nitkkr.ac.innitkkr.ac.in/docs/3-Geometrical Applications of Integration.pdfnitkkr.ac.in

Engineering Mathematics through Applications200

Geometrical Applications of Integration

3aaaaa

3.1 INTRODUCTION

In general, we consider the integration as the inverse of differentiation. In the expression of

the sum, ( ) ,b

af x x∆∑ f is considered continuous on a ≤ x ≤ b and we find that limit of S as ∆x

approaches to zero is the number = −∫ ( ) ( ) ( ),b

af x dx F b F a where F is any anti-derivative of f. We

apply this contention in finding the area between the x-axis and the curve y = f(x), a ≤ x ≤ b. Weextend the application to compute distances, volumes and volumes of revolution, length ofcurves, area of surface of revolution, average value of function, centre of mass, centroid, etc.

3.2 AREA OF BOUNDED REGIONS (QUADRATURE)

I. Areas of Cartesian CurvesThe area bounded by the curve y = f(x), X-axis and the

ordinates x = a, x = b is ∫b

ay dx , when f (x) is continuous single

valued and finite function of x, and y does not change sign in theinterval [a, b].

If AB is the curve y = f(x) between the ordinatesLA(x = a) and MB(x = b) with a condition that y is strictlyincreasing (or strictly decreasing) function of x in theinterval [a, b].

Let P(x, y) and Q(x + δx, y + δy) be two neighbouringpoints on this continously increasing curve y = f (x) andNP, N’Q be their respective ordinates.

Here clearly the ALNP i.e. A depends on the position ofthe point P(x, y) whose abscissa is ‘x’ and area PNN’Q = δA lies between the areas of the

rectangles PN’ and NQ, i.e. δA lies between area yδx and (y + δy)δx or Ax

δδ lies between y and

(y + δy).

On taking the limits as Q → P, i.e. δx → 0 (meaning thereby δy → 0), =dA ydx

Integrating both sides between the limits x = a to x = b, we get

Fig. 3.1

A( , )x y P

P´

yB

x-axisO L N N´ M

y f x = ( )

x δx

Q

Q´( + , + )x x y yδ δ

y y y + δ

Geometrical Applications of Integration 201

Area b

ba

aALMB A ydx= = ∫ .

However, if x and y are interchanged in the above formula, wesee that the area bounded by the curve x = f (y), Y-axis and the abscissa

y = a, y = b is .b

axdy∫

Observations:(i) The area bounded by the curve y = f(x), the two ordinates at A and B and the X-axis is often called the

area under the curve AB and the process of calculating the area bounded by the curve is called quadrature.(ii) An area whose boundary is described in anti-clockwise direction is considered positive otherwise

negative. Or in other words, for the portion of the curve (underconsideration) above X-axis for which y is positive, area enclosed ispositive, whereas for the portion of the curve (under consideration)below X-axis for which y is negative, area is negative.

But in case of area with negative sign, we mean numerical value ofthe area.

(iii) If in the interval, a ≤ x ≤ c, the curve y = f(x) is above the x-axis and in theinterval c ≤ x ≤ b, the curve y = f(x) is below the x-axis then we write thearea

= +∫ ∫ ∫b c b

a a cy dx y dx y dx

or in otherwords, if x-changes sign from a to b, y = f(x) changes sign atsome interval point x = c (say), then the area for x from a to c andc to b, are calculated seperately and then their numerical valueare added (see Fig. 3.4)

Similarly, the result can be extended if y changes sign at morethan one intermediate point in the interval [a, b].

(iv) Area of the region bounded between two continuous curves f(x)and g(x) on [a, b] and the vertical lines x = a, x = b is given by

= −∫ [ ( ) ( )]ba

A f x g x dx ,

where f(x) ≥ g(x) in [a, b]. (see Fig. 3.5)In the region under consideration, representative rectangle isshown with height: f(xi) – g(xi), width: ∆x and P(xi g(xi)); Q(xi f(xi))If f(x) ≥ g(x) in [a, c] and f(x) ≤ g(x) in [c, b], then we write the area as

= − + −∫ ∫[ ( ) ( )] [ ( ) ( )]c b

a cA f x g x dx g x f x dx

as shown in Fig. 3.6.

g x( )i

∆x

Y

f( )xi

f x( )

g x( )

O a xi bX

P

QQ

P

x c = x a = x b =

Y

f x( ) g x( )

X

Fig. 3.5 Fig. 3.6

L MX

O

A

B

x a = x b =

y f x = ( )

Y

Fig. 3.3

Y

A

+ve area

x a =

O–vearea

x b =

B

Mx c = XNL

Fig. 3.4

Fig. 3.2

Y

BMx f y = ( )

y b =

LAy a =

OX


Example 1:(i) Find the area bounded by the parabola y2 = 4ax and its latus-ractum.

(ii) Show that the area cut off from a parabola by any double ordinate is two-third ofthe corresponding rectangle obtained by the double ordinates and its distance fromthe vertex.

Solution (i): For the parabola y2 = 4ax, let the double ordinate PP' be x = cSince the curve is symmetrical about X-axis, therefore, for part of the curve above X-axis,

y be taken as positive, i.e. = 4y axThus, the area bounded by the parabola with its double ordinate

PMP’ (i.e. Latus ractum) = Area P’APP’ = 2Area PAMP

31 1 322 2 2

0 0

0

82 2 4 4 3 32

c

c c xydx ax dx a a c

= = = =

∫ ∫

(ii) Again for P(x, y), 1 12 24 4 2y MP ax ac a c= = = = so that

1 12 22 4PP' MP a c= =

Now the area of the rectangle formed by the double ordinate(PMP’) and its distance from the vertex A (i.e. AM) = PP' × AM

( )1 1 1 32 2 2 24 4a c c a c= ⋅ =

Now the area cut of from the parabola by double ordinates, i.e. area P’APP’

( )1 3 1 12 2 2 28 2 24 area of the rectangle formed by and .

3 3 3a c a c PP' AM= = =

Hence, the area formed by the parabola and its latus ractum is two third of the area of therectangle formed by the double ordinates with its distance from the vertex, A.

Note: Vice versa, the area of the rectangle is 32

times the area bounded by the parabola with latus ractum.

Example 2: Find the area between the curvex2y2 = a2(y2 – x2) and its asymptotes.

Solution: The given curve x2y2 = a2(y2 – x2) issymmetrical about both the axis and at the origin,

y = ±x as the tangents. Further, x = ± a are the twoasymptotes parallel to Y-axis. The curve no whereintersects with the axis except at (0, 0). Whence thecurve does not enclose area with its abscissa orordinates (see Fig. 3.8).

Due to its symmetry about both the axis, the wholearea between the curve and its asymptotes is

2 2004 4

a a axy dx dxa x

= =−

∫ ∫ (Putting x = a sinθ)Fig. 3.8

Fig. 3.7

XA a( , 0)

x a = – x a=

Y

O

y

x = –

yx

=

B a(– , 0)

Y

A

B P

x = 0

x c =

CP´

(4 , 4 )a a y ax2 = 4

M


2

2 2 2

0 0cos 4sin4 sin 4

cosa daa a d a

a

ππ

⋅ θ θ =θ= ⋅ θ θ =θ ∫∫

Example 3: Find the area of the curve a2y2 = x3(2a – x).

Solution: Without going into geometrical details of the curve, the area enclosed by it in thefirst quadrant is

32 2 2

0 02

a aA xydx a x dx

a= = −∫ ∫

(i) The curve intersects X-axis at x = 0, x = 2a.(ii) Axis of X is the tangent at the origin.

(iii)30 at 2 , at 2

dy dyx a x a

dx dx= = = ∞ =

Put x = 2a sin2θ

32 22 2

0

(2 sin ) 2 2 sin 4 sin cosAa a a a d

a

π

= ⋅θ − θ ⋅ θ θ θ∫ 2 2 4 2

016 sin cosa d

π

= θ θ θ∫

⋅ π π = = ⋅ ⋅

22 3 116

6 4 2 2 2aa

2

0

( 1)( 3) ( 1)( 3)using sin cos

( )( 2) 2p p q q

p q dp q p q

π − − … − − … πθ θ θ = + + − ∫Hence the total area is πa2.

Example 4: For the curve y2(a – x) = x2(a + x) [NIT Kurukshetra, 2004](i) Find the area of the loop

(ii) Area of the portion bounded by the curve andits asymptote.

