of i− 2j? Give reason for youranswer.

Page|46

CHAPTER 4

Extreme Values

Introduction :A continuous function defined on closed bounded region in XY -plane take absolute maximum andminimum values on the domains. It is important to find these values and to know where theyoccur.For function of one variable, to find local extreme points we see the points where the graph hashorizontal tangent line. At such points we look for local maxima, local minima and point ofinflection.For function of two variables, we see the points where the surface z = f(x, y) has a horizontaltangent plane. At such points we look for local maxima, local minima and saddle points.

4.1: Local (Relative) Maxima and Minima of a Function of Two Variables:-

Definition 1: Local (Relative) Maximum Value :-

Suppose f(x, y) is defined on region R. (a, b) is a point in R and in domain of f(x, y). f(a, b) iscalled a local (relative) maximum value of function f(x, y) if there exists some neighbourhood of(a, b) such that for every point (a+ b, b+ k) of this neighbourhood f(a, b) ≥ f(a+ h, b+ k).The point (a, b) is called Local (Relative) Maximum Point.

Definition 2: Local (Relative) Minimum Value :-

Suppose f(x, y) is defined on a region R. (a, b) is a point in R and in the domain of f(x, y), f(a, b)is called a local (relative) minimum value of function f(x, y) if there exists some neighbourhood of(a, b) such that for every point (a+ h, b+ k) of this neighbourhood f(a, b) ≤ f(a+ h, b+ k). Thepoint (a, b) is called Local (Relative) Minimum Point.

Definition 3: f(a, b) is said to be a local (relative) extreme value of the function f(x, y) if it iseither a local (relative) maximum or local (relative) minimum value.

? First Derivative Test : (Necessary condition for extremum):-Theorem 1: If f(x, y) has a local maximum or minimum value at an interior point (a, b) of it’sdomain and if the first partial derivatives fx(x, y) and fy(x, y) exists in a neighbourhood of (a, b)then fx(a, b) = 0 and fy(a, b) = 0.

Proof: We shall prove this result by taking one variable constant.Take y = constant = b (say). Then the function of two variables f(x, y) becomes f(x, b) which isof one variable.Here f(x, b) will have an extreme value of (a, b) if its derivatives w.r.t. x at (a, b) (i.e. at x = a) iszero.i.e. (fx(x, b))(a,b) = 0 or (fx(x, b))x=a = 0⇒ fx(a, b) = 0.Similarly we can prove that fy(a, b) = 0 by taking x constant.The theorem is proved.

Remark : The converse of the above theorem is not true e.g. consider f(x, y) = x2 − y2.Here fx(x, y) = 2x, fy(x, y) = −2y.Take the point (a, b) = (0, 0). Then fx(0, 0) = 0 and fy(0, 0) = 0.This shows that both first order partial derivatives at (0, 0) vanish but f has neither maxima norminima at (0, 0).For consider any neighbourhood of (0, 0) for small values of h both (2h, h) and (h, 2h) points arein neighbourhood of (0, 0) and we have f(2h, h) > 0, f(h, 2h) < 0 ⇒ f has neither maxima norminima at (0, 0).Thus, it is clear that vanishing of the first order partial derivatives is a necessary condition but nota sufficient condition.

Definition 4: A point (a, b) is said to be a critical point or a stationary point of a function f(x, y)if fx(a, b) = 0 = fy(a, b).

Note that: Every point of extremum is a stationary point but every stationary point may not be

47

Chapter 4. Extreme Values

a point of extremum.

Definition 5: A point (a, b) is said to be a saddle point of a function f if in every neighbourhoodof (a, b) there are points (x, y) for which f(x, y) > f(a, b) and also the points for which f(x, y) <f(a, b).? Second Derivative Test For Extrema :-

Theorem 2: Suppose f(x, y) is a function of two variable x and y defined in a region R such thatits first and second order partial derivatives are continuous in some neighbourhood of (a, b) of theregion R and fx(a, b) = 0 = fy(a, b) then(i) f has local maximum at (a, b) iffxx(a, b) < 0, fxx(a, b) · fyy(a, b)− f2

xy(a, b) > 0.(ii) f has local minimum at (a, b) iffxx(a, b) > 0, fxx(a, b) · fyy(a, b)− f2

xy(a, b) > 0.(iii) f has saddle point at (a, b) iffxx(a, b) · fyy(a, b)− f2

xy(a, b) < 0.(iv) Test is inconclusive at (a, b) iffxx(a, b) · fyy(a, b)− f2

xy(a, b) = 0.

The expression fxx · fyy − f2xy is called descriminant of f and

fxx · fyy − f2xy =

∣∣∣∣ fxx fxyfxy fyy

∣∣∣∣example 1: exampleine for extreme values the functionf(x, y) = 2(x2 − y2)− x4 + y4.Solution :fx = 4x− 4x3, fxx = 4− 12x2,fy = −4y + 4y3, fyy = −4 + 12y2, fxy = 0

For extremum we have fx = 0, fy = 0.∴ 4x− 4x3 = 0 and −4y + 4y3 = 0∴ 4x(1− x2) = 0 and 4y(−1 + y2) = 0⇒ x = 0, 1− x2 = 0 and y = 0,−1 + y2 = 0⇒ x = 0, x = ±1 and y = 0, y = ±1.∴ The critical /stationary points are (0, 0), (0,±1), (±1, 0), (1, 1) and (−1,−1).[At (0, 0) :] fxx(0, 0) = 4, fyy(0, 0) = −4, fxy(0, 0) = 0.

∴ fxx · fyy − f2xy = 4(−4)− 02 = −16 < 0 ⇒ At (0, 0)f has a saddle point.

[At (0,±1) :] fxx(0,±1) = 4 > 0, fyy(0,±1) = 8, fxy = 0

∴ fxx · fyy − f2xy = (4)(8)− 0 = 32 > 0 ⇒ f has minimum at (0,±1) and f(0,±1) = −1.

[At (±1, 0) :] fxx(±1, 0) = −8 > 0, fyy(±1, 0) = −4, fxy = 0

∴ fxx · fyy − f2xy = (−8)(−4)− 0 = 32 > 0 ⇒ f has maximum at (±1, 0) and f(±1, 0) = 1.

[At (1, 1) :] fxx(1, 1) = −8, fyy(1, 1) = 8, fxy = 0.

∴ fxx · fyy − f2xy = (−8)(8)− 0 = −64 < 0 ⇒ f has saddle point at (1, 1).

[At (−1,−1) :] fxx(−1,−1) = −8, fyy(−1,−1) = 8, fxy = 0.

∴ fxx · fyy − f2xy = (−8)(8)− 0 = −64 < 0 ⇒ f has saddle point at (−1,−1).

example 2: Investigate the maximum and minimum values off(x, y) = (x+ y − 1)(x2 + y2).

Solution : fx = x2 + y2 + 2x(x+ y − 1), fy = x2 + y2 + 2y(x+ y − 1),fxx = 6x+ 2y − 2, fyy = 2x+ 6y − 2, fxy = 2y + 2x.For extremum, we have fx = 0 = fy∴ x2 + y2 + 2x(x+ y − 1) = 0 · · · (i) andx2 + y2 + 2y(x+ y − 1) = 0 · · · (ii)subtracting (i) and (ii), we get (x+ y − 1)(x− y) = 0.⇒ x = y or x = 1− y.Case (1): With x = y (i) becomes

x2 + x2 + 2x(x+ x− 1) =0

∴ 6x2 − 2x =0

i.e. 2x(3x− 1) =0

Page|48


⇒ x = 0 or x =1

3

As x = y ∴ we get the points as (0, 0) and

(1

3,1

3

).

Case (2): With x = 1− y in (i) we get 1− 2y + 2y2 = 0.This has imaginary roots.

∴ The stationary points are (0, 0) and

(1

3,1

3

).

[At (0, 0) :] fxx = −2 < 0, fyy = −2, fxy = 0

∴ fxx · fyy − f2xy = (−2) · (−2)− 0 = 4 > 0 ⇒ f has maximum at (0, 0) and f(0, 0) = 0.

[At

(1

3,1

3

):] fxx =

2

3> 0, fyy =

2

3, fxy =

4

3

∴ fxx · fyy − f2xy =

(2

3

)(2

3

)− 16

9=−4

3< 0 ⇒ f has saddle point at

(1

3,1

3

)example 3: Find extreme value of a function.f(x, y) = xy − x2 − y2 − 2x− 2y + 4.Solution : fx = y − 2x− 2, fy = x− 2y − 2fxx = −2 < 0, fyy = −2, fxy = 1.For extremum we have fx = 0 = fy∴ y − 2x− 2 = 0 and x− 2y − 2 = 0 solving these for x and y we get x = y = −2.∴ The point (−2,−2) is the only point where f may have extreme values.Now fxx · fyy − f2

xy at (−2,−2) = (−2)(−2)− (1)2 = 3 > 0.⇒ f has local maximum at (−2,−2), and f(−2,−2) = 8.example 4: Find and classify the extreme points of the functionf(x, y) = x4 − 3x2y + y3.Solution : fx = 4x3 − 6xy, fy = −3x2 + 3y2

fxx = 12x2 − 6y, fyy = 6y, fxy = −6xFor extremum, we have fx = 0 = fy∴ 4x3 − 6xy = 0, 3x2 + 3y2 = 02x(2x2 − 3y) = 0, ∴ y = ±x⇒ x = 0 or 2x2 = 3y ∴ x =

3

2or x =

−3

2.

∴ The critical points are

(3

2,3

2

),

(−3

2,3

2

)and (0, 0).

[At

(3

2,3

2

):] fxx = 12

(3

2

)2

− 6

(3

2

)= 18 > 0 fyy = 6

(3

2

)= 9 fxy = −6

(3

2

)= −9

∴ fxx · fyy − f2xy = (18)(9)− (−9)2 = 81 > 0. ⇒ f has local minimum at

(3

2,3

2

).

Similarly, f has also local minimum at

(−3

2,3

2

)[At

(−3

2,3

2

):] fxx = 12

(−3

2

)2

− 6

(3

2

)= 18 > 0 fyy = (6)

3

2= 9 fxy = −6

(−3

2

)= 9

∴ fxx · fyy − f2xy = (18)(9)− (9)2 = 81 > 0. ⇒ f has local minimum at

(−3

2,3

2

).

[At (0, 0) :] fxx = 0, fyy = 0, fxy = 0.

∴ fxx · fyy − f2xy = 0 ∴ Test Fails. But f(x, x) = x3(1− 2x)

∴ For 0 < x <1

2, f(x, x) > 0 and for

−1

2< x < 0, f(x, x) < 0.

⇒ f has saddle point at (0, 0).example 5: Find the extreme values of the function

f(x, y) = xy +50

x+

20

y

Solution : fx = y − 50

x2, fy = x− 20

y2

Page|49


fxx =100

x3, fyy =

40

y3, fxy = 1

For extremum, we have fx = 0 = fy

∴ y − 50

x2= 0 and x− 20

y2= 0.

⇒ y =50

x2with this x− 20

y2= 0 becomes x− 200x4

2500= 0

⇒ x

(1− x3

2500

)= 0 ⇒ x = 0 or 1− x3

2500= 0.

∴ x = 5. putting this in y − 50

x2= 0 gives y = 2

∴ (5, 2) is the only point where f take extreme value

fxx(5, 2) =4

5> 0, fyy(5, 2) = 5, fxy(5, 2) = 1.

∴ fxx · fyy − f2xy =

(4

5

)(5)− (1)2 = 3 > 0

⇒ f has minimum at (5, 2) and f(5, 2) = 30.

example 6: Find extreme values of the functionf(x, y) = 3x2(y − 1) + y2(y − 3) + 1.

Solution : fx = 6x(y − 1), fy = 3(x2 − 2y + y2)fxx = 6(y − 1), fyy = 6(y − 1), fxy = 6xFor extremum, we have fx = 0 = fy∴ 6x(y − 1) = 0 and 3(x2 − 2y + y2) = 0 ⇒ x = 0 or y = 1.When x = 0, x2 − 2y + y2 = 0 ⇒ y = 0 or y = 2when y = 1, x2 − 2y + y2 = 0 ⇒ x = ±1.∴ The stationary points are (0, 0), (0, 2), (1, 1), (−1, 1).

[At (0, 0) :] fxx = −6 < 0, fyy = −6, fxy = 0.

∴ fxx · fyy − f2xy = (−6)(−6)− 0 = 36 > 0 ⇒ f has maximum at (0, 0) and f(0, 0) = 1.

[At (0, 2) :] fxx = 6 > 0, fyy = 6, fxy = 0

∴ fxx · fyy − f2xy = 36− 0 = 36 > 0. ⇒ f has minimum at (0, 2) and f(0, 2) = −3.

[At (1, 1) :] fxx = 0, fyy = 0, fxy = 6

∴ fxx · fyy − f2xy = 0− 36 = −36 < 0. ⇒ f has saddle point at (1, 1).

[At (−1, 1) :] fxx = 0, fyy = 0, fxy = −6

∴ fxx · fyy − f2xy = 0− (−6)2 = −36 < 0. ⇒ f has saddle point at (−1, 1).

example 7: A rectangular box open at the top is to have a volume of 32m3. What must be thedimensions so that the total surface area is minimum?

Solution: Let the length, breadth and height of the rectangular box be x, y, z respectively, withsurface S and volume V.Here, V = 32m3 ⇒ xyz = 32 · · · (i)We want to minimize the surface area of the rectangular box.

But surface area = S is given by S = 2z(x+ y) + xy. But from (i) z =32

xy

∴ S = xy + 64

(1

x+

1

y

)= f(x, y) say,

Now Sx = y − 64

x2, Sy = x− 64

y2, Sxx =

128

x3, Syy =

128

y3, Sxy = 1.

For extremum, Sx = 0 = Sy

∴ y − 64

x2= 0 and x− 64

y2= 0

∴ x2y = 64 and y2x = 64

⇒ x2y = y2x ⇒ x = y ∴ y − 64

x2= 0

⇒ x− 64

x2= 0 ⇒ x3 = 64⇒ x = 4 and hence y = 4.

∴ The point (4, 4) is only point at which S may take extreme value. [At (4, 4) :] Sxx(4, 4) =128

64= 2 > 0, Syy(4, 4) = 2, Sxy = 1

Page|50


∴ Sxx · Syy − S2xy = (2)(2)− (1)2 = 3 > 0 ⇒ S has minimum at (4, 4).

we have V = xyz = 32 ∴ (4)(4)z = 32 ⇒ z = 2.

∴ At (4, 4, 2) : S has minimum value∴ (S)min = 2z(x+ y) + xy

= 2(2)(4 + 4) + (4)(4)(S)min = 32 + 16 = 48

Hence, Length = 4m, breadth = 4m, Height = 2m.

In example (4), we have obtained the minimum of the function x4− 3x2y+ y3 and in example (7),we have found the minimum of the function 2z(x+ y) + xy subject to the condition xyz = 32.Here we observe that these two problems are of different types. example (4) is a problem of freeextrema where as example (7) we have an additional condition called constraint or side conditioni.e. problem is of constrained extrema.To solve example (7) we have obtained the function S in terms of two variables x and y by replac-ing the value of z from the side condition. Another method to solve the problems of constrainedextrema is given by ’Lagrange’. The method is known as ’Lagrange’s method of multipliers.

4.2: Lagrange’s Method of undetermined multiplier(s) :-

M-(1): Let f(x, y, z) be a function of three variables x, y, z which is to be exampleined for ex-tremum and let the variables x, y, z are connected by the relation φ(x, y, z) = 0 · · · (1) Since

f(x, y, z) is to have extremum ∴∂f

∂x= 0,

∂f

∂y= 0,

∂f

∂z= 0 so that

∂f

∂xdx +

∂f

∂ydy +

∂f

∂zdz =

0 · · · (2) Differentiating the relation (1) we have∂φ

∂xdx+

∂φ

∂ydy+

∂φ

∂zdz = 0 · · · (3) Multiply equa-

tion (3) by a parameter λ and adding in equation (2) we get

(∂f

∂x+ λ

∂φ

∂x

)dx+

(∂f

∂y+ λ

∂φ

∂y

)dy+(

∂f

∂z+ λ

∂φ

∂z

)dz = 0. This equation will be satisfied identically if coefficients of dx, dy, dz are 0.

i.e. if∂f

∂x+ λ

∂φ

∂x= 0 · · · (4)

∂f

∂y+ λ

∂φ

∂y= 0 · · · (5)

∂f

∂z+ λ

∂φ

∂z= 0 · · · (6)

The equation (1), (4), (5) and (6) will determine the values of x, y, z and λ for which f(x, y, z) isstationary.

M-(2): Suppose f(x, y, z) and g(x, y, z) are differentiable functions. To find the local maximumand minimum values of f(x, y, z) subject to the constraint g(x, y, z) = 0, find the values of x, y, zand λ that satisfy the edquations Of = λOg and g(x, y, z) = 0 simulteneously.

Remark :

1. Inspite of usefulness in applications, Lagrange’s method fails to determine the nature ofthe stationary point.

2. It is used in finding extreme values of constrained functions.

example 1: Divide the number 36 into three parts so that the continued product of the first,square of the second and cube of the third may be maximum.

Solution : Let the numbers be x, y, z respectively.Let f(x, y, z) = xy2z3 and g(x, y, z) ≡ x+ y + z = 36.Construct the auxiliary function F as

F = f(x, y, z) + λg(x, y, z)

F = xy2z3 + λ(x+ y + z − 36)

Differentiating F partially w.r.t. x, y, z and λ, and then equating to 0, we getFx = y2z3 + λ = 0 · · · (1)Fy = 2xyz3 + λ = 0 · · · (2)Fz = 3xy2z2 + λ = 0 · · · (3)Fλ = x+ y + z − 36 = 0 · · · (4)

Page|51


Now multiply equation (1) by x, (2) by y, (3) by z and adding,we get

6xy2z3 + λ(x+ y + z) = 0

∴ 6xy2z3 + 36λ = 0

⇒ λ =−xy2z3

6

Putting this value in

xy2z3 + λx = 0

∴ xy2z3 − xy2z3

6· x = 0

∴ xy2z3(

1− x

6

)= 0

⇒ 1− x

6= 0,

... xy2z3 6= 0

⇒ x = 6.

Similarly, putting the value of λ in 2xy2z3 +λy = 0 and 3xy2z3 +λz = 0 respectively we get y = 12and z = 18.∴ The three numbers are 6, 12, & 18.and f(6, 12, 18) = 6(12)2(18)3 = 5038848.

example 2: Obtain the shortest distance of the point (1, 2,−3) from the plane 2x− 3y+ 6z = 20.

Solution : Suppose A(1, 2,−3) and let p(x, y, z) be any point on the plane (say) φ(x, y, z) =2x− 3y + 6z − 20 = 0.The distance = d2 = Ap = (x− 1)2 + (y − 2)2 + (z + 3)2 ≡ f(x, y, z). Which is to be minimize.Construct the auxiliary function

F = f(x, y, z) + λφ(x, y, z)

∴ F = (x− 1)2 + (y − 2)2 + (z + 3)2 + λ(2x− 3y + 6z − 20).

Differentiating F w.r.t. x, y, z and λ, equate to zero

Fx = 2(x− 1) + 2λ = 0 · · · (1)Fy = 2(y − 2)− 3λ = 0 · · · (2)Fz = 2(z + 3) + 6λ = 0 · · · (3)Fλ = 2x− 3y + 6z − 20 = 0 · · · (4)

Multiply equation (1) by (x−1), (2) by (y−2), (3) by (z+3) we get2(x− 1)2 + 2λ(x− 1) = 0 · · · (5)2(y − 2)2 − 3λ(y − 2) = 0 · · · (6)2(z + 3)2 + 6λ(z + 3) = 0 · · · (7)

Adding (5), (6) and (7) we get 2[(x − 1)2 + (y − 2)2 + (z + 3)2] + λ(2x − 3y + 6z) + 22λ = 0

∴ 2d2 + 42λ = 0 ⇒ λ =−d2

21. Taking value of λ in equation (1), (2) and (3) we get

x− 1 =d2

21, y − 2 =

−d2

14, z + 3 =

d2

7

∴ d2 =

(d2

21

)2

+

(−d2

14

)2

+

(d2

7

)2

⇒ d = 6.

∴ The shortest distance is 6 unit.

example 3: Show that the greatest value of 8xyz under the conditionx2

9+y2

16+z2

4= 1 is

64√3.

Solution : Let f(x, y, z) = 8xyz, g(x, y, z) =x2

9+y2

16+z2

4− 1 = 0.

Page|52


We construct the auxiliary function F as F = 8xyz + λ

(x2

9+y2

16+z2

4− 1

)Differentiate F par-

tially w.r.t. x, y, z and λ, then equating to zero, we get

Fx = 8yz +2λx

9= 0 · · · (1)

Fy = 8xz +2λy

16= 0 · · · (2)

Fz = 8xy +2λz

4= 0 · · · (3)

Fλ =x2

9+y2

16+z2

4− 1 = 0 · · · (4)

Multiply equation (1) by x, (2) by y, (3) by z we get

8xyz +2λx2

9= 0 · · · (5)

8xyz +2λy2

16= 0 · · · (6)

8xyz +2λz2

4= 0 · · · (7)

Adding these

equations we get 24xyz+ 2λ

(x2

9+y2

16+z2

4

)= 0⇒ 24xyz+ 2λ = 0 ⇒ λ = −12xyz. Putting the

value of λ in equation (5), we get 8xyz +2(−12xyz) · x2

9= 0 ⇒ 8xyz

(1− 3x2

9

)= 0

⇒ 9− 3x2 = 0... xyz 6= 0

⇒ x =√

3.

Similarly putting the value of λ in equation (6) and (7) respectively, we get y =4√3

and z =2√3.

∴ The point

(√3,

4√3,

2√3

)is the stationary point.

∴ Maximum value of 8xyz is 8(√

3)

(4√3

)(2√3

)=

64√3.

example 4: Find the greatest and smallest values of the function f(x, y) = xy takes on the ellipsex2

8+y2

2= 1.

Solution : We have f(x, y) = xy, Suppose g(x, y) =x2

8+y2

2− 1 = 0.

Of =∂f

∂xi+

∂f

∂yj and Og =

∂g

∂xi+

∂g

∂yj

Of = yi+ xj and Og =2x

8i+

2y

2j =

x

4i+ yj.

Now consider

Of = λOg

∴ yi+ xj = λ(x

4i+ yj

)yi+ xj =

λ

4xi+ λyj

⇒ y =λ

4x and x = λy

∴ y =λ

4(λy)⇒ y =

λ2

4y i.e. y

(λ2

4− 1

)= 0

∴ y = 0 orλ2

4− 1 = 0 ⇒ λ = ±2.

Case 1.: If y = 0 then x = 0 ∴ we get the point (0, 0). But (0, 0) is not on the given ellipse.∴ y 6= 0.

Case 2.: If y 6= 0 then λ = ±2 ∴ x = ±2y with this g(x, y) = 0 gives(±2y)2

8+y2

2= 1 ⇒

4y2 + 4y2 = 8 ⇒ y = ±1.∴ The ciritical points are (±2, 1) and (±2,−1).∴ The greatest value of function f(x, y) = xy = 2. and the smallest value of function f(x, y) =xy = −2.

Page|53


example 5: Find the extreme value of the function f(x, y) = 3x+ 4y on the circle x2 + y2 = 1.

Solution : We have f(x, y) = 3x+ 4y. Suppose g(x, y) = x2 + y2 − 1 = 0.