Solution: The curve passing through the origin issymmetrical about X-axis and has x = a as itsasymptote. It intersects the axis of X at (–a, 0) as shownin Fig. 3.10.

For the area of the loop, x varies from –a to 0.Further, the loop is symmetrical about X-axis

∴ 0 0 0

2 22Area of the loop 2 2

a a a

a x a xy dx x dx x dxa x a x− − −

= + += =− −∫ ∫ ∫ (on rationalization)

0 0 2

2 2 2 22

a adx dxax x

a x a x− −+

= − − ∫ ∫

1 2 2 20 02 2 2

2 2

( )2 ( )a a

dx dxa a xax a x

a x

−

− −+

− −= − − ∫ ∫

Y

X´ XA a(2 , 0)

O

( , 0)a(0, 0)

3, 0

2

a

Fig. 3.10

x a=

yx

=

y

x = –

O A

(– , 0)a

Y

X

Y´

X´

Fig. 3.9


10 0 0

2 2 2 2 222 2

12 ( )( )a a a

a dx a dxx a x dx a xa x

−

− − −− +

= − − − − − ∫ ∫ ∫

010 2 2 22 2 2 1 122 ( ) sin sin

2 2aa

a a x x a x a xd a x dxa a

− −−

−

− + −= − − +

∫

0 001 2 2 22 2 122 ( ) sin

2 2a aa

a a x x a xa xa

−

− −−

− + − = − −

( ) ( ) ( )2

2 12 0 0 sin ( 1) 0 0 ,2aa −+ = − − − − − −

where − −π− =1sin ( 1)2

π = − + = ⋅ − + π = π −

2 2 222 2 [ 4 ] ( 4) numerically

4 4 2a a aa

Alternately:

0

2 ,a

a xA x dxa x−

+=−∫ Put x = a sinθ, dx = acosθdθ

Limits for x = 0, θ = 0; x = –a, −πθ =2

0

2

(1 sin )2 sin cos(1 sin )

aA a a da−π

+ θ= θ θ θ− θ∫

0

2

1 sin 1 sin2 sin cos1 sin 1 sin

a a d−π+ θ + θ= θ θ θ− θ + θ∫

0 0

2 2 22

2 2

2sin cos (1 sin )

2 [sin sin ]1 sin

d aa d−π −πθ =θ θ + θ

= θ + θ θ− θ∫ ∫

002 2

2 2

21 sin 22 sin (1 cos 2 ) cos2 2 4

d aa −π −πθ = θ θ = θ + − θ − θ + − ∫

= π − = − π2 2

( 4) (4 ) numerically2 2a a

(b) Area between the curve and its asymptote

+= =−∫ ∫0 0

2 2a a a xydx x dx

a xThis integrand is same as in the case (a) simply limits are 0 to a.

1 2 2 2 2

2 2 12

0

Area 2 ( ) sin ( 4)2 2 2

aa x x a x aa a x

a− −= − − + − = π +

Example 5: Show that the area of the loop of the curve y2(a + x) = x2(3a – x) is equal to thearea between the curve and its asymptote. [KUK, 2002; NIT Kurukshetra, 2003]


Solution: The curve y2(a + x) = x2(3a – x) issymmetrical about X-axis and passes through the

origin with = ± 3y x as the tangent. The line x =0 or x = –a is the asymptote parallel to Y-axis.Further the curve intersects X-axis at (0, 0), (3a, 0)making one loop in the positive direction of X-axisas shown in the geometry.

Here 3a xy x a x−= +

Clearly the area of the loop is the area which thecurve encloses between the ordinatesx = 0, x = 3a and the X-axis, viz.

= + −= =− −∫ ∫

3 3

0 02

2 ( )2(Area ) 22 ( )

a a a a xA ABO y dx x dxa a x

Take (a – x) = 2a cosθ so that dx = 2a sinθdθ and the limits 0, ;33 ,

xx a

π= θ == θ = π

Implying 3

1 cos2 (1 2cos ) 2 sin1 cos

A a a dπ

π+ θ= − θ θ θ− θ∫

2

23

(1 cos )4 (1 2cos ) sin(1 cos )

a dπ

π+ θ= − θ θ θ− θ∫

2 2 2

3 3

4 (1 2cos )(1 cos ) 4 (1 2cos cos )a d a dπ π

π π= − θ + θ θ = − θ − θ θ∫ ∫

02 2

3 3

sin 24 (cos2 cos ) 4 sin2

a d aπ

π π

θ = − θ + θ θ = − + θ ∫

2 21 3 34 3 32 2 2

a a = − − − =

Further, the area between the curve and its asymptote is approximately,

0 0

,2 ( )2 22 ( )a a

dxa a xA y dx xa a x− −

+ −= =− −∫ ∫

Same substitution ( ) 2 cos ,

and limits: , 0; 0, 3

a x a

x a x

− = θ π = − θ = = θ =

32 2

0

sin 24 sin 3 32

A a a

πθ = − + θ = −

which is numerically the same as the area of the loop of the curve.

2π/3θ π = /3

( , )a a

B

A(3 , 0)a

x a = –

( , – )a a

O

x = –

Fig. 3.11


Alternatively, y can also be expressed as 4 ( )3 a a xa xy x xa x a x

− +−= =+ + and if (a + x) = 4asin2θ

then

3 2

2 22 2 2 2 220

6 6

cos(4 sin ) 8 sin cos 8 (4 sin cos cos )sin

aydx a a a d a a d

π π

π π= θθ − ⋅ θ θ θ = θ θ − θ θθ∫ ∫ ∫

2 22 2 2 2 2 2

6 6

8 (2 sin cos ) cos 8 sin 2 cosa d a dπ π

π π= θ θ − θ θ = θ − θ θ ∫ ∫

( )2 22 2

6 6

8 1 cos 4 1 cos 2 4 cos 4 cos 22 2

a d a dπ π

π π= − θ + θ − θ = − θ + θ θ ∫ ∫

π

π π= − θ θ θ θ + = +

22 2

6 6

4 sin 4 sin 2 sin 4 sin 244 2 4 2

a a as at π2 , limit value is zero.

= + =

2 21 3 1 34 3 34 2 2 2

a a

Example 6: Find the area included between the circle x2 + y2 = 2ax and the parabolay2 = ax. [NIT Kurukshetra, 2005]

Solution: The circle x2 + y2 = 2ax is symmetricalabout X-axis, and passes through the origin withcentre as (a, 0), radius ‘a’, Y-axis (x = 0) as thetangent at the origin. Further, x2 + y2 – 2ax = 0or (x – a)2 + y2 = a2 means it intersects X-axis atx – a = ±a or x = 0, 2a.

In case of parabola, y2 = ax, it intersects the circlex2 + y2 – 2ax = 0 at (0, 0), (a, a).

Hence the area outside the parabola and insidethe circle is covered under limits x = 0 and x = a.(see geometry).

Hence the desired area = 2Area OBCO

( ) ( )0

2 of the circle of the parabolaa

y y dx = − ∫ 2

0 02 2

a adxax x dx ax = − − ∫ ∫

I II

22 2 2 2 2 2

0 0 0I 2 ( ) cos sin

a aax x dx a a x dx a a a d

π

= − = − − = − θ ⋅ θ θ∫ ∫ ∫

Fig. 3.12

y ax2 = y-axis

( , )a aC

O( , 0)a

A

B

x-axis(2 , 0)a

(– , )a a


4

22 2

0sin

4a ad

π

= = πθ θ∫ (Taking a – x = acosθ)

32 2

0

0

2II 3 32

a

a xax dx a a= = =

∫

∴ π π = − = − = π −

4 22 22 42 (3 8)

4 3 2 3 6a aA a a

Example 7: Find the area included between the parabola x2 = 4ay and the curve ayx a

=+

3

2 28

4.

Solution: The curve, x2 = 4ay represents an upward parabola symmetrical about Y-axis withthe origin as its vertex.

The curve 3

2 28

4ay

x a=

+ which is symme-

trical about Y-axis, does not pass throughthe origin. Further, y = 0, i.e. axis of X isan asymptote to it. It meets the Y-axisat (0, 2a).

To find the points of intersection,equating the two values of y, i.e.