Of =∂f

∂xi+

∂f

∂yj and Og =

∂g

∂xi+

∂g

∂yj

Of = 3i+ 4j and Og = 2xi+ 2yj.

Consider

Of = λOg

∴ 3i+ 4j = λ(2xi+ 2yj)

⇒ 2xλ = 3 and 2yλ = 4

Since λ 6= 0, x =3

2λand y =

2

λWith this, g(x, y) = 0 becomes(

3

2λ

)2

+

(2

λ

)2

− 1 = 0 ⇒ 4λ2 = 25 ⇒ λ = ±5

2

∴ x = ±3

5and y = ±4

5.

∴ The stationary points are

(±3

5,±4

5

).

The extreme values of f(x, y) = 3x+ 4y are 5 and −5.

example 6: Find the extreme values of f(x, y, z) = x− 2y + 5z on x2 + y2 + z2 = 30.

Solution : We have f(x, y, z) = x− 2y + 5z.Suppose g(x, y, z) = x2 + y2 + z2 − 30 = 0.

Of =∂f

∂xi+

∂f

∂yj +

∂f

∂zk and Og =

∂g

∂xi+

∂g

∂yj +

∂g

∂zk

∴ Of = i− 2j + 5k and Og = 2xi− 2yj + 2zkConsider

Of = λOg

∴ i− 2j + 5k = λ(2xi+ 2yj + 2zk)

⇒ 2xλ = 1, 2yλ = −2, 2zλ = 5

⇒ x =1

2λ, y =

−1

λ, z =

5

2λ.

∴ g(x, y, z) = 0 becomes

(1

2λ

)2

+

(−1

λ

)2

+

(5

2λ

)2

− 30 = 0 ⇒ λ = ±1

2.

Putting this value of λ in x, y, z we get x = ±1, y = ±2, z = ±5.∴ The stationary point is (x, y, z) = (±1,±2,±5). So that the extreme values of function f(x, y, z)are 22 and −20.

Exercise (A)

1. Find all local maximum, minimum and saddle points of functions(i) f(x, y) = 2x3 + 2y3 − 9x2 + 3y2 − 12y.(ii) f(x, y) = 8x3 + y3 + 6xy.(iii) g(x, y) = x2 + xy + y2 + 3x− 3y + 4.(iv) h(x, y) = 2xy − 5x2 − 2y2 + 4x+ 4y − 4.(v) h(x, y) = x2 − y2 − 2x+ 4y + 6.(vi) g(x, y) = 3 + 2x+ 2y − 2x2 − 2xy − y2.

(vii) g(x, y) = 6x2 − 2x3 + 3y2 + 6xy.(viii) f(x, y) = x2 + 3xy + 3y2 − 6x+ 3y − 6.

(ix) f(x, y) = x2 + xy + 3x+ 2y + 5.(x) h(x, y) = y2 + xy − 2x− 2y + 2.(xi) w(x, y) = x2 − 4xy + y2 + 6y + 2.

(xii) p(x, y) = 4x2 − 6xy + 5y2 − 20x+ 26y.(xiii) w(x, y) = x2 + xy.(xiv) z(x, y) = 2xy − 5x2 − 2y2 + 4x− 4.(xv) f(x, y) = 3y2 − 2y3 − 3x2 + 6xy.

2. Find the extreme values of the functions.(i) f(x, y) = xy on the ellipse x2 + 2y2 = 1.

Page|54


(ii) f(x, y) = 49− x2 − y2 on the line x+ 3y = 10.(iii) f(x, y) = x2y on the line x+ y = 3.(iv) f(x, y) = x2 + y2, g(x, y) = x2 − 2x+ y2 − 4y = 0.(v) f(x, y) = 3x− y + 6, g(x, y) = x2 + y2 − 4 = 0.

(vi) f(x, y, z) = x+ 2y + 3z on x2 + y2 + z2 = 25.(vii) a) f(x, y) = x+ y, g(x, y) = xy − 16 = 0, x > 0, y > 0

b) f(x, y) = xy, on the line x+ y = 16.(viii) f(x, y) = xy, g(x, y) = x2 + y2 − 10 = 0.(ix) Find the three real numbers whose sum is 9 and the sum of whose squares is as

small as possible.(x) Find the largest product of positive numbers

x, y, z if x+ y + z2 = 16.3. Find the maximum value of xyz subject to the conditionx+ y + z = 1.

4. Find the extreme values of the function x2+y2+z2 subject to the condition ax+by+cz = p.

5. If xyz = 8, find the values of x, y, z for which5xyz

x+ 2y + 4zis maximum.

6. If x2y2(x+ y + 1) then show that origin is a point of minima.7. Determine the shortest distance from the point (1, 0) to the parabola y2 = 4x.8. Show that the shortest distance from the point (a, b, c) to the plane Ax+By+Cz+D = 0

is

∣∣∣∣Aa+Bb+ Cc+D√A2 +B2 + C2

∣∣∣∣ .9. If a, b, c are positive numbers, find the extreme value off(x, y, z) = xaybzc subject to the condition x+ y + z = 1.

10. If4

x+

9

y+

16

z= 25. Obtain the values of x, y, z which makes x + y + z minimum. Use

Lagrange’s Method.

Page|55


4.3: Taylor’s Formula For Functions of Two Variables:

Theorem 3: If f(x, y) and its partial derivatives of order (n+1) are continuous in neighbourhoodof a point (a, b) and if (a+ h, b+ k) is any point in this neighbourhood then there exists a positivenumber c, 0 < c < 1 such that

f(a+h, b+k) = f(a, b)+

(h∂

∂x+ k

∂

∂y

)f(a, b)+

1

2!

(h∂

∂x+ k

∂

∂y

)2

f(a, b)+· · ·+ 1

n!

(h∂

∂x+ k

∂

∂y

)nf(a, b)

+1

(n+ 1)!

(h∂

∂x+ k

∂

∂y

)n+1

f(a+ ch, b+ ck).

Proof: Let x = a+ ht, y = b+ kt; where 0 ≤ t ≤ 1 is a parameter∴ f(x, y) = f(a+ ht, b+ kt) = F (t).Since f(x, y) possesses continuous partial derivatives of order (n+1) in any neighbourhood of pointthe (a, b)., F (t) is continuous in [0, 1] and

F ′(t) =∂f

∂x· dxdt

+∂f

∂y· dydt

F ′(t) = h∂f

∂x+ k

∂f

∂y=

(h∂

∂x+ k

∂

∂y

)f

∴ F ′′(t) =∂f ′

∂x· dxdt

+∂f ′

∂y· dydt

=∂

∂x

(h∂f

∂x+ k

∂f

∂y

)· h+

∂

∂y

(h∂f

∂x+ k

∂f

∂y

)· k

=

(h∂2f

∂x2+ k

∂2f

∂x∂y

)h+

(h∂2f

∂y∂x+ k

∂2f

∂y2

)k

=h2∂2f

∂x2+ 2hk

∂2f

∂x∂y+ k2∂

2f

∂y2

=

(h∂

∂x+ k

∂

∂y

)2

f

F ′′′(t), · · · , Fn+1(t) =

(h∂

∂x+ k

∂

∂y

)n+1

f

By Maclaurin’s theorem, we have F (1) = F (0)+F ′(0)+1

2!F ′′(0)+· · ·+ 1

n!Fn(0)+

1

(n+ 1)!Fn+1(c) · · · ?

But

F (1) = f(a+ h, b+ k),

F (0) = f(a, b),

F ′(0) =

(h∂

∂x+ k

∂

∂y

)f(a, b),

F ′′(0) =

(h∂

∂x+ k

∂

∂y

)2

f(a, b), · · · ,

Fn(0) =

(h∂

∂x+ k

∂

∂y

)nf(a, b),

Fn+1(c) =

(h∂

∂x+ k

∂

∂y

)n+1

f(a+ ch, b+ ck)

Putting all these values in equation ? we get

f(a+ h, b+ k) = f(a, b) +

(h∂

∂x+ k

∂

∂y

)f(a, b)

+1

2!

(h∂

∂x+ k

∂

∂y

)2

f(a, b)

+ · · ·+ 1

n!

(h∂

∂x+ k

∂

∂y

)nf(a, b)

+1

(n+ 1)!

(h∂

∂x+ k

∂

∂y

)n+1

f(a+ ch, b+ ck);

0 < c < 1.

Page|56


First n derivative terms are evaluated at point (a, b). The last term is evaluated at some point(a+ ch, b+ ck) on the line segment joining (a, b) and (a+ h, b+ k).

Remark :

1, The last term is called the remainder and the theorem is called Taylor’s expansion aboutthe point (a, b).

2. Another form of Taylor’s Formula / theorem is

f(x, y) = f(a, b) +

[(x− a)

∂

∂x+ (y − b) ∂

∂y

]f(a, b)

+1

2!

[(x− a)

∂

∂x+ (y − b) ∂

∂y

]2

f(a, b)

+ · · ·+ 1

(n+ 1)!

[(x− a)

∂

∂x+ (y − b) ∂

∂y

]n+1

f(a+ (x− a)c, b+ (y − b)c),

0 < c < 1.

This is called Taylor’s expansion of f(x, y) about the point (a, b) in powers of (x− a) and(y − b).

3. If a = 0, b = 0 and h, k are independent variables i.e. h = x, k = y then we get

f(x, y) = f(0, 0) +

(x∂

∂x+ y

∂

∂y

)f(0, 0)

+1

2!

(x∂

∂x+ y

∂

∂y

)2

f(0, 0)

+ · · ·+ 1

(n+ 1)!

(x∂

∂x+ y

∂

∂y

)n+1

f(cx, cy), 0 < c < 1.

This is called Maclaurin’s expansion.4. Taylor’s Formula gives polynomial approximation of two variables. First three terms give

the functions linearization. The last term gives approximation error.

Working Rule : To workout the examples, we will be interested in the expansion of some functionf(x, y) upto and including certain terms.For this we need to write first few terms of the expansion:

f(x, y) = f(a, b) + (hfx(a, b) + kfy(a, b))

+1

2![h2fxx(a, b) + 2hkfxy(a, b) + k2fyy(a, b)]

+1

3!

[h3fxxx(a, b) + 3h2kfxxy(a, b) + 3hk2fxyy(a, b) + k3fyyy(a, b)

]OR

f(x, y) = f(a, b) + [(x− a)fx(a, b) + (y − b)fy(a, b)]

+1

2![(x− a)2fxx(a, b) + 2(x− a)(y − b)fxy(a, b) + (y − b)2

fyy(a, b)] + · · ·

example 1: Expand f(x, y) = x3 + xy2 in powers of (x− 2) and (y − 1).

Solution : We have f(x, y) = x3 + xy2 a = 2, b = 1.∴ f(2, 1) = 10.

fx = 3x2 + y2 ∴ fx(2, 1) = 13, fy = 2xy ∴ fy(2, 1) = 4,

fxx = 6x ∴ fxx(2, 1) = 12, fyy = 2x ∴ fyy(2, 1) = 4,

fxy = 2y ∴ fxy(2, 1) = 2, fyx = 2y ∴ fyx(2, 1) = 2,

fxx = 6 ∴ fxxx(2, 1) = 6, fyyy = 0 ∴ fyyy(2, 1) = 0,

fxxy = 0 ∴ fxxy(2, 1) = 0, fxyy = 2 ∴ fxyy(2, 1) = 2.

Page|57


Now, we have the Taylor’s Formula

F (x, y) = f(a, b) + [(x− a)fx(a, b) + (y − b)fy(a, b)]

+1

2[(x− a)2fxx(a, b) + 2(x− a)(y − b)fxy(a, b) + (y − b)2fyy(a, b)]

+1

6[(x− a)3fxxx(a, b) + 3(x− a)2(y − b)fxxy(a, b) + 3(x− a)(y − b)2

fxyy(a, b) + (y − b)3fyy(a, b)] + · · ·

Putting all the values in this, we get

x2 + xy2 = 10 + 13(x− 2) + 4(y − 1)

+1

2[12(x− 2)2 + 4(x− 2)(y − 1) + 4(y − 1)2]

+1

6[6(x− 2)3 + 2(x− 2)(y − 1)2].

example 2: Expand f(x, y) = sinxy in powers of (x− 1) and(y − π

2

)upto and including second

degree terms.