2 3

2 28

4 4x aa x a

= +or x4 + 4a2x2 – 32a4 = 0

or (x2 + 8a2)(x2 – 4a2) = 0Rejecting (x2 + 8a2) = 0 (which gives imaginary values of ‘x’), we get x2 – 4a2 = 0, i.e. x = ± 2a, and y = a.Thus, these two curves intersects at (2a, a) and (–2a, a).

∴2 3 2

2 20

8The required area 2 24 4

a a xOABC OBC dxx a a

= × = − +∫

233 1

0

1 12 8 tan2 2 4 3

ax xa

a a a− = −

( ) π = − − − 2 312 4 0 8 0

4 12a a

a

( ) = π − = π − 2 2 22 22 3 2

3 3a a a

Example 8: Find the area between the curve x3 + y3 = 3axy and its asymptote x + y + a = 0.[KUK, 2001; NIT Kurukshetra, 2007]

Y

(2 , )a a

(0, 2 )aB

x ay2 = 4

A

O

C(–2 , )a a

X

x a2 2 + 4

8a3y =

Fig. 3.13


Solution: Clearly from the Fig. 3.14, which has already been explained in detail under tracingof curves, the product of the slops of the line of symmetrical, y = x and the equation ofasymptote, x + y + a = 0 is –1, i.e. they are at right angle to each other. Whence if the axes ofreferences are turned through an angle of 45°, the new X-axis coincides with the symmetricalline y = x with a changed equation of curve havingasymptote parallel to the new Y-axis instead the obliqueasymptote x + y + a = 0.

After transformation, new x and y are

−= ° − ° = ( )

X cos 45 sin 452

x yx y

+= ° + ° = ( )

sin 45 cos 452

x yY x y

and hence the new equation of curve becomes

− + − + + = ⋅

3 3( ) ( ) ( ) ( )3

2 2 2 2x y x y x y x y

a

3 31 1 3( ) ( ) ( )( )2 2 2 2 2

ax y x y x y x y− + + = − +

or + = −3 2 2 23( 3 ) ( )2ax xy x y

+ − − =

2 2 33 02 2

a ay x x x

3y2(x + b) – x2(3b – x) = 0, where =2

ab

Clearly, the asymptote of this equation parallel to Y-axis is x = –b.Hence, the area between the curve and its asymptote is given by

0

2b

A ydx−

= ∫

−

− − = +∫1

0 2323b

dxx b xx b

(Taking negative sign)

−

−−=+ −∫

0 (3 )23 ( )(3 )b

dxx b xx b b x

Putting x = b(1 – 2cosθ) so that dx = 2bsinθ dθ

and for x = –b, θ = 0; x = 0, πθ =3

∴ ( )

( )( )3

0

(1 2cos ) 3 ( 2 cos ) 2 sin23 ( 2 cos ) 3 ( 2 cos )

b b b b bA d

b b b b b b

π − θ − − − θ ⋅ θ −= θ − θ + − − θ∫

3 2 2

0

2 2 (1 cos 2cos )3

b dπ

−= − θ − θ θ∫

Fig. 3.14

45°X

x´Y' Y

yx m

= ( = 1)

1

O(90) AX'

B

xy

a

m

+ + = 0(

= –1)

2

–Y


2

3

0

4 (cos cos 2 )3b d

π

= θ + θ θ∫

2 3 2 2

0

4 sin 2 3sin 33 2 2b b a

πθ = θ + = =

Example 9: Show that area common to the two parabolas y2 = 4a(x + a) and

y2 = 4b(b – x) is a b ab+8 ( )3

.

Solution: Both the parabolas, = + …

= − …

2

24 ( ) ( )4 ( ) ( )

y a x a iy b b x ii

are

symmetrical about X-axis. Parabola, y2 = 4a(x + a), has itsvertex at A(–a, 0) and latus ractum as 4a, whereas theparabola y2 = 4b(b – x), has its vertex at B(b, 0) and latusractum as 4b.

For intersection of these two, we get 4a(x + a) = 4b(b – x)

or x(a + b) = b2 – a2

or x = (b – a)Hence the two parabolas intersects at L and M for x = b – a.

∴ Area included between them = area ALBMA = 2 × area ALBNA = 2(Area ALNA + Area LNBL)

−

− −+ = + − ∫ ∫2 4 ( ) 4 ( )

b a b

a b adx dxa x a b b x

3 32 2( ) ( )4 43 3

2 2

b a b

a b a

x a b xa b

−

− −

+ − = + −

3 3 1 3 3 12 2 2 2 2 28 8 8 80 0 ( )

3 3 3 3a bb a a b a b a b ab

= − − − = + = +

Example 10: Find the area common to the circle x2 + y2 = 4 and the ellipse x2 + 4y2 = 9.

Solution: The equation x2 + y2 = 4 represents circle with centre (0, 0) and radius 2 units

where the equation x2 + y2 = 9 or

+ =

22

221

3 32

yx

represents an ellipse with semi-major axis as 3

units and semi-minor axis as 32

units and for intersection of these two, we get

Y

Ly a x a2 = 4 ( + )

y2 = 4b(b – x)

A(– , 0)aO N B b( , 0)

M

X

Fig. 3.15


x2 + 4(4 – x2) = 9 or 3x2 = 7 i.e. = 73

x

Since the ellipse and the circle both are symmetrical about both the axis∴ Required common area to the circle and the ellipse = 4 × area common to them in the Ist quadrant = 4[area OAPD] = 4[Area OLPD + area LAPL]

7 23

703

4 (value of ellipse) (value of circle)y dx y dx = + ∫ ∫

= − + − ∫ ∫

7 23 2 2

703

14 9 42

x dx x dx

−

= − + +

732 1

0

94 9 sin2 2 3x xx

−

− +

22 1

73

44 sin2 2 2x xx

1 1 17 7 7 7 7 72 9 9sin 0 0 4sin (1) 4 4sin

3 3 3 3 3 3 2 3− − − = − + − + + − − −

1 17 20 7 7 5 72 9sin 4 4sin

3 3 27 2 3 3 12− − π= + + × − ⋅ −

1 17 20 5 7 72 2 9sin 4sin

3 3 3 27 12− − = − + π + −

1 135 7 72 2 9sin 4sin

3 27 12− − = + π + −

ASSIGNMENT 1

1. Find the area of the ellipse 22

2 21

yxa b

+ =

2. Find the area of the circle x2+ y2 = a2

3. Find the area of the loop of the curve ay2 = x2(a – x)

[ ]HINT : ( )x a x dx a a x a xdx − = − − − ∫ ∫4. Calculate the area of the curve a2x2 = y3(a – y)5. Find the whole area of the curve x2(x2 + y2) = a2(x2 – y2) [HINT : Put x2 = a2cosθ]

x y + = 422

B P(0, 2)

Px y + 4 = 922

(0, 3/2)D

O L

A(2, 0)

C(3, 0)73

A

(2, 0)

Fig. 3.16


6. Find the area between the curve y2(2a – x) = x3 and its asymptote.7. Find the area bounded by the curve xy2 = 4a2(2a – x) and its asymptote.8. Compute the area bounded by the parabola y = x2 + 2 and the straight line x = 0, x = 1

and x + y = 0.9. Find the area enclosed between the curve x2 = 4y and the straight line x = 4y – 2.

10. Find the area of the segment cut off from the parabola x2 = 8y by the line x – 2y + 8 = 0.11. Find the area bounded by the parabola y2 = 4ax and the line x + y = 3a.

12. Prove that the area common to the parabolas x2 = 4ay and y2 = 4ax is 216

3a

.

II. Area of Curves Given in Parametric Form(i) The area bounded by the curves x = f(t), ψ = φ(t), the X-axis and the ordinates at the

points where, t = a, t = b is given by ∫

b

a

dxy dtdt

.

(ii) The area bounded by the curve x = f(t), ψ = φ(t), the Y-axis and the abscissac at the

points where, t = c, t = d is given by ∫

d

c

dyx dt

dt.

Example 11: Find the area included between the cycloid x = a(θθθθθ + sinθθθθθ), y = a(1 – cosθθθθθ);and its base. Also find the area between the curve and X-axis.

Solution: For the inverted cycloid }= θ + θ= − θ

( sin ),(1 cos )

x ay a which is symmetrical about Y-axis, the

point O (the farthest point on it) is its vertex and the line AB which is parallel to X-axis is itsbase.