ORFind the quadratic approximation near the point

(1,π

2

)for the function f(x, y) = sinxy.

Solution ; Here a = 1, b =π

2, f(x, y) = sinxy, ∴ f

(1,π

2

)= 1.

fx = y cosxy ∴ fx

(1,π

2

)= 0,

fy = x cosxy ∴ fy

(1,π

2

)= 0

fxx = −y2 sinxy ∴ fxx

(1,π

2

)=−π2

4,

fyy = −x2 sinxy ∴ fyy

(1,π

2

)= −1

fxy = −xy sinxy + cosxy ∴ fxy

(1,π

2

)=−π2

Now, we have Taylor’s Formula

f(x, y) = f(a, b) + hfx(a, b) + kfy(a, b)

+1

2![h2fxx(a, b) + 2hkfxy(a, b) + k2fyy(a, b)]

∴ sinxy = 1 +1

2!

[(x− 1)2(−π2/4) + 2(x− 1)(y − π/2)(−π/2) + (y − π/2)2(−1)

]sinxy ≈ 1− π2

8(x− 1)2 − π

2(x− 1)(y − π/2)− 1

2(y − π/2)2.

example 3: Find the cubic approximation for f(x, y) = xy near the point (1, 1)

ORExpand f(x, y) = xy in powers of (x− 1) and (y − 1) upto third degree terms.

Solution : Here a = 1, b = 1. f(x, y) = xy ∴ f(1, 1) = 1.

fx = yxy−1 ∴ fx(1, 1) = 1, fy = xy · log x ∴ fy(1, 1) = 0

fxx = y(y − 1)xy−2 ∴ fxx(1, 1) = 0, fyy = log x · xy · log x ∴ fyy(1, 1) = 0

fxy = xy · 1

x+ log x · y · xy−1 ∴ fxy(1, 1) = 1,

fxxx = y(y − 1)(y − 2)xy−3 fxxx(1, 1) = 0,

fyyy = (log x)2 · xy · log x ∴ fyyy(1, 1) = 0.

fxxy = (y2 − y)xy−2 · log x+ xy−2(2y − 1) ∴ fxxy(1, 1) = 1,

fxyy = xy−1 · log x+ y(xy−1 · 1

x+ log x · xy−1 · log x) + xy−1 · log x, ∴ fxyy(1, 1) = 1.

Page|58


Putting these values in the Taylor’s Formula, we get

f(x, y) = xy = 1 + (x− 1)1 + (y − 1) · 0

+1

2[(x− 1)2 · 0 + 2(x− 1)(y − 1) · 1 + (y − 1)2 · 0]

+1

6[(x− 1)3 · 0 + 3(x− 1)2(y − 1) · 1 + 3(x− 1)(y − 1)2 · 1 + (y − 1)2 · 0]

xy ≈ 1 + (x− 1) + (x− 1)(y − 1) +1

2(x− 1)2(y − 1) +

1

2(x− 1)(y − 1)2.

example 4: Show that sinx sin y = xy− 1

6[(x3 +3xy2) cos θx ·sin θy+(y3 +3x2y) sin θx ·cos θy], 0 <

θ < 1. Solution : Here, f(x, y) = sinx sin y, The point (a, b) = (0, 0).By Maclaurin’s expansion, upto three terms, we have

f(x, y) = f(0, 0) + xfx(0, 0) + yfy(0, 0)

+1

2![x2fxx(0, 0) + 2xyfxy(0, 0) + y2fyy(0, 0)]

+1

3![fxxx(θx, θy) · x3 + 3x2yfxxy(θx, θy) + 3xy2fxyy(θx, θy)

+y3fyyy(θx, θy)] · · · ?f(x, y) = sinx sin y ∴ f(0, 0) = 0, fx = cosx sin y ∴ fx(0, 0) = 0,

fy = sinx cos y ∴ fy(0, 0) = 0, fxx = − sinx sin y ∴ fxx(0, 0) = 0,

fyy = − sinx sin y ∴ fyy(0, 0) = 0, fxy = cosx cos y fxy(0, 0) = 1,

fxxx = − cosx sin y ∴ fxxx(θx, θy) = − cos θx sin θy,

fyyy = − sinx cos y ∴ fyyy(θx, θy) = − sin θx cos θy,

fxyy = − cosx sin y ∴ fxyy(θx, θy) = − cos θx sin θy.

With all these values, equation ? becomes

sinx sin y = xy +1

6[x3(− cos θx sin θy) + 3x2y(− sin θx cos θy)

+3xy2(− cos θx sin θy) + y3(− sin θx cos θy)], 0 < θ < 1.

∴ sinx sin y = xy − 1

6[(x3 + 3xy2) cos θx sin θy + (y3 + 3x2y) sin θx cos θy], 0 < θ < 1.

Exercise (B)

1. Expand x2 + 3y − 2 in powers of (x− 1) and (y − 2).2. Expand x2y + 3y − 2 in powers of (x− 1) and (y + 2).

3. Show that eax cos by = 1 + ax+1

2(a2x2 − b2y2).

4. Expand f(x, y) = exy about the point (2, 3) upto second degree terms.5. Show that

eax sin by = by + abxy +1

6eaθx[(a3x3 − 3ab2xy2]

sin bθy + (3a2bx2y − b3y3) cos bθy, 0 < θ < 1.

6. Expand f(x, y) = ex+y about (0, 0) upto and including the third degree terms in x+ y.7. Find a quadratic approximation of f(x, y) = sinx sin y near the origin.8. Find quadratic and cubic approximations near the origin.

(i) f(x, y) = xey, (ii) f(x, y) = ex cos y, (iii) f(x, y) = y sinx,

(iv) f(x, y) = cosx cos y, (v) f(x, y) = sin(x2 + y2),

(vi) f(x, y) = cos(x2 + y2) (vii) f(x, y) = ex log(1 + y),

(viii) f(x, y) = log(2x+ 2y + 1) (ix) f(x, y) =1

1− x− y,

(x)1

1− x− y + xy, (xi) f(x, y) = sinx cos y.

Page|59

CHAPTER 5

Multiple Integrals

5.1. Double Integrals over Rectangles

We consider a function f(x, y) defined on rectangular region R R : a ≤ x ≤ b, c ≤ y ≤ d.We sub divide R into small rectangles using a network of lines parallel to the x-axis and y-axis asshown in figure. The lines divide R into n rectangular pieces, where the number of such pieces ngets large as the width and height of each piece gets small. These rectangles form a partion of R. Asmall rectangular piece of width 4x and height 4y has area 4A = 4x4y. If we number the smallpieces partitioning R in some order, the their areas are given by numbers4A1,4A2,4A3, · · ·4An,where 4Ak is the area of the kth small rectangle.To form a Riemann sum over R, we choose a point (xk, yk) in the kth small rectangle, multiply the

value of f at the point by the area 4Ak, and add togehter the products: Sn =

n∑k=1

f(xk, yk)4Ak.

Depending on how we take (xk, yk) in kth small rectangle, we may get different values for Sn.We are interested in what happens to these Riemann sums as the widths and heights of all thesmall rectangles in the partition of R approach zero. The norm of a partition P written ‖P‖, isthe largest width or height of any rectangle in the partition. If ‖P‖ = 0 · 1, then all the rectanglesin the partion of R have width at most 0 · 1 and height at most 0 · 1. Sometimes the Riemannsums converge as the norm of P goes to zero, written ‖P‖ → 0. The resulting limits is written as

lim‖P‖→0

n∑k=1

f(xk, yk)4Ak. As ‖P‖ → 0 and the rectangles get narrow and short, their number n

increases, so we can also write this limit as limn→∞

n∑k=1

f(xk, yk)4Ak. with the understanding that

4Ak → 0 as n→∞ and ‖P‖ → 0.If f is continuous throughout R, then as we refine the mesh width to make both 4x and 4y go tozero, the sums in (1) approach a limit called the double integral of f over R.

The notation is

∫ ∫Rf(x, y)dA or

∫ ∫Rf(x, y)dxdy.Thus∫ ∫

Rf(x, y)dA = lim

n→∞

n∑k=1

f(xk, yk)4Ak = limn→∞

sn.

Fubini’s theorem for calculating double integrals :-

Theorem 1 Fubini’s Theorem (First Form) :-

If f(x, y) is continuous on the rectangular region R : a ≤ x ≤ b, c ≤ y ≤ d, then

∫ ∫Rf(x, y)dA =∫ d

c

∫ b

af(x, y)dxdy =

∫ b

a

∫ d

cf(x, y)dydx. Fubini’s theorem state that double integrals over rect-

angles can be calculated as iterated integrals. This means that we can evaluate a double integralby integrating w.r.t. one variable at a time.

Example 1: Calculate

∫ ∫Rf(x, y)dA for

f(x, y) = 4− y2 and R : 0 ≤ x ≤ 3, 0 ≤ y ≤ 2.

60

§ 5.1. Double Integrals over Rectangles Chapter 5. Multiple Integrals

Solution : By Fubini’s theorem∫ ∫Rf(x, y)dA =

∫ 3

0

∫ 2

0(f − y2)dydx

=

∫ 3

0

[4y − y3

3

]2

y=0

dx

=

∫ 3

0

16

3dx

=16

3[x]30

= 16.

Reversing the order of integration∫ ∫Rf(x, y)dA =

∫ 2

0

∫ 3

0(4− y2)dxdy

=

∫ 2

0(4− y2)[x]3x=0dy

= 3

∫ 2

0(4− y2)dy = 3

[4y − y3

3

]2

0

= 3(16)

3= 16.

By reversing the order of integration we sec that the value of integral remains the same.

Example 2: Evaluate

∫ 3

0

∫ 0

−2(x2y − 2xy) dydx.

Solution : by Fubini’s Theorem,∫ 3

0

∫ 0

−2(x2y − 2xy) dydx =

∫ 3

x=0

[x2

∫ 0

y=−2ydy − 2x

∫ 0

y=−2ydy

]dx

=

∫ 3

x=0

{x2

[y2

2

]0

y=−2

− 2x

[y2

2

]0

y=−2

}dx

=

∫ 3

x=0

{x2

[0− 4

2

]− 2x

[0− 4

2

]}dx

=

∫ 3

0{−2x2 + 4x} dx

=

[−2

x3

3+ 2x2

]3

x=0

= [−12 + 18] = 6.

Reversing the order of integration∫ 0

−2

∫ 3

0(x2y − 2xy) dxdy =

∫ 0

−2

[d3

3y − 2x2

2y

]3

x=0

dy

=

∫ 0

−2(6y − 9y) dy

=

∫ 0

−2−3ydy

=

[−3

y2

2

]0

y=−2

= [0 + 6]

= 6.

Page|61

Chapter 5. Multiple Integrals § 5.1. Double Integrals over Rectangles

Theorem 2: Fubini’s Theorem (Stronger Form) :-

Let f(x, y) be continuous on a region R.(1). If R is defined by a ≤ x ≤ b, g1(x) ≤ y ≤ g2(x), with g1 and g2 are continuous on [a, b], then∫ ∫

Rf(x, y) dA =

∫ b

a

∫ g2(x)

g1(x)f(x, y) dydx.

(2). If R is defined by c ≤ y ≤ d, h1(y) ≤ x ≤ h2(y). with h1 and h2 are continuous on [c, d], then∫ ∫Rf(x, y) dA =

∫ d

c

∫ h2(y)

h1(y)f(x, y) dxdy.

Example 1: Evaluate∫ ∫

R

sinx

xdA where R is the triangle in the xy-plane bounded by the x-axis,

the line y = x and the line x = 1.Solution: The region of integration is shown in figure we integrate first w.r.t. y and then withrespect to x.