For half of the cycloid, θ varies from 0 to π.∴ Area between the curve and its base BCA

= area BOA = 2 area OCA

π

= = θθ∫ ∫

2

0 02

a dyxdy x d

d

0

2 ( sin )( sin )a a dπ

= θ + θ θ θ ∫

2 2

02 ( sin sin )a d

π= θ θ + θ θ∫

2 2

0 0

1 cos 22 sin 22

a d a dπ π − θ = θ θ θ + θ ∫ ∫

2 20

0

sin 22 ( cos ) 1( cos )2

a d aπ

π θ = θ − θ − − θ θ + θ − ∫

[ ]2

2 2 sin2 cos sin0

a a π = −π π + π + π −

Y

C A( = )θ πB( = )θ π

2a

L

dy2a

M O P

dx

Fig. 3.17


= 2a2(– π) + a2π = 3πa2.Area between the curve and X-axis = 2 area OAL

π

= = θθ∫ ∫

2

0 02 2

a dxydx y dd

0

2 (1 cos ) (1 cos )a a dπ

= − θ + θ θ∫

/2

2 2 2 2 2

0 02 sin 4 sina d a d a

π π= θ θ = θ θ = π∫ ∫

π

∫ ⋅ − − … πθ θ = − …

2

0

( 1)( 3)using sin ( 2) 2

p pd p p

Example 12: Show that the area of the hypocycloid x = acos3t, y = bsin3t is abπ38

. Hence

deduce the area of the asteriod x = acos3t, y = asin3t.

Solution: The given curve x = a cos3t, y = b sin3t (hypocycloid) shown in the figure, 3.18

meets x-axis at t = 0 and t = π and to y-axis at π=2

t and π= 32

t , and is symmetrical about

both the axis.Y

t = /2π(0, )b

t = π (– , 0)a t = 0

( , 0)a

(0, – )b

X

Fig. 3.18

∴ The required area 2

0 04 4

a dxydx y dtdt

π

= =∫ ∫

0

3 2

/2

4 sin .3 cos . sinb t a t t dtπ

= −∫

/2

4 2

012 sin cosab t t dt

π= ⋅∫

(4 1) (4 3) (2 1) 312

6 4 2 2 8abab − ⋅ − ⋅ − π π= =

⋅ ⋅

π − − … − − … π = + + − ……

∫2

0

( 1)( 3) 1( 1)( 3) 1using sin cos

( )( 2) 2p q p p q q

x x dxp q p q

Note: Area in case of asteriod 2 2 23 3 3x y a+ = which is a particular case of hypocycloid, when a and b are equal

becomes π 238a .


Example 13 For any real t, t t t te e e ex y

− −+ −= =,2 2

is a point

of the x2 – y2 = 1. Show that the area bounded by thisparabola and the lines joining its centre to the pointscorresponding to t' and –t' is t'.

Solution: Let P(t') and Q(– t') be two points on the hyperbolax2 – y2 = 1 (Fig. 3.19). Then the area bounded by the hyperbolaand the two lines OP and OQ is shown by the shaded portion.

The required area is the difference of the area of the ∆ OPQand area PAQBP.

Now

2 21Area of ( )2 2 4

t' t' t' t't' t'e e e eOPQ PB OB e e

− −−+ −∆ = ⋅ = ⋅ = −

0 0

Area 2 2 22 2

B t' t' t t t t

A

dx e e e ePAQBP y dx y dt dtdt

− −− −= = = ⋅∫ ∫ ∫

−

− = + − = + − − ∫2 2

2 2

0 0

1 1( 2) 22 2 2 2

t't' t tt t e ee e dt t

−= − −2 21 ( 4 ')4

t' t'e e t

∴ 2 2 2 21 1The required area ( ) ( 4 )4 4

t' t' t' t'POQAP e e e e t t'− −= − − − − =

ASSIGNMENT 2

1. Show that the area of the loop of the curve }sin 2 ,sin

x a ty a t

== is 24

3a .

2. Show that the area bounded by the cissiod = =

2

3sin ,sincos

x a tty at

and its asymptote is π 234

a .

3. Find the whole area of the curve

− = + = +

2

2

2

112

1

tx at

aty

t(It is the parametric form of the circle)

4. Find the area included between the cycloid }= −= −

( sin )(1 cos )

x a t ty a t and its base.

Fig. 3.19

Y

P t( )

BA

Q t´( )

x y + = 122

XO

OB x = = e et t' – ' + 2

PB y = = e et t' – ' – 2


5. Find the area included between the inverted cycloid }( sin ),(1 cos )

x ay a

= θ − θ= + θ and its base.

6. Prove that the whole area between the four infinite branches of the tractrix

= + =2cos log tan , sin2 2a tx a t y a t is πa2.

III. Area for Curves Given in Polar Form, r = f(θθθθθ).Area bounded by the curve r = f(θ) and radii vectors θ = α, θ = β is

β

αθ∫ 2 ,1

2r d where r = f(θ) is continuous and single valued.

Let AB be the curve r = f(θ), and OA, OB be the radii vectorsfor θ = α and θ = β respectively Fig. 3.20

Let P(r, θ) be any general point on the curve such thatQ(r + δr, θ + δθ) as its neighbouring point.

Let the area OAP (which is a function of θ) be A so that sectorialarea OPQ is δA.

Evidently area OPQ lies between the sectors OPS and ORQ, i.e.

δA lies between δθ212

r and 2

21Using Area of circular sector (radius)1 ( ) , 2

2 circular measure of the angler r

= ×+ δ δθ

∴δδθA

lies between 212

r and + δ 21 ( )2

r r

Proceeding to limits as δθ → 0 or δr → 0, we get =θ

212

dA rd

Integrating both sides from θ = α to θ = β, we get

[A]αβ = 21(value of for ) (value of for )

2A A r d

β

αθ = β − θ = α = θ∫

A = 21(sectorial area 0)2

OAB r dβ

α− = θ∫

Hence the area 212

OAB r dβ

α= θ∫

Note 1: In the above result, we have supported that r is an increasing function of θ in the interval [α, β]. Thesame formula is valid even if the radius vector r decreases as θ varies from α to β. However, the same formulais not necessarily valid if r = f(θ), takes both positive and negative values in the interval [α, β].

If there are finite number of points of Maximum and Minimum radii vectors in the interval [α, β] say at θ =θ1, θ2, … θn, then we divide the sectorial area OAB into (n + 1) sectors with the corresponding limits as follows

−

+ … + +∫ ∫ ∫ ∫1 2 n

n 1 n

2 2 2 2

1

1 1 1 1A = Area OAB r d + r d r d r d2 2 2 2

=θ θ θ β

α θ θ θθ θ θ θ

Note 2: In case of area bounded by two polar curves r = f(θ) and r = ψ(θ) and the radii vectors

θ = α and θ = β is −∫ 2 21 2

1 (r r )d2

β

αθ

where r1 is the radius vector for outer curve and r2 is the radius vector for inner curve.

Fig. 3.20

θβ

=

B

Q r r( + , + )δ θ δθ

SRP r( , )θ

6θ

θ

r A

θ α =

θ = 0 initialaxis


Example 14: Find the area of the loop of the curve r = acos2θθθθθ and hence find the total areaof the curve.

Solution: As we know that the curve r = a cos nθ or r = asinnθhave equal loops if n is odd and 2n equal loops if n is even.

In our problem, r = acos2θ, n is even means the curve has 4equal loops.

Further to find the limits of integration for a loop, wegenerally put r = 0 and find two consecutive values of θ.

Thus, there r = 0 implies cos2θ = 0 i.e. πθ = ±22

or 4πθ = ±

i.e. for the first loop of the curve θ varies from π π to 4 4 as shown

in Fig. 3.21.

Area of one loop of the curve /4 /4

2 2 2

/4 /4

11 cos 22 2

r d a dπ π

−π −π= θ = θ θ∫ ∫

/42

2 2

02 cos 2

2a a d

π= ⋅ θ θ∫

2

0

As cos 2 is an even function of and for an even function

( ) 2 ( )a aa f x dx f x dx−

θ θ = ∫ ∫

Putting 2θ = t, we get

Area π π π= = =∫

/2 2 22 2

0

1cos 2 2 2 2 8dt a aa t

The total area of the curve is 4 times the area of the single loop, i.e. π×

24

8a

or π 2

2a

Example 15: Find the area outside the circle r = 2acosθθθθθ and inside the cardiod r = a(1 + cosθθθθθ).[NIT Kurukshetra, 2008]

Solution: For the circle, r = 2a cosθ, θ = ⇒ = π θ = ⇒ =

0 2

02

r a

r.

Further, the circle r = 2acosθ is symmetrical about the initial axis.2 2

2 2 2Otherwise also in cartisan coordinates, it becomes 2 cos or ( ) 2( ) ( 0) , i.e. circle with centre ( , 0) and radius .

r r ar x y axx a y a a a

⋅ = θ + = − + − =

For cardiod, r = a(1 + cosθ);

0, 2 ;

, ;2, 0

r a

r a

r

θ = = πθ = = θ = π =

θπ

= 3/4 θ π = /4

θπ

= 5

/4 θπ

π

= 7/4 = – /4

O

Fig. 3.21


Further, the cardiod also is symmetrical about the initial line.For intersection, the geometry is as shown in Fig. 3.22 (III)

Y-axis

O a aθ = 0 X-axis

A a(2 , 0)

0 ,2π

θ π = /2( , /2)a π

θ = 0θ π = O (0, )π A a(2 , 0)

A

O

θ π = /2

B

D

θ π = θ(2 , 0)a

dθ

(I) (II) (III)

Fig. 3.22

Above the initial axis, the cardiod is traced from θ = 0 to π, whereas in case of circle, θ goes

from 0 to π2

.

So the area outside the circle inside the cardiod, i.e. Area OABDO.= 2 [Area of the cardiod above X-axis – Area enclosed by circle above X-axis]

/2

2 21 2

0 0

1 122 2

r d r dπ π = θ − θ ∫ ∫

/2

2 2 2 2

0 0(1 cos) 4 cosa d a d

π π = + θ − θ θ ∫ ∫

2 /2

2 2 2

0 02cos 4 cos

2a d d

π π θ = θ − θ θ ∫ ∫

/2

2 4 2

0 04 cos cos

2a d d

π π θ= θ − θ θ ∫ ∫I II

/2 /22 4 2

0 04 cos cos In I, putting 2

a t dt d tπ π θ = − θ θ = ∫ ∫

2

2 3 1 144 2 2 2 2 2

aa ⋅ π π π = − = ⋅

/2

0

( 1)( 3) using cos

( 2) 2p p p

dp p

π − − … πθ θ = − … ∫

Example 16: Show that the area included between the cardiod r = a(1 + cosθθθθθ) and

r = a(1 – cosθθθθθ) is a π −2(3 8)2

. [KUK, 2001]


Solution: Both the curves r = a(1 + cosθ) and r = a(1 – cosθ) are symmetrical about the initial

axis and they intersects at πθ = ±2

as a(1 + cosθ) = a(1 – cosθ)

implies cos θ = 0, i.e. πθ = ±2

From the geometry it is clear that areaincluded between the two curves is 2 times thearea OPBQO above the initial axis for θ = 0 to π.Further, this area OPBQO is taken up as the

breakup of the area OPBO for πθ = 0 to2

and

area BQOB for πθ = πto2

Whence the area included between the two curves,

π π

π

= θ + θ ∫ ∫/2

2 21 2

0 /2

1 122 2

r d r d

I II (r1 = a(1 – cosθ) and r2 = a(1 + cosθ))

/2

2 2 2

0 /2(1 cos 2cos ) (1 cos 2cos )a d d

π π

π

= + θ − θ θ + + θ + θ θ ∫ ∫

/2

2

0 /2

3 cos 2 3 cos 22 cos 2 cos2 2 2 2

a d dπ π

π

θ θ = + − θ θ + + + θ θ ∫ ∫

/2

2

0 /2

3 sin 2 3 sin 22sin 2sin2 4 2 4

aπ π

π

θ θ = θ + − θ + θ + + θ

2 3 3 32 1 2 12 2 2 2 2

a π π = ⋅ − ⋅ + π − − ⋅

2

2 (3 8)3 42 2

aaπ −π = − =

Example 17: Prove that the area enclosed by one loop of the curve x3 + y3 = 3axy is threetimes the area enclosed by the curve r2 = a2cos2θθθθθ.

Solution: The curve x3 + y3 = 3axy is symmetrical about the line y = x and insect this line

y = x at points (0, 0) and

3 3,2 2

a a . For details see the Fig. 2.61 under tracing. It transforms

this curve to polar coordinates, }= θ= θ

cos ,sin

x ry r resulting in, r3(cos3θ + sin3θ) = 3ar2sinθ cosθ

or 3 33 sin cossin cos

ar θ θ = θ + θ

B

θ π = /2r a = (1 + cos )θ

θ π = C A θ = 0(2 , 0)a

r a = (1 – cos )θ

PQ

O

Fig. 3.23


Clearly the loop is bounded for r = 0, i.e. sinθcosθ = 0 or 0,2πθ = .

∴ /2 / 2 2 2

2 23 3 20 0

1 1 sin cosThe required area of the loop 92 2 (sin cos )

r d a dπ π θ θ= θ = θ

θ + θ∫ ∫

/22 2 2

3 20

9 tan sec2 (1 tan )a d

π θ θ= θ+ θ∫

Put tan3θ = t, so that 3tan2θ sec2θdθ = dt and limits are t = 0 to ∞

∞

=+∫

2

20

9 12 (1 ) 3a dt

t

∞ ∞− + − = = = − +

12 2 2

00

(1 )3 3 1 32 1 2 1 2

ta a at

For the curve r2 = a2cos2θ, one of the loop is bounded in between the radii vectors −πθ =4

to π4 , as r = 0 gives θ = ±π/4.

For details, see the Fig. 2.66 under tracing.

∴ Area of the loop /4/4 /4 2

2 2 2

0 0 0

1 sin 22 cos 22 2 2

ar d a d aππ π θ = θ = θ θ = = ∫ ∫

Whence the area of the loop of the curve x3 + y3 = 3axy is three times the area of the one ofthe loops of r2 = a2cos2θ.

Example 18: Find the ratio of the two parts into which the parabola 2a = r(1 + cosθθθθθ)divides the area of the cardiod r = 2a(1 + cosθθθθθ).

Solution: The curve r = 2a(1 + cosθ) is a standard cardiod with valuesθ = 0, r = 4a, point A(4a, 0)

πθ =2 , r = 2, point B 2 , 2a π

θ = π, r = 0, point O (0, π)

πθ = 32 , r = 2a, point C 2 ,

2a π −

Likewise, for parabola =+ θ

21 cos

ar ,

θ = 0, r = 2a, pointA (a, 0)

πθ =2 , r = 2a, k point B

π 2 ,

2a

πθ = 32 , r = 2a, k point C 2 ,

2a π −


Clearly the two curves intersects at ±πθ =2

, otherwise also,

= = + θ+ θ

2 2 (1 cos )1 cos

ar a ⇒ (1 + cosθ)2 = 1 implying cosθ = 0, i.e. 2πθ = ±

yθ π = /2

A

r a= 2 (1 + cos )θB a(2 , /2)π

E(0, )π O θ = 0θ π =

C a(2 , – /2)π

θ π = /2

y

21 cos

ar = + θ

F

D

Fig. 3.24

2 2 2

0 0

1 1The whole area of the cardiod 2 2 4 (1 cos )2 2

r d a dπ π

= θ = + θ θ∫ ∫ 2 2

0= 4 (1 2 cos cos )a d

π+ θ + θ θ∫

2 2 2

0 0

1 cos2 3 1= 4 1 2cos 4 2sin sin 2 62 2 4

a d a aππ + θ + θ + θ = θ + θ + θ = π ∫

… (1)Area of the unshaded region between the two curves = 2[Area OABO – area ODBO]

π π = θ − θ ∫ ∫

/2 /22 22 1

0 0

1 122 2

r d r d ,

where r1 = 2a(1 + cosθ), cardiod and =+ θ2

21 cos

ar , the parabola

/2/2

2 220 0

14 (1 cos )(1 cos )

a d dππ

= + θ θ − θ + θ ∫ ∫

/2 /22

20 0 2

3 cos2 14 2cos2 2

2cos2

a dπ π

θ = + θ + θ − θ

∫ ∫

/2 /2

2 2 2

00

3 sin 2 14 2sin sec sec2 4 4 2 2

a dπ π θ θ θ = + θ + − ⋅ θ ∫


/2

2 2 2

0

3 14 2 1 tan sec4 4 2 2

a dπ θ θ = π + − + θ ∫ ,

Put tan2

tθ = so that 2 1sec2 2

d dtθ ⋅ θ =

π = + − ⋅ +

∫1

2 2

0

3 14 2 2 (1 )4 4

a t dt

13

2 2

0

(9 16)3 14 24 2 3 3

ta t a π +π = + − + =

… (2)

∴ The area of the shaded region = The whole area of the cardiod – the area of theunshaded region of the cardiod

2 2 2(9 16) (9 16)63 3

a a aπ + π −= π − = … (3)

Hence the required ratio viz. the ratio of shaded region to that to unshaded region

π − = π +9 169 16 units … (4)

ASSIGNMENT 3

1. Find the whole area of the cardiod r = a(1 + cosθ)2. Find the whole area of the curve r = 3 + 2cosθ3. Find the area of one loop of the curve r = asin2θ.4. Find the area common to the circle r = a and the cardiod r = a(1 + cosθ).