∫ ∫R

sinx

xdA =

∫ 1

x=0

∫ x

y=0

sinx

xdxdy =

∫ 1

x=0

(∫ x

y=0

sinx

xdy

)dx =

∫ 1

x=0

sinx

x[y]xy=0 dx∫ ∫

R

sinx

xdA =

∫ 1

x=0

sinx

x(x− 0) dx =

∫ 1

0sinxdx = [− cosx]10 = − cos 1 + cos 0 = 1− cos 1 .

Example :- Sketch the region of integration for the integral∫ 20

∫ 2xx2 (4x+ 2) dydx and write an equivalent integral with the order of integration reversed.

Solution : The region of integration is given by the inequalities x2 ≤ y ≤ 2x and 0 ≤ x ≤ 2.Therefore the region is bounded by the curve parabola y = x2 and the lines y = 2x, x = 0, x = 2.

Page|62

§ 5.1. Double Integrals over Rectangles Chapter 5. Multiple Integrals∫ 2

0

∫ 2x

x2(4x+ 2) dydx =

∫ 2

0(4x+ 2)[y]2xy=x2dx

=

∫ 2

0(4x+ 2)(2x− x2) dx

=

∫ 2

0(6x2 − 4x3 + 4x) dx

= [2x3 − x4 + 2x2]20

= 24 − 24 + 23 = 8.

To find limits for integrating in the reverse order, we consider a horizontal line passing from left toright through the region.

This line enters at x =y

2and leaves at x =

√y. To cover all such lines we let y run from y = 0 to

y = 4. Now by changing the order of integration.∫ ∫R

sinx

xdA

∫ 2

0

∫ 2x

x2(4x+2) dydx =

∫ 4

0

∫ √yy

2

(4x+2) dxdy =

∫ 4

y=0[2x2+2x]

√y

xy

2

dy =

∫ 4

0

[2y + 2

√y − y2

2− y]dy

∫ ∫R

sinx

xdA

∫ 2

0

∫ 2x

x2(4x+ 2) dydx =

∫ 4

0

(y + 2

√y−y2

2

)dy =

[y2

2+

2y3/2

3/2− y3

6

]4

0

= 8.

Example : Evaluate∫ ∫

R f(x, y) dA where f(x, y) = y cosxy over the rectangle 0 ≤ x ≤ π, 0 ≤y ≤ 1.Solution : ∫ ∫

Rf(x, y) dA =

∫ π

x=0

∫ 1

y=0y cosxy dydx.

=

∫ 1

y=0

∫ π

x=0y cosxy dxdy

=

∫ 1

y=0y

[sinxy

y

]πx=0

dy

=

∫ 1

y=0sinπy dy

=[− cosπy]1y=0

=− cosπ + cos 0

=− (1) + 1

=2.

Example : Integrate f over the given region f(u, v) = v −√u over the triangular region cut off

from the first quadrant of uv plane by the line u+ v = 1.Solution :

Page|63


Consider the strip parallel to v-axis. This strip varies from u = 0 to u = 1. The limits of u areu = 0 to u = 1. Now the values of v interms of u are v = 0 to v = 1− u. The limits of v are v = 0to v = 1− u. ∫ ∫

Rf(u, v) dA =

∫ ∫R

(v −√u) dA

=

∫ 1

u=0

∫ 1−u

v=0(v −

√u) dvdu

=

∫ 1

u=0

[v2

2−√uv

]1−u

v=0

du

=

∫ 1

u=0

[(1− u)2

3−√u(1− u)

]du

=

∫ 1

u=0

[(1− u)2

2− u1/2 + u3/2

]du

=

[(1− u)3

6− u3/2

3/2+u5/2

5/2

]1

u=0

=0− 2

3+

2

5− 1

6

=−10 + 6

15− 1

6

=−4

15− 1

6

=−8− 5

30

=−13

30.

Example : Evaluate∫ 1

0

∫ √1−s20 8tdtds (the st plane).

Solution : ∫ 1

0

∫ √1−s2

08tdtds =

∫ 1

s=0[4t2]

√1−s2

t=0 ds

=

∫ 1

s=04(1− s2) ds

=4

[s− s3

3

]1

0

=4

(1− 1

3

)=

8

3.

Page|64


Example : Evaluate∫ 0−2

∫ −vv 2dpdv (the pv plane)

Solution : We have the limits of pare p = v and p = −v, v = −2, v = 0.

∫ 0

−2

∫ −vv

2dpdv =

∫ 0

−22[p]−vp=v dv =

∫ 0

−22[−v − v] dv =

∫ 0

−2−4vdv =

[−4v2

2

]0

−2

= 8.


R f(x, y) dA where R is the region in the first quadrant bounded by the

lines y = x, y = 2x, x = 1, x = 2 and f(x, y) =x

y.

Solution : Draw the lines y = x, y = 2x, x = 1, x = 2.

Page|65

Chapter 5. Multiple Integrals § 5.1. Double Integrals over Rectangles∫ ∫Rf(x, y) dA =

∫ 2

x=1

∫ 2x

y=x

x

ydydx

=

∫ 2

x=1x[log y]2xy=x dx

=

∫ 2

x=1x[log 2x− log x] dx

=

∫ 2

x=1x log 2dx

= log 2

[x2

2

]2

1

=3

2log 2.


R

1

xydxdy where R is the square 1 ≤ x ≤ 2, 1 ≤ y ≤ 2.

Solution :

∫ ∫R

dxdy

xy=

∫ 2

x=1

∫ 2

y=1

1

xydydx

=

∫ 2

x=1

1

xdx

∫ 2

y=1

1

ydy

= [log x]2x=1[log y]2y=1

= [log 2− log 1][log 2− log 1]

= [log 2]2.


R f(x, y) dA where f(x, y) = x2 + y2 and R is triangular region withvertices (0, 0), (1, 0) and (0, 1).

Solution : Let O(0, 0), A(1, 0), B(0, 1).

Slope of AB = m =y2 − y1

x2 − x1=

1− 0

0− 1= −1. Equation of AB is

y − 0 = m(x− 1)y = −1(x− 1)y = −x+ 1.

Consider the strip parallel to y axis. This strip varies from x = 0 to x = 1.The limits of x are from x = 0 to x = 1. Now the values of y interms of x at the end point of the

Page|66


strip are y = 0 to y = 1− x.∫ ∫Rf(x, y) dA =

∫ 1

x=0

∫ 1−x

y=0(x2 + y2) dydx

=

∫ 1

x=0[x2y + y3]1−xy=0 dx

=

∫ 1

0[x2(1− x) + (1− x)3] dx

=

∫ 1

0[x2 − x3 + (1− x)3] dx

=

[x3

3− x4

4+

(1− x)4

−4

]1

x=0

=1

3− 1

4− 0 + 0 +

1

4

=1

3.

Example : Sketch the region of integration and evaluate the integrals.

(a)∫ 2ππ

∫ π0 (sinx+ cos y) dxdy (b)

∫ π0

∫ x0 x sin y dydx

(c)∫ π

0

∫ sinx0 y dydx.

Solution : a) We have the limits of x are x = 0, x = π and limits of y are y = π to y = 2π. Drawthe lines x = 0, x = π, y = 0, y = 2π.

∫ 2π

π

∫ π

0(sinx+ cos y) dxdy =

∫ 2π

y=π

{∫ π

x−0sinxdx+ cos y

∫ π

x=01dx

}dy

=

∫ 2π

y=π{[− cosx]πx=0 + cos y[x]πx=0} dy

=

∫ 2π

y=π{− cosπ + cos 0 + cos y[π − 0]} dy

=

∫ 2π

π{2 + π cos y} dy

= [2y + π sin y]2πy=π

= 4π − 2π + π[sin 2π − sinπ]

= 2π + π = 3π.

(b) We have the limits of y are y = 0 to y = x and the limits of x are x = 0 to x = π. Draw thelines y = 0, y = x, x = 0, x = π.

Page|67


∫ π

0

∫ x

0x sin y dydx =

∫ π

0x[− cos y]xy=0 dx

=

∫ π

0x(1− cosx) dx

=

∫ π

0xdx−

∫ π

0x cosx dx

=

[x2

2

]π0

{[x sinx]π0 −

∫ π

0sinx dx

}=

π2

2+ [− cosx]π0

=π2

2+ 2.

(c)∫ π

0

∫ sinx0 y dydx

∫ π

0

∫ sinx

0y dydx =

∫ π

0

[y2

2

]sinx

y=0

dx

=

∫ π

0

sin2 x

2dx

=1

2

∫ π

0

1− cos 2x

2dx

=1

4

[x− sin 2x

2

]π0

=1

4[π]

=π

4.

Example : Change the order of integration and evaluate∫ 1/16

0

∫ 1/2y/4 cos(16πx5) dxdy.

Solution : We have the limits of x are x = y1/4 and x =1

2i.e. x4 = y and x =

1

2. The limits of y

Page|68


are y = 0 and y =1

16. Draw the curve x4 = y and the lines x =

1

2, y = 0, y =

1

16.

Consider the strip parallel to y-axis. This strip varies from x = 0 to x =1

2. The limits of x are

x = 0 to x =1

2. Now the values of y interms of x are y = 0 to y = x4

∴ The limits of y are y = 0 to y = x4.∫ 1/16

0

∫ 1/2

y1/4cos(16yx5) dxdy =

∫ 1/2

x=0

∫ x4

y=0cos(16πx5) dydx

=

∫ 1/2

x=0cos(16πx5)x4 dx

=

∫ π/2

0cos t

dt

80π

=1

80π[sin t]

π/20

=1

80π[sin

π

2− sin 0]

=1

80π.

Put 16πx5 = t, 80πx4dx = dt.

when x =1

2, t =

π

2when x = 0, t = 0.

Example : Change the order of integration and evaluate∫ 8

0

∫ 23√x

dydx

y4 + 1.

Solution : We have the limits of y are y = 3√x to y = 2 that is y3 = x to y = 2 and limits of x

are x = 0 to x = 8.Draw the curve y3 = x and the straight lines y = 2, x = 0, x = 8.

Page|69


Consider the strip parallel to x-axis. This strip varies from y = 0 to y = 2. Therefore limits of yare y = 0 to y = 2. Now the values of x interms of y at the end points on strip are x = 0 to x = y3.∫ 8

0

∫ 2

3√x

dydx

y4 + 1=

∫ 2

y=0

∫ y3

x=0

dxdy

y4 + 1=

∫ 2

y=0

1

y4 + 1[x]y

3

x=0 dy =

∫ 2

y=0

y3

y4 + 1dy.

Substitute y4 + 1 = t∫ 8

0

∫ 2

3√x

dydx

y4 + 1=

∫ 17

1

dt

4t=

1

4log 17.


0

∫ 1y x2exy dxdy.

Solution : We have the limits of x are x = y to x = 1 and the limits of y are y = 0 to y = 1.Draw the lines y = 0, y = 1, x = 1, y = x.

Consider the strip parallel to y-axis. This strip varies from x = 0 to x = 1.∴ The limits of x are x = 0 to x = 1. Now the values of y interms of x at the end point of strip arey = 0 to y = x.∴ The limits of y are y = 0 to y = x.By changing order of integration∫ 1

0

∫ 1

yx2exy dxdy =

∫ 1

x=0

∫ x

y=0x2exy dydx

=

∫ 1

x=0x2

[exy

x

]xy=0

dx

=

∫ 1

0x[ex

2 − e0] dx

=

∫ 1

0xex

2dx−

∫ 1

0xdx

=

∫ 1

0etdt

2−[x2

2

]1

0

=1

2[et]10 −

1

2+ 0

=1

2[e− 1]− 1

2

=e

2− 1.

Put x2 = t, xdx =dt

2.

Example : Change the order of integration and evaluate it∫ 3

0

∫ 1√x/3

ey3dydx.

Solution : We have the limits of y are y =

√x

3and y = 1. The limits of x are x = 0 and x = 3.