5. Find the area common to the circle 2 .r a= and r = 2acosθ. [NIT Kurukshetra, 2002]

6. Show that the area between the cissiod θ=

θ

2sincos

ar and its asymptote is π 234a

.

HINT : Here as increase from 0 to , increases from 0 to and as increases2

from to , increases from to 0. Asymptote is cos i.e. 2

r

r r a x a

π θ ∞ θ π π − ∞ θ = =

7. Show that the area between the curve r = a (secθ + cosθ) and its asymptote is π 254

a .

HINT : Here we see, = 0, = 2 ; = , i.e. as increases from 0 to ,2 2

increases from 2 to and as increases from to , increases to 2 .2

Otherwise also, in cartisan form, the given curve is

r a r

r a r a

r xr ax r

±π πθ θ = ±∞ θ

π∞ θ π − ∞

= +2

2 (2 ) or x a xyx a

− = −

8. Find the area lying between the cardiod r = a(1 + cosθ) and its double tangent.


3HINT : Here and for all position P( , ), . If 2 2 2 2

then . i.e. tangent at P will be perpendicular to the initial line.3

prθ θ θ π φ = θ ψ = θ + φ = θ + = ψ = πθ =

3.3 LENGTH OF CURVES (RECTIFICATION)

The process of finding the length of an arc of a curve between the two points on it is calledrectification.

I. Length of Curves in Cartisan Coordinate System and its Intrinsic EquationLength of the arc of the curve y = f(x), between two points x = a and x = b (as two abscissae)is given by

+ ∫

2

1b

a

dydx

dx

where in y and dydx

are continuous and single valued

functions in the internal [a, b] and the integrand eitherpositive (or negative) throughout the interval.

Let AB be the curve y = f(x) between the two pointA(x = a) and B(x = b) as the two abscissae points and CAand DB their respective ordinates.

Let P(x, y) be any point on the curve with MP asperpendicular on x-axis. If s denotes the length of the arcAP measured from a fixed point A to the variable point P,then s clearly is a function of ‘x’ and

= +

2

1dyds

dx dx

[ ]2

(value of s for ) (value of s for ) (Arc ) 1bb

aa

dyS x b x a AB O dx

dx = = − = = − = + ∫

Hence the arc length 21AB 1

b

ay dx= +∫

Observations: As 2

1dyds

dx dx = ± + , but we have assumed that s increases with increase in x (taking

+ sign of the radical). Negative sign means, s decreases with increase in x, throughout.

If dsdx or the integrand changes sign at some intermediate value x = c, then we divide the interval a to b into

two parts one from a to c and the another c to b and, value of dsdx is taken positive(or negative) throughout in

each sub interval accordingly.

Fig. 3.25

By

P x y( , )

sA

y

O C x a( = ) M D x b( = )

x


If the equation of the curve is off the form x = F(y), then the length of the arc between points

y = c and y = d is + ∫

2

1d

cdy

dydx

by taking x as an explicit function of y(means x = F(y) =

containing only terms of y in right hand side).

Intrinsic Equation: A relation between the variable s (the arc) and ψ(the tangent at variablepoint P(x, y) makes with the X-axis) is called intrinsic equation of a curve. For practical

purposes, eliminate x between equation of = + ∫2

1 dxdy

Sdx

and ψ =tandydx

.

Example 19: Show that the whole length of the curve x2(a2 – x2) = 8a2y2 is 2aπ .[KUK, 2001]

Solution: The curve x2(a2 – x2) = 8a2y2 is symmetrical about both the axis and 1

2 2y x= ± are

the tangents at the origin. Further, it intersects x-axis at (a, 0); (–a, 0) and,

( )( )2 2

xy a x a xa

= ± + − . Means the curve wholly lies between x = –a and x = a, and the

origin is a node.

Here 2 2

2 2

1 22 2

dy a xdx a a x

−= −

so that − + = + −

2 2 2 2

2 2 2

( 2 )1 18 ( )

dy a xdx a a x

4 2 2 4

2 2 29 12 4

8 ( )a a x x

a a x− +=

−

2 2

2 2

(3 2 )12 2 ( )

a xa a x

−=−

2 2 2

2 2

2( )12 2 ( )

a x aa a x

− += −

2

2 22 2

1 22 2

aa xa a x

= − + − and hence the length of the curve,

2 2

2 22 20 0 0

24 1 2a a a

dxdy as dx a x dxdx a a x

= + = − + − ∫ ∫ ∫

2 2 2

1 2 1

0

2 2 sin sin2 2

ax a x a x xa

a a a− −

−= + +

2 1 22 22 sin 1 2 22

a a aa a

− π= = ⋅ = π

Y

B(– , 0)a( , 0)aA X

O

Fig. 3.26


y

x( , 0)a

x = 0 x a =

Fig. 3.27

y = x3

y = –x3

Example 20: Find the parmeter of the loop 3ay2 = x2(a – x).

Solution: 3ay2 = x2(a – x), on differentiation, implies 26 2 3dy

ay ax xdx

= − or 2

12 3

6ax xy

ay−=

so that 2 2 2 2 22

221

36 (2 3 )2 31 1 6 (6 )a y x a xax xy ay ay

+ −− + = + =

⋅ − + −=

−

2 2 2

2

12 ( ) (2 3 )12 ( )

a x a x x a xa x a x

− + − −=

−

2 2 2 2

2[12 12 4 12 9 ]

12 ( )x a ax a ax x

a x a x

2 2 2

2[16 24 9 ]

12 ( )x a ax x

a x a x− +=

−

2 2 2

2

(4 3 ) (4 3 )12 ( ) 12 ( )x a x a x

a x a x a a x− −= =

− −

So the desired perimeter of the loop, s 21

0 0

(4 3 )2 1 22 3 ( )

a adxa xy dx

a a x−= + =

−∫ ∫ 0

3( )22 3

adxa x a

a a x− +=

−∫

Let (a – x) = t so that – dx = dt and the limits are }0, ;, 0

x t ax a t

= == =

s 1 12 2

0 0

3 22 32 3 2 3

a adtt a dt t at

a t a

− += × = + ∫ ∫

3 13 1 32 22 2 2

3 12 2

0

1 1 1 43 2 2 43 3 3 3

a

t t aa a a a aa a a

= ⋅ + = + ⋅ = ⋅ =

Example 21: Determine the length of the curve x

xeye

−= +1log1

from x = 1 to x = 2.

Solution: The given equation of the curve

−= = − − ++

1log log( 1) log( 1)1

xx x

xey e ee

∴ = − =− + −2

21 1 1

x x x

x x x

dy e e edx e e e

Required length of the curve, 22

1S 1

dydx

dx = + ∫


Y

θ π = /2

A

B

O θ = 0

Fig. 3.28

2 2 222 2

2 2 21 1

( 1) 441

( 1) ( 1)

x xx

x xdx

e eedx

e e

− += + =

− −∫ ∫

2 2 22 2

2 21 1

( 1) ( 1)( 1) ( 1)

x x

x xdx

e e dxe e

+ += =− −∫ ∫

2 2

11

( ) log( ) ,( )

x xx x

x xe e dx e ee e

−−

−+= = − −∫ ( )

using log ( )( )

f' xdx f x

f x

= ∫

−

− −−

−= − − − =−

2 22 2 1

1log( ) log( ) log e ee e e e

e e

− = + = + 1 1log( ) loge e e

e

II. Arc Length of Curves in Parametric Form:If the parametric form of a curve is given by x = φ(t), y = ψ(t), a ≤ t ≤ b where φ(t) and ψ(t) arecontinuously differentiable function on [a, b] then the arc length of the curve is given by

22

2 2( ) (bb

a a

dydxs ') dt dtdt dt

= φ + ψ = +′ ∫ ∫Note: For intrinsic equation of the curves, eliminate ‘t’ between 2( ') ( ')2s x y dt= +∫ and ψ = =tan

dy y'dx x'

.