Page|70


Draw the curve y2 =x

3, y = 1, x = 0, x = 3.

Consider the strip parallel to x-axis. This strip varies from y = 0 to y = 1.∴ The limits of y are y = 0 to y = 1. Now the values of x in terms of y at the end point of stripare x = 0 and x = 3y2.∴ The limits of x are x = 0 to x = 3y2.By changing order of integration∫ 3

0

∫ 1√x

3

ey3dydx =

∫ 1

y=0

∫ 3y2

x=0ey

3dxdy

=

∫ 1

y=0ey

3[x]3y

2

x=0 dy

=

∫ 1

y=0ey

3(3y2) dy

=

∫ 1

0et dt

= [et]10= e− 1.

Example : Evaluate∫ ln 8

1

∫ ln y0 ex+y dxdy.

Solution : ∫ ln 8

1

∫ ln y

0ex+y dxdy =

∫ ln 8

1

{∫ ln y

x=0ex dx

}eydy

=

∫ ln 8

1[ex]ln yx=0 e

y dy

=

∫ ln 8

1(y − 1)ey dy

= [(y − 1)ey]ln 81 −

∫ ln 8

1ey dy

= (ln 8− 1)8− [ey]ln 81

= (ln 8− 1)8− [8− e]= 8 ln 8− 8− 8 + e

= 8 ln 8− 16 + e.

Example : Sketch the region of integration, determine the order of integration and evaluate the

integral.∫ π

0

∫ πx

sin y

ydydx.

Solution : We have the limits of y are y = x and y = π and the limits of x are z = 0 to z = π.Draw the lines x = 0, x = π, y = x, y = π.

Page|71


We cannot integrate∫ πx

sin y

ydy w.r.t. y. Therefore we change the order of integration.

Consider the strip parallel to x-axis. This strip varies from y = 0 to y = π. The limits of y arey = 0 to y = π. Now the values x interms of y at the end point of strip are x = 0 to x = y.∴ The limits of x are x = 0 to x = y. Hence by changing order of integration we get

∫ π

0

∫ π

x

sin y

ydydx =

∫ π

y=0

∫ yπ

x=0

sin y

ydxdy

=

∫ π

y=0

sin y

y[x]yx=0 dy

=

∫ π

0sin y dy

= [− cos y]π0= − cosπ + cos 0

= 2.


∫ 2x 2y2 sinxy dydx.

Solution : We have the limits of y are y = x and y = 2 and the limits of x are x = 0 and x = 2.Draw the lines x = 0, x = 2, y = x, y = 2.

Consider the strip parallel to x-axis. This trip varies from y = 0 to y = 2. The limits of y are y = 0to y = 2. Now the values of x interms of y are x = 0 to x = y.∴ The limits of x are x = 0 to x = y.

Page|72


∴ By changing order of integration.∫ 2

0

∫ 2

x2y2 sinxy dydx =

∫ 2

y=0

∫ y

x=02y2 sinxy dxdy

=

∫ 2

y=02y2

[− cosxy

y

]yx=0

dy

=

∫ 2

y=02y[− cos y2 + cos 0] dy

=

∫ 2

y=0− cos y2(2ydy) +

∫ 2

02dydy

=

∫ 4

0− cos t dt+ [y2]20

= − sin t

∫ 4

0+4

= 4− sin 4

= 4− (−0 · 7568)

= 4 · 7568.

Definition : The Jacobian determinant or Jacobian of the co-ordinates transformation

x = g(u, v) y = h(u, r) is J(u, v) =

∣∣∣∣∣∣∣∂x

∂u

∂x

∂v∂y

∂u

∂y

∂v

∣∣∣∣∣∣∣ =∂x

∂u

∂y

∂v− ∂x

∂v

∂y

∂u. (1) The Jacobian is also

denoted by J(u, v) =∂(x, y)

∂(u, v). e.g. for polar co-ordinates we have

x = r cos θ, y = r sin θ.The Jacobian is

J(r, θ) =∂(x, y)

∂(r, θ)=

∣∣∣∣∣∣∣∂x

∂r

∂x

∂θ∂y

∂r

∂y

∂θ

∣∣∣∣∣∣∣ =

∣∣∣∣ cos θ −r sin θsin θ r cos θ

∣∣∣∣= 4 cos θ2 + r sin2 θ

= r

The Jacobian of co-ordinate transformation

x = g(u, v, w), y = h(u, v, w), z = k(u, v, w). is defined by J(u, v, w) =∂(x, y, z)

∂(u, v, w)=

∣∣∣∣∣∣∣∣∣∣

∂x

∂u

∂x

∂v

∂x

∂w∂y

∂u

∂y

∂v

∂y

∂w∂z

∂u

∂z

∂v

∂z

∂w

∣∣∣∣∣∣∣∣∣∣e.g. for cylindtrical co-ordinates we have x = r cos θ, y = r sin θ, z = z The Jacobian transforma-tion is

J(r, θ, z) =∂(x, y, z)

∂(r, θz)=

∣∣∣∣∣∣∣∣∣∣

∂x

∂r

∂x

∂θ

∂x

∂z∂y

∂r

∂y

∂θ

∂y

∂z∂z

∂r

∂z

∂θ

∂z

∂z

∣∣∣∣∣∣∣∣∣∣=

∣∣∣∣∣∣cos θ −r sin θ 0sin θ r cos θ 0

0 0 g

∣∣∣∣∣∣= cos θ[r cos θ − 0] + r sin θ[sin θ − 0] + 0

= r cos2 θ + r sin2 θ

= r(cos2 θ + sin2 θ)

= r.

Example : Find the Jacobian∂(x, y)

∂(u, v)for the transformation

(a) x = u cos v, y = u sin v (b) x = u sin v, y = u cos v.

Page|73


Solution : (a) Since x = u cos v, y = u sin v.

∂(x, y)

∂(u, v)=

∣∣∣∣∣∣∣∂x

∂u

∂x

∂v∂y

∂u

∂y

∂v

∣∣∣∣∣∣∣ =

∣∣∣∣ cos v −u sin vsin v u cos v

∣∣∣∣= u cos2 v + u sin2 v

= u.

(b) Since we have

x = u sin vy = u cos v

∂(x, y)

∂(u, v)=

∣∣∣∣∣∣∣∂x

∂u

∂x

∂v∂y

∂u

∂y

∂v

∣∣∣∣∣∣∣ =

∣∣∣∣ sin v u cos vcos v −u sin v

∣∣∣∣= −u sin2 v − u cos2 v

= −u.

Example : Solve the system u = x− y, v = 2x+ y for x and y interms of u and v. Then find the

value of the Jacobian∂(x, y)

∂(u, v).

Solution : We haveu = x− y (1)v = 2x+ y (2)

Solving (1) and (2) we get x =u+ v

3, y =

v − 2u

3.

J =∂(x, y)

∂(u, v)=

∣∣∣∣∣∣∣∂x

∂u

∂x

∂v∂y

∂u

∂y

∂v

∣∣∣∣∣∣∣ =

∣∣∣∣∣∣∣1

3

1

3−2

3

1

3

∣∣∣∣∣∣∣=

1

9+

2

9

=1

3.

Example : Find the Jacobian∂(x, y, z)

∂(u, v, w)of the transformation

(a) x = u cos v, y = u sin v, z = w

(b) x = 2u− 1, y = 3v − 4, z =1

2(w − 4).

Solution : (a)

J =∂(x, y, z)

∂(u, v, w)=

∣∣∣∣∣∣∣∣∣∣

∂x

∂u

∂x

∂v

∂x

∂w∂y

∂u

∂y

∂v

∂y

∂w∂z

∂u

∂z

∂v

∂z

∂w

∣∣∣∣∣∣∣∣∣∣=

∣∣∣∣∣∣cos v −u sin v 0sin v u cos v 0

0 0 1

∣∣∣∣∣∣= cos v[u cos v − 0] + u sin v[sin v − 0] + 0[0− 0]

= u cos2 v + u sin2 v

= u

Page|74


(b)

J =∂(x, y, z)

∂(u, v, w)=

∣∣∣∣∣∣∣∣∣∣

∂x

∂u

∂x

∂v

∂x

∂w∂y

∂u

∂y

∂v

∂y

∂w∂z

∂u

∂z

∂v

∂z

∂w

∣∣∣∣∣∣∣∣∣∣=

∣∣∣∣∣∣∣2 0 00 3 0

0 01

2

∣∣∣∣∣∣∣= 2

(3

2− 0

)= 3.

Example : Solve the system u = x + 2y, v = x − y for x and y in terms of u and v. Then find

the value of the Jacobian∂(x, y)

∂(u, v).

Solution : (1) and (2) y =u− v

3, x =

u+ v

3.

J =∂(x, y)

∂(u, v)=

∣∣∣∣∣∣∣1

3

−1

31

3

1

3

∣∣∣∣∣∣∣=

1

9+

1

9

=2

9


0

∫ y2+1

x=y

2

2x− y2

dxdy by applying the transformation u =2x− y

2, v =

y

2

and integrating over an appropriate region in the uv-plane.

Solution : We sketch the region R of integration in the xy-plane and identify its boundaries. We

solve two equations u =2x− y

2, v =

y

2for x and y in terms of u and v. x = u + v, y = 2v. We

find the boundaries of G by substituting these expressions into the equations for boundaries of R.

xy equations for Corresponding uv-equations Simplifiedthe boundary of R for the boundary of G uv-equations

x =y

2u+ v =

2v

2= v u = 0

x =y

2+ 1 u+ v = v + 1 u = 1

y = 0 2v = 0 v = 0y = 4 2v = 4 v = 2.

We use the formula in change of variable∫ ∫Rf(x, y) dxdy =

∫ ∫Gf(g(u, v), h(u, v)) |J | dudv

Page|75


∫ 4

0

∫ y

2+1

x=y

2

2x− y2

dxdy = 2

∫ 1

u=0

∫ 2

v=0u dvdu = 4.


0

∫ 1−x0

√x+ y(y − 2x)2 dydx.

Solution : We sketch the region R of integration in the xy plane and identify its boundaries. Theintegrand suggest the transformation u = x+ y and v = y − 2x. Solving these two equation for x

and y in terms of u and v. x =u

3− v

3, y =

2u

3+v

3We can find the boundaries of the uv-region G.

xy equations for Corresponding uv-equations Simplifiedthe boundary of R for the boundary of G uv-equations

y = 1− x 2u

3+v

3= 1− u

3+v

3u = 1

y = 0 2u

3+v

3= 0 2u+ v = 0 i.e. v = −2u

x = 0u

3− v

3= 0 v = u

x = 1u

3− v

3= 1 u− v = 3, v = u− 3

J =

∣∣∣∣∣∣∣1

3−1

32

3

1

3

∣∣∣∣∣∣∣ =1

9+

2

9=

3

9=

1

3. By using change of variable in double integration

∫ 1

0

∫ 1−x

0

√x+ y(y − 2x)2 dydx =

∫ 1

u=0

∫ u

v=−2u

√uv2

(1

3

)dvdu

=1

3

∫ 1

u=0u

1

2

[v3

3

]uv=−2u

du

=1

9

∫ 1

u=0u

1

2 [u3 + 8u3] du

=1

9

∫ 1

09u

7

2

=

u9

2

9

2

1

0

=2

9.

Page|76


Area in Polar Co-ordinates :-The area of a closed and bounded region R in the polar co-ordinate plane is A =

∫ ∫R rdrdθ.

Example : Find the area enclosed by the lemniscate r2 = 4 cos 2θ.

Solution : We sketch the graph of the lemniscate to determine the limits of integration. We seethat the total area is 4 times the first-quadrant portion.

A = 4

∫ π

4

θ=0

∫ √4 cos 2θ

r=0rdrdθ

= 4

∫ π

4

θ=0

[r2

2

]√4 cos 2θ

r=0

dθ

= 2

∫ π

4

04 cos 2θ dθ

= 8

[sin 2θ

2

]π4

0

= 4

[sinπ

2− sin 0

]= 4.