Example 22: Rectify the ellipse x = acos θθθθθ and y = b sin θθθθθ, 22

2 2or 1 .

yxa b

+ =

Solution: The ellipse x = a cosθ, y = b sinθ, 22

2 2i.e. 1

yxa b

+ = is symmetrical about both the

axes.On differentiating with respect to ‘θ’,

= − θθ = θ

θ

sin

cos

dx addy

bd

∴ The perimeter of the ellipse

π = × = + θ θ∫

22/2

04 Arc AB 4

dydxd d

/2 2 2 2 2

04 sin cos a b d

π= θ + θ θ∫

/2 2 2 2 2 2

04 sin (1 )cos a a e d

π= θ + − θ θ∫

(Since for an ellipse b2 = a2(1 – e2), where e is the ecentricity of the ellipse)


(– , 0)a

B

Y(0, )b

A a( , 0)X

C

D b(0, – )

Fig. 3.29

/2 2 2 2 2

04 (sin cos ) cosa e d

π= θ + θ − θ θ∫

/2 12 2 2

04 (1 cos )a e d

π= − θ θ∫

π −

= + − θ + − θ

∫/2

2 2 2 2 2

0

1 1 11 2 24 1 ( cos ) ( cos )2 2

a e e

2 2 3

1 1 11 22 2 2 ( cos )

3e d

− − + − θ + …… θ

2( 1)By Binomial theorem, (1 ) 12

n n nx nx x −+ = + + + ……

/2 2 2

4 4 6 6

0

1 · 31 cos 14 1 cos cos2 2 2 · 4 2 · 4 · 6

ea e e dπ θ= − − θ − θ… θ

∫/2

0

( 1)( 3)using cos /2, for even

( 2)p p p

d pp p

π − − …θ θ = π − … ∫

2 4 63 · 1 1 · 3 5 · 3 · 11 1 14

2 2 2 2 2 · 4 4 · 2 2 2 · 4 · 6 6 · 4 · 2 2a e e e

π π π π= − − − − ……

2 22 4 62 1 · 3 1 · 3 · 512 1

2 2 · 4 3 2 · 4 · 6 5e ea e

= π − − − − ……

Example 23: Find the perimeter (full length) of the hypocycloid x = acos3θθθθθ, y = bsin3θ, θ, θ, θ, θ, i.e.

yx

a b + =

2233

1 [KUK, 2001]

Solution: The given equation of the curve is

+ =

2233

1yx

a bi.e. = −

2 23 3

1y xb a

if x > a,

23y

b is negative, i.e.

2

2

yb

or y2 is negative.

Thus, the curve does not lie beyond x = ±a.Similarly, the curve does not lie beyond y = ±b.Hence the shape of the curve is as shownThereore, whole length of the curve = 4 × the length of the arc in Ist quadrant = 4 × Arc ABClearly from arc AB, x varies from 0 to a.


On differentiating the given curve with respect to x, we get

−− + =

11332 1 2 1 0

3 3y dyx

a a b b dxor

= −

13y

dy b bxdx aa

Thus,

− + = + = +

2233

2 2 2

22 23

11 1 1

xydy b b ab

xdx a a xaa

Therefore, ( )

( )

22 32

220 0 3

1the required length 4 1 4 1

a a x ady bdx dxdx a x a

− = + = + ∫ ∫

Let x = asin3 θ, so that dx = 3a sin2θcosθdθ and for 0, 0;

,2

x

x a

= θ = π = θ =

∴2/2 2

22 20

(1 sin )Required length 4 1 · 3 sin cossin

b a da

π − θ= + θ θ θθ∫

/2 2 2 2 2

012 sin cos sin cosa b d

π= θ + θ θ θ θ∫

Again taking a2sin2θ + b2cos2θ = t2, we get

{a2 · 2sinθcosθ + b2 · 2cosθ(– sinθ)}dθ = 2t dt

i.e. 2(a2 – b2) · sinθcosθdθ = 2t dt

i.e. 2 2sin cos

( )td dt

a bθ θ θ =

−

Also, for

0, ;

,2

t a

t b

θ = = π θ = = Thus with above substractions, the required integral becomes,

2

2 2 2 212S 12 ·

( ) ( )

b b

aa

t dtt t dt

a b a b= =

− −∫ ∫

2 233 3

2 2 2 2

( 2 )12 4· ( ) 4( ) 3 ( ) ( )

b

a

a b abt a ba b a b a b

+ += = − =− − +

Alternatively: In parametric form x = a cos3θ, y = b sin3θ, so that


∴ 22

2 4 2 2 4 29 cos sin 9 sin cosdydx a b

d d + = θ θ + θ θ θ θ

= 9sin2θcos2θ(a2cos2θ + b2sin2θ)Thus, the required length in the first quadrant is

π = + θ θ∫

22/2

0

dydxsd d

/2

2 2 2 2

03cos sin ( cos sin )a b d

π= θ θ θ + θ θ∫

Putting a2cos2θ + b2sin2θ = t2 so that 2(b2 – a2) cosθ sinθdθ = 2tdt

and for

0, ;

,2

t a

t b

θ = = π θ = =

∴ 2 23

b

a

tdts tb a

=−∫

22 2

3 b

at dt

b a=

− ∫

2 23 3 3

2 2 2 2( )( )3 3 ( )( )

b

a

b a b ab at b ab a b ab a b a

− + +−= = = − +− −

2 2( )( )

a ab ba b

+ += +

Hence the full length of the curve + +=+

2 2( )4 ·( )

a ab ba b

Example 24: Find the length of the tractrix cos log tan , sin2tx a t y a t = + = from the

fixed point (0, a) on the curve. Also find the intrinsic equation of the curve.

Solution: From the given equation,

= − + = − + =

22sec1 1 cos2sin sin

2 sin sintan2

tdx a ta t a ttdt t t

and = cosdy

a tdt

∴ 2cos tancossin

dydy a tdt tdx a tdx

dt t

= = = … (1)

At the point (0, a), π=2

t and /2/2

(tan ) .tt

dyt

dx =π=π

= = ∞ Therefore, y-axis is the tangent to the curve at the point (0, a).


Whence

π ψ = = = −

1tan cot tan2

t tdydx

and, therefore

.

2tπ ψ = − … (2)

Now 22 2 4

2 22/2 /2

cosS cossin

t tdydx a tdt a t dtdt dt tπ π

= + = + ∫ ∫ 2

/2cos cot 1

ta t t dt

π= +∫

/2/2cot log(sin )

t ta t dt a t ππ

= = ∫ = a log(sint) … (3)

Therefore, the intrinsic equation,

S log sin log(cos )2

a aπ = − ψ = ψ … (4)

(using (2))Example 25: Find length of one full arc of the cycloid x = a(t + sint), y = a(1 – cost) andshow that the intrinsic equation of the cycloid is s2 + ρ2 = 16a2.

Solution: The equation of cycloid }= += −

( sin )(1 cos )

x a t ty a t implies

= + =

(1 cos ),

sin

dx a tdtdy

a tdt

Step: I Measuring S from the point where t = 0

22

2 2 20 00

S (1 cos ) sin 2(1 cos )t t tdydx dt a t a t dt a t dt

dt dt = + = + + = + ∫ ∫∫

2

0 0

0

sin22 · 2cos 2 cos 2 4 sin12 2

2

t

t tt

t t ta dt a dt a aa

= = = =

∫ ∫ … (1)

Here for full branch, θ varies from – π to 0 and 0 to π, for two halves.