Changing Cartesian Integrals into Polar Integrals :-The procedure for changing a Cartesian integral

∫ ∫R f(x, y) dxdy into a polar integral has two

steps.

Step 1: Substitute x = r cos θ and y = r sin θ and replace dxdy by rdrdθ in the cartesian integral.

Step 2: Supply polar limits of integration for boundary of R.The cartesian integral then becomes

∫ ∫R f(x, y) dxdy =

∫ ∫G f(r cos θ, r sin θ) rdrdθ. where G G

denotes the region of integration in polar co-ordinates.


0

∫ x0 ydydx.

Solution : ∫ 2

0

∫ x

0ydydx =

∫ 2

0

[y2

2

]xy=0

dx

=

∫ 2

0

[x2

2

]dx

=

[x3

6

]2

x=0

=8− 0

6

=4

3.

Page|77


Reversing the order the integration gives the, same answer.

∫ 2

0

∫ x

0ydydx =

∫ 2

y=0

∫ 2

x=yydxdy

=

∫ 2

y=0y[x]2x=y dy

=

∫ 2

y=0y(2− y) dy

=

∫ 2

y=0(2y − y2) dy

=

[2y2

2− y3

3

]2

y=0

= 4− 8

3

=4

3

Transforming to polar co-ordinates by the substitutionx = r cos θ, y = r sin θ, dxdy = rdrdθ.when x = 0, r = 0

when x = 2, 2 = r cos θ ⇒ r =2

cos θ= 2 sec θ.

The limits of r are r = 0 2 sec θ.when y = 0, θ = 0.when y = x, r sin θ = r cos θ i.e. tan θ = 1.

∴ θ =π

4.

∫ 2

0

∫ x

0ydydx =

∫ π

4

θ=0

∫ 2 sec θ

r=0r sin θ rdrdθ

=

∫ π

4

θ=0sin θ

[r3

3

]2 sec θ

r=0

dθ

=

∫ π

4

0sin θ

8 sec3 θ

3dθ

=8

3

∫ π

4

0

sin θ

cos θsec2 θ dθ

=8

3

∫ π

4

0tan θ sec2 θ dθ

=8

3

[(tan θ)2

2

]π4

θ=0

=8

3

[1− 0

2

]=

4

3.


R ex2+y2 dydx. where R is the semicircular region bounded by the x-axis

and the curve y =√

1− x2.Solution :

Page|78


Transforming to polar co-ordinates by the substitutionx = r cos θ, y = r sin θ (0 ≤ r ≤ 1, 0 ≤ θ ≤ π). dxdy = |J |drdθ = rdrdθ.∫ ∫

Rex

2+y2 dydx =

∫ π

θ=0

∫ 1

r=0er

2rdrdθ =

π

2(e− 1).


0

∫√4−y20 (x2 + y2) dxdy.

Solution :Transforming to polar co-ordinates by the substitution

x = r cos θ, y = r sin θ (0 ≤ r ≤ 2, 0 ≤ θ ≤ π

2)∫ 2

0

∫ √4−y2

0(x2 + y2) dxdy =

∫ 2

0

∫ π

2

θ=0r2rdrdθ =

∫ 2

r=0r3[θ]

π

20 dr = 2π.

Example : Evaluate∫ a−a∫ √a2−x2−√a2−x2 dydx.

Solution : Transforming to polar coorodinates by the substitutionx = r cos θ, y = r sin θ (0 ≤ r ≤ a; 0 ≤ θ ≤ 2π)∫ a

−a

∫ √a2−x2−√a2−x2

dydx =

∫ a

r=0

∫ 2π

θ=0rdrdθ = πa2.

OR∫ a

−a

∫ √a2−x2−√a2−x2

dydx = 4

∫ a

x=0

∫ √a2−x2y=0

dydx = 4

∫ a

x=0

√a2 − x2 dx = πa2.

Example : Find the area of the region R bounded by y = x and y = x2 in the first quadrant.Solution :

We sketch the region R bounded by the line y = x and parabola y = x2.

Required area=

∫ ∫RdA =

∫ 1

x=0

∫ x

y=x2dydx =

1

6.

OR

Required area =

∫ 1

y=0

∫ √yx=y

dxdy =1

6.

Example : Find the area of the region R enclosed by the parabola y = x2 and the line y = x+ 2.

Solution :

Page|79


If we divide R into the region R1 and R2 shown in figures (A)

A =

∫ ∫R1

da+

∫ ∫R2

dA

=

∫ 1

y=0

∫ √yx≡√y

dxdy +

∫ 4

y=1

∫ √yx=y−2

dxdy

=

∫ 1

y=0[x]√y

x≡√y dy +

∫ 4

y=1[x]√y

x=y−2 dy

=

∫ 1

y=0(y

1

2 + y

1

2 ) dy +

∫ 4

y=1(y

1

2 − y + 2) dy

=

2y

3

2

3

2

1

0

+

y3

2

3

2

− y2

2+ 2y

4

1

=4

3+

14

3− 15

2+ 6

=12− 15

2

=9

2.

On the other hand, reverting the order of integration figure B gives.

A =

∫ 2

−1

∫ x+2

x2dydx

=

∫ 2

−1[y]x+2

y=x2dx

=

∫ 2

−1[x+ 2− x2] dx

=

[x2

2+ 2x− x3

3

]2

−1

=

[4− 1

2+ x(2 + 1)− (8 + 1)

3

]=

3

2+ 3

=9

2.

Example : Find the area bounded by the co-ordinate axes and the line x+ y = 2.

Solution : Draw the co-ordinate axes and the line x+ y = 2.

Required Area =

∫ 2

x=0

∫ x−2

y=0dydx = −2.

Evaluating Polar Integrals :-

Example : Change the cartesian integral into an equivalent polar integral. Then evaluate thepolar integral

Page|80


(a)∫ 1−1

∫ √1−x20 dydx. (b)

∫ 1−1

∫ √1−x2−√

1−x2 dydx.

Solution : (a) Transforming to polar co-ordinates by the substitutionx = r cos θ, y = r sin θ (0 ≤ r ≤ 1 0 ≤ θ ≤ π) dxdy = rdrdθ.∫ 1

−1

∫ √1−x2

0dydx =

∫ 1

r=0

∫ π

θ=0rdrdθ = π

∫ 1

0rdr =

π

2.

OR∫ 1

−1

∫ √1−x2

0dydx =

∫ 1

−1

√1− x2 dx = 2

∫ 1

0

√1− x2 dx (putx = sin θ) = 2

∫ π

2

0cos2 θdθ =

π

2.

(b) Transforming to polar Co-ordinates by substitutionx = r cos θ, y = r sin θ, (0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π) dxdy = rdrdθ.

∫ 1

−1

∫ √1−x2

−√

1−x2dydx =

∫ 1

r=0

∫ 2π

θ=0rdrdθ

=

∫ 1

r=0r[θ]2π0 dr

= 2π

∫ 1

0rdr

= 2π

[r2

2

]1

0= π.

Example : Change the cartesian integral into an equivalent polar integrals. Then evaluate thepolar integrals∫ 1

0

∫√1−y20 (x2 + y2) dxdy.

Solution : Transforming to polar co-ordinates by the substitution

x = r cos θ, y = r sin θ (0 ≤ r ≤ 1, 0 ≤ θ ≤ π

2)

∫ 1

0

∫ √1−y2

0(x2 + y2) dxdy =

∫ 1

r=0

∫ π

2

θ=0r2 rdrdθ

=

∫ 1

r=0r3[θ]

π

20 dr

=

∫ 1

r=0r2(π

2−)dr

=π

2

∫ 1

0r3 dr

=π

2

[r4

4

]1

r=0

=π

2

[1− 0]

4

=π

8.


0

∫ √1−x20 e−(x2+y2) dydx.


Page|81


x = r cos θ, y = r sin θ (0 ≤ θ ≤ π

2, 0 ≤ r ≤ 1) dxdy = rdrdθ.

∫ 1

0

∫ √1−x2

0e−(x2+y2) dydx =

∫ 1

r=0

∫ π

2

θ=0e−r

2rdrdθ

=

∫ 1

r=0e−r

2rdr[θ]

π

20

=π

2

∫ 1

0e−r

2rdr

Put r2 = t

2rdr = dt

rdr =dt

2when r = 0, t = 0

when r = 1, t = 1

=π

2

∫ 1

t=0e−t

dt

2

=π

2

[e−t

2

]1

t=0

=π

2

[e−1 − e−0]

2

=π

4(1− e−1)

=π(e− 1)

4e.


∫ √1−x2−√

1−x22

(1 + x2 + y2)2dydx.

Solution : Transforming to polar co-ordinates by the substitutionx = r cos θ, y = r sin θ (0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π) dxdy = rdrdθ x2 + y2 = r2.∫ 1

−1

∫ √1−x2

−√

1−x2

2

(1 + x2 + y2)2dydx =

∫ 1

r=0

∫ 2π

θ=0

2

(1 + r2)2rdrdθ

=

∫ 1

r=0

2rdr

(1 + r2)2[θ]2πθ=0

= 2π

∫ 1

r=0

2rdr

(1 + r2)2

Put 1 + r2 = t

2rdr = dt

when r = 0, t = 0

when r = 1, t = 2.

= 2π

∫ 2

t=1

dt

t2

= 2π

[t−1

−1

]2

t=1

= 2π

[−1

2+

1

1

]= 2π

(1

2

)= π.


∫√1−y2

−√

1−y2log(x2 + y2 + 1) dxdy.


Page|82


x = r cos θ, y = r sin θ (0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π)

∫ 1

−1

∫ √1−y2

−√

1−y2log(x2 + y2 + 1) dxdy =

∫ 1

r=0

∫ 2π

θ=0log(r2 + 1) rdr dθ

=

∫ 1

r=0log(r2 + 1) rdr[θ]2πθ=0

= 2π

∫ 1

r=0log(r2 + 1) rdr

Put r2 + 1 = t

2rdr = dt

when r = 0, t = 1

when r = 1, t = 2

= π

∫ 2

t=1log t dt

= π

{[t log t]2t=1 −

∫ 2

1t1

tdt

}= π{2 log 2− 0− [t]21}= π(2 log 2− 1)

= π(log 4− 1).

Part 3

Triple Integrals in Rectangular Co-ordinates :-We use triple integrals to find the volumes of three dimentional shapes, moments of solids and themasses and the average value of functions of three variables.

Triple Integrals :-If F (x, y, z) is a function defined on a closed bounded region D in space - the region occupied bya solid ball, for example, or a lump of clay - then the integral of F over D may be defined inthe following way. We partition a rectangular region containing D into rectangular cells by planesparallel to the co-ordinate planes. We number the cells that lie inside D from 1 to n in some order, atypical cell having dimensions 4xk by 4yk by 4zk and volume 4Vk. We choose a point (xk, yk, zk)

in each cell and form the sum Sn =n∑k=1

F (xk, yk, zk)4Vk (1) If F is continuous and the bounding

surface of D is made of smooth surfaces joined along continuous curves, then as 4xk,4yk and 4zkapproach zero independently the sums Sn approach a limit lim

n→∞Sn =

∫ ∫ ∫DF (x, y, z) dv. We

call this limit the triple integral of F over D. The limit also exists for some discontinuous functions.

Properties of Triple Integrals :-If F = F (x, y, z) and G = G(x, y, z) are continuous, then

1.∫ ∫ ∫

D kFdv = k∫ ∫ ∫

D Fdv (any number k).2.∫ ∫ ∫

D (F ±G) dv =∫ ∫ ∫

D Fdv ±∫ ∫ ∫

DGdv.3.∫ ∫ ∫

D Fdv ≥ 0 if F ≥ 0 on D.4.∫ ∫ ∫

D Fdv ≥∫ ∫ ∫

D Gdv if F ≥ G on D.5. If the domain D of a continuous function F is partitioned by smooth surfaces into a finite

number of nonoverlapping cellsD1, D2, D3, · · · , Dn, then

∫ ∫ ∫D Fdv =

∫ ∫ ∫D1

Fdv +∫ ∫ ∫

D2Fdv + · · ·+

∫ ∫ ∫Dn

Fdv.

Definitions :- The volume of a closed, bounded region D in space is V =∫ ∫ ∫

D dv.

Example : Find the volume of the region D enclosed by the surfaces z = x2 + 3y2 and z =8− x2 − y2.

Solution : The volume is v =∫ ∫ ∫

D dzdydx z = x2 + 3y2 = 8− x2 − y2

2x2 + 4y2 = 8x2 + 2y2 = 4.

Page|83


2y2 = 4− x2

y2 =4− x2

2

y = ±√

4− x2

2

Required volume V =

∫ 2

x=−2

∫ √√√√4− x2

2

y=−

√√√√4− x2

2

∫ 8−x2−y2

z=x2+3y2=

∫ 2

x=−2

∫ √√√√4− x2

2

y=−

√√√√4− x2

2

(8− x2 − 4y2) dydx = 4

∫ 2

x=0

∫ √√√√4− x2

2

y=0(8− x2 − 4y2) dydx

Areas, Moments and Centers of Mass :-Area of Bounded Regions in the Plane :-If we take f(x, y) = 1 in the definition of the double integral over a region R, then the partial

sums reduces to Sn =

n∑k=1

f(xk, yk)4Ak =

n∑k=1

4Ak (1) As 4x and 4y approach zero, the coverage

of R by the 4Ak’s becomes increasingly complete and we define the area of R to be the limit.

Area = limn→∞

n∑k=1

4Ak =

∫ ∫RdA. (2)

Definition :- The area of a closed, bounded plane region R is A =∫ ∫

R dA. (3) To evaluate theintegral in (3), we integrate the constant function f(x, y) = 1 over R.

Example : Find the area of the region R bounded by y = x and y = x2 in the first quadrant.

Solution : Draw the line y = x and parabola y = x2.

Required area =

∫ 1

0

∫ x

y=x2dydx =

∫ 1

0[y]xy=x2 dx =

∫ 1

0[x− x2] dx =

1

6.

Example : A thin plate covers the triangular region bounded by the x-axis and the lines x = 1and y = 2x in the first quadrant. The plate’s density at the point (x, y) is δ(x, y) = 6x + 6y + 6.Find the plate’s mass first moments, center of mass, moments of intertia, and radii of gyrationabout the co-ordinate axes.

Solution : We sketch the plate and put in enough detail to determine the limits of integration forthe integrals we have to evaluate.The plate’s mass is

M =

∫ 1

0

∫ 2x

y=0δ(x, y) dydx =

∫ 1

x=0

∫ 2x

y=0(6x+ 6y + 6) dydx. = 14.

The first moment about the x-axis is

Mx =

∫ 1

0

∫ 2x

0yδ(x, y) dydx =

∫ 1

0

∫ 2x

0(6xy + 6y2 + 6y) dydx11.

Similarly the first moment about the y-axis is

My =

∫ 1

0

∫ 2x

0xδ(x, y) dydx = 10.

The moment of intertia about the x-axis is

Ix =

∫ 1

0

∫ 2x

0y2δ(x, y) dydx =

∫ 1

0

∫ 2x

0(6xy2 + 6y3 + 6y2) dydx = 12.

Similarly, the moment of inertia about the y-axis is

Iy =

∫ 1

0

∫ 2x

0x2δ(x, y) dydx =

39

5

∴ I0 = Ix + Iy = 12 +39

5=

60 + 39

5=

99

5.

The three radii of gyration are

Page|84


Rx =√Ix/M =

√12

14=

√6/7

Ry =√Iy/M =

√39

5/14 =

√39

70

R0 =√I0/M =

√99

4/14 =

√99

70.

Example : (1) Evaluate

∫ 1

0

∫ 1

0

∫ 1

0(x2 + y2 + z2) dzdydx.

Solution :∫ 1

0

∫ 1

0

∫ 1

0(x2 + y2 + z2) dzdydx =

∫ 1

x=0

∫ 1

y=0

[(x2 + y2)z +

z3

3

]1

z=0

dydx

=

∫ 1

x=0

∫ 1

y=0

[(x2 + y2) +

1

3

]dydx

=

∫ 1

x=0

[(x2 +

1

3

)y +

y3

3

]1

y=0

dx

=

∫ 1

x=0

(x2 +

2

3

)dx =

1

3.

(2) Evaluate

∫ e

1

∫ e

1

∫ e

1

1

xyzdxdydz.

Solution : ∫ e

1

∫ e

1

∫ e

1

1

yz

1

xdxdydz =

∫ e

1

∫ e

1

1

yz[log x]e1 dydz

=

∫ e

1

1

z

∫ e

1

1

ydydz = 1

Example : Evaluate

∫ 1

0

∫ π

0

∫ π

0y sin z dxdydz.

Solution : ∫ 1

z=0

∫ π

y=0

∫ π

x=0y sin z dxdydz =

∫ 1

z=0

∫ π

y=0y sin z dydz

= π

∫ 1

z=0

[y2

2

]πy=0

sin z dz =π3

2[1− cos 1].

Example : Evaluate

∫ 1

0

∫ 2−x

0

∫ 2−x−y

0dzdydx.

Solution :∫ 1

0

∫ 2−x

0

∫ 2−x−y

0dzdydx =

∫ 1

0

∫ 2−x

0[z]2−x−yz=0 dydx =

∫ 1

0

∫ 2−x

y=0(2− x− y) dydx

=

∫ 1

x=0

[(2− x)y − y2

2

]2−x

y=0

dx =

∫ 1

x=0

[(2− x2)− (2− x)2

2

]dx =

1

2

∫ 1

x=0(2− x)2 dx =

7

6.

Example : Find the volume of the region D enclosed by the surfacesz = x2 + 3y2 and z = 8− x2 − y2.

Solution : The volume is V =

∫ ∫ ∫Ddzdydx.

z = x2 + 3y2 and z = 8− x2 − y2

The limits of z. are z = x2 + 3y2 to z = 8− x2 − y2.x2 + 3y2 = 8− x2 − y2

x2 + 2y2 = 4

y2 =4− x2

2

y = ±√

(4− x2

2)

Page|85


The limits of y are from y = −√

4− x2

2to y =

√4− x2

2.

4− x2 = 0 ∴ x = ±2The limits of x are from x = −2 to x = 2.

V =

∫ 2

x=−2

∫ √√√√4− x2

2

y=−

√√√√4− x2

2

∫ 8−x2−y2

z=x2+3y2dzdydx

= 2

∫ 2

x=02

∫ √√√√4− x2

2

y=0

∫ 8−x2−y2

z=x2+3y2dzdydx

= 4

∫ 2

x=0

∫ √√√√4− x2

2

y=0(8− 2x2 − 4y2) dydx

= 4

∫ 2

x=02

[(4− x2)y − 2y3

3

]√√√√4− x2

2

y=0

= 8

∫ 2

x=0

(4− x2)

(4− x2

2

)1

2 − 2

3

(4− x2

2

)3

2

dx

= 8

∫ 2

x=0

(4− x2)

3

2√

2− 2(4− x2)

3

2

3 · 2√

2

dx=

(8)2

3√

2

∫ 2

0(4− x2)

3

2 dx

Put x = 2 sin θ dn = 2 cos θdθ

when x = 0, θ = 0

when x = 2, θ =π

2

=16

3√

2

∫ π

2

08 cos3 θ2 cos θdθ.

=8√

2

3

∫ π

2

016

(1 + cos 2θ

2

)2

dθ

=8√

2

3

∫ π

2

04(1 + 2 cos 2θ + cos2 2θ) dθ

=8√

2

3

∫ π

2

04

[1 + 2 cos 2θ +

1 + cos 4θ

2

]dθ

=8√

2

34

[θ +

2 sin 2θ

2+θ

2− sin 4θ

8

]π2

0

=8√

2

3· 4[pi

2+π

4

]=

8√

2

34

(3π

4

)= 8

√2π.

Example : Evaluate the cylindrical co-ordinate integrals∫ 2π0

∫ 10

∫ √2−r2r dz rdrdθ.

Page|86


Solution : ∫ 2π

0

∫ 1

0

∫ √2−r2

rdz rdrdθ =

∫ 2π

θ=0

∫ 1

r=0[z]√

2−r2z=r r drdθ

=

∫ 2π

θ=0

∫ 1

r=0[√

2− r2 − r] rdrdθ

=

∫ 2π

θ=0

∫ 1

r=0[(√

2− r2)rdr − r2dr] dθ

=

∫ 2π

θ=0

{∫ 1

r=0

√2− r2rdr −

∫ 1

r=0r2dr

}dθ

Put 2− r2 = t

−2rdr = dt

rdr = −dt2

=

∫ 2π

θ=0

∫ 1

t=2t

1

2(−dt)

2−[r3

3

]1

0

dθ

=

∫ 2π

θ=0

1

2

t3

2

3

2

2

t=1

− 1

3

dθ

=1

3

∫ 2π

θ=0[2

3

2 − 1− 1] dθ

=1

3(2√

2− 2)[θ]πθ=0

=2(√

2− 1)

3[2π − 0]

=4π

3(√

2− 1).


0

∫ 3−3x0

∫ 3−3x−y0 dzdydx.

Solution : ∫ 1

0

∫ 3−3x

0

∫ 3−3x−y

0dzdydx =

∫ 1

x=0

∫ 3−3x

y=0

∫ 3−3x−y

z=0dzdydx

=

∫ 1

x=0

∫ 3−3x

y=0[z]3−3x−y

z=0 dydx

=

∫ 1

x=0

∫ 3−3x

y=0(3− 3x− y) dydx

=

∫ 1

x=0

[(3− 3x)y − y2

2

]3−3x

y=0

dx

=

∫ 1

x=0

[(3− 3x)2 − (3− 3x)2

2

]dx

=

∫ 1

0(3− 3x)2 dx

=

[(3− 3x)3

3(−3)

]1

x=0

=(3− 3)3 − (3)3

−9

=27

9= 3.


∫ 1−1

∫ 1−1 (x+ y + z) dydxdz.

Page|87


Solution :∫ 1

−1

∫ 1

−1

∫ 1

−1(x+ y + z) dydxdz =

∫ 1

z=−1

∫ 1

x=−1

∫ 1

y=−1(x+ y + z) dydxdz

=

∫ 1

z=−1

∫ 1

x=−1

[(x+ z)y +

y2

2

]1

y=−1

dxdz

=

∫ 1

z=−1

∫ 1

x=−1[2(x+ z) + 0] dxdz

=

∫ 1

z=−1[x2 + xz]1x=−1 dz

=

∫ 1

z=−12zdz

= 2

[z2

2

]1

z=−1

= 0.

Example : Evaluate∫ e

1

∫ e1

∫ e1 ln r ln s ln t dt drds.

Solution :∫ e

1

∫ e

1

∫ e

1ln r ln s ln t dt drds =

∫ e

s=1

∫ e

r=1ln r ln s

{[t log t]e1 −

∫ e

1t · 1

tdt

}drds

=

∫ e

s=1

∫ e

r=1ln r ln s {e log e− 0− [t]e1} drds

=

∫ e

s=1

∫ e

r=1ln r ln s {e log e− e+ 1} drds

=

∫ e

s=1ln s

{[r ln r]er=1 −

∫ e

1r

1

rdr

}(e log e− e+ 1) ds

=

∫ e

s=1ln s[e log e− 0− e+ 1](e log e− e+ 1) ds

= (e log e− e+ 1)2

{[s log s]es=1 −

∫ e

1ds

}= (e log e− e+ 1)2 {e log e− e+ 1}= (e log e− e+ 1)3.

Page|88

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