∴ Length of one full branch is π= × =2 4 sin 82

a a

Step II 2

2sin cossin 2 2tan tan(1 cos ) 22cos

2

dy t tdy a t tdt

tdxdx a tdt

ψ = = = = =+

implying ψ =2t

… (2)

Step III Eliminating t by putting 2t

as Ψ

From (1), = = ψ4 sin 4 sin2ts a a or 4 cosds a

dρ = = ψ

ψ …(3)


θ π =

B

θ π = /2

P

θ π = /3O θ = 0x

A

Fig. 3.30

∴ s2 + ρ2 = (4asin Ψ)2 + (4a cos Ψ)2 = 16a2

III Length of Polar CurvesThe arc length of a curve r = f(θ) between the points θ = α and θ = β is given by

2

2 2 21

drr d r r dd

β β

α α

+ θ = + θ θ∫ ∫where θ

drd is a continuous and single valued in [α, β].

When the equation is of the form θ = f(r), the arc length between two radii vectors r1 andr2 is given by

2 2

1 1

22

1

1 1r r

r r

d rr dr drdr r

θ + = + ∫ ∫where

θddr is continuous and single valued in [α, β].

Note: For intrinsic equation of the curve, eliminate θ and φ between r = f(θ), 2 21s r r d

β

α= + θ∫ and φ =

1

tan rr

(where ψ = θ + φ)

Example 26: Find the whole length of the cardiodr = a(1 + cos θθθθθ). Also show that the upper half is bisected by

πθ =3

. [NIT Kurukshetra, 2004, 2005]

Solution: The cardiod r = a(1 + cosθ) is symmetrical about theinitial axis and for the upper half, θ = goes from 0 to π. Here

sindr ad

= − θθ

∴ Length of the whole arc 2

2 2 2 2 2

0 02 2 (1 cos ) sindrr d a a d

d

π π = + θ = + θ + θ θ θ∫ ∫

0 02 2(1 cos ) 2 2 cos

2a d a d

π π θ= + θ θ = θ∫ ∫

( )0

sin24 8 1 0 81

2

a a aπ

θ = = − =

∫Thus, the length of the upper half of the curve is = 4a.

Here length of the arc AP(for θ varying from 0 to π/3) /3

02(1 cos ) 2 ,a d a

π= + θ θ =∫

is half the length of the upper half of the cardiod.

Example 27: For the curve r = aeθθθθθ · cotααααα, prove that sr

= constant , s being measured from theorigin.


Solution: For given curve r = a eθ · cotα, we get · cot cot · cotdr ae rd

θ α= α = αθ

So the length of the curve r = f(θ) = a eθ · cotα between two points is given by

2

2 2 2 2

00

cot cosec ( cosec )drs r d r r d r d rd

θβ θ

α

= + θ = + α θ = α θ = θ α θ∫ ∫ ∫ (s being measured from the origin)

or · cosec ,sr

= θ α a constant number (say λ)

Hence the proof.

Example 28: Find the length of the arc of the equiangular spiral r = aeθθθθθ · cotααααα between thetwo points for which radii vectors are r1 and r2.

Solution: Given r = a eθ · cotα so that · cot · cot cotdr ae rd

θ α= α = αθ

Here length of the curve between two points for which radii vectors are r1 and r2 is givenas:

2 2

1 1

2 2

1 1cot

r r

r r

d rs r dr drdr rθ = + = + α∫ ∫

2

1

21 tanr

rdr= + α∫

2

1

secr

rdr= α ∫

2 1( )secr r= − α

Example 29: Find the length of the arc of the hyperbolic spiral (Reciprocal Spiral) rθθθθθ = afrom the point r = a to r = 2a.

Solution: Rewrite the given equation as θ = ar , so that

θ −=2

d adr r

The required length from r = a to r = 2a is given by

22 2 22 2 2

24

1 1 ·a a a

a a a

d a a rS r dr r dr drdr r rθ + = + = + = ∫ ∫ ∫

Put a2 + r2 = t2 so that 2rdr = 2tdt and for r = a, 2t a= ; r = 2a, 5t a=

Implying 5 52 2

2 2 2 22 21

a a

a a

t aS dt dtt a t a

= = + − −∫ ∫

52

2

1· log2

a

a

t at aa t a

− = + +


( ) 5 25 2 log log2 5 2a a a a aa

a a a a − −= − + − + +

( ) − += − + × + − 5 1 2 15 2 log

2 5 1 2 1aa

( ) ( )( ) +− = − +

− +

2

2

2 15 15 2 log ·2 2 15 1

aa

( ) ( )( )

2 15 2 log

5 1a a

+= − +

+

Example 30: Find the whole length of lemniscate r2 = a2cos2θθθθθ.

Solution: The curve r2 = a2cos2θ is symmetrical about the initial line and makes two loops intotal. One loop is enclosed between radius vectors θ = –π/4 to π/4, below and above theinitial line.

∴ Full length /4 /42 2 2 2

1 1/4 02 4r r d r r d

π π

−π= + θ = + θ∫ ∫

Here r2 = a2cos2θ ⇒ 2rr1 = a2(– sin2θ) × 2 or − θ=2

1sin 2arr

/4 4 2

220

sin 24 ar dr

π θ= + θ∫

/4 /44 2 4 2

20 0

cos 2 sin 2 14 4cos2 cos2

a a a da

π πθ + θ= = θθ θ∫ ∫

Take 2θ = t so that 2dθ = dt and for

= θ = π π= θ =

0, 0;

,4 2

t

t

Implying, length /2 /2

0 0

1 14 2cos 2 cos

dtl a a dtt t

π π= =∫ ∫

Again, take cos t = cos2φ so that – sint dt = – 2cosφ sin φ dφ, (Limits remains unchanged)

∴ π φ φ

= × φφ∫

/2

0

2cos sin12cos sin

l a dt

/2

120 2

sin4(sin )

a dt

π φ= φ∫

/2 /2

1 12 40 02 2

sin sin4 4(1 cos ) (1 cos )

a d a dt

π πφ φ= φ = φ− − φ∫ ∫

/2

1 12 20 2 2

sin4(1 cos ) (1 cos )

a dπ φ= φ

− φ + φ∫


12 2/2 /2

120 02

sin4 2 2 12(2 sin )

da a d

−π πφ φ= = − φ − φ∫ ∫

Now,

12 2 2 31

2sin 1 1 3 1 3 51 (1 ) 1 · · · ,

2 2 2 2 2! 2 2 2 3!x xx x

−− φ− = − = + + + + … where φ=

2sin2

x

322 2 21 · 3 · 5sin sin sin1 1 31 ·2 2 2 4 2 2 · 4 · 6 2

r φ φ φ= + + + + ……

∴ 2 32/2

4 6

0

1 · 3 1 · 3 · 5sin1 1 1 12 2 1 sin sin2 2 2 2 · 4 2 2 · 4 ·6 2

l a dπ φ = + + φ + φ + … φ ∫

/2

0

( 1)( 3)using sin for even

( 2) 2p p p

d pp p

π − − … πφ θ = − … ∫

2 2 31 · 3 3 · 1 1 · 3 · 5 5 · 3 · 11 1 1 12 2

2 2 2 2 2 · 4 4 · 2 2 2 2 · 4 ·6 6 · 4 · 2 2 2a

π π π π = + + + + …

= π + + + + …

2 22 2 31 · 3 1 · 3 · 51 1 1 12 12 2 2 · 4 2 2 · 4 ·6 2

a

Example 31: Show that the whole length of the lemicon r = a + bcos θθθθθ (a > b) is equal tothat of an ellipse whose semi-axis are equal to length to the maximum and minimum radiivectors of the lemicon.

Further, prove that the perimeter of the lemicon r = a + b cosθθθθθ, if ba be small, is

approximately, baa

π + 2

212 14

.

Solution: The equation of the lemicon is r = a + bcosθ (a > b).

(i) The curve is symmetrical about the initial line.

(ii) Further r = 0 implies −θ = >cos 1ab

numerically

(as given a > b) which is impossible, since always|cosθ| < 1. Whence, r is never zero and the curvedoes not pass through the pole, though it goeson decreasing from r = (a + b) to r = (a – b) uptoθ = π.

(iii) Some of the broad values of r for the values of(a) are as:

B

θ π = /2

A

O

θ = 0C

θ π =

Fig. 3.31


θ : θ π3

π2

π23 π

r : a + b +2ba a −

2ba a – b

Thus, clearly for upper half of the curve, θ varies from 0 to π.

Whence the whole length of the lemicon π = + θ θ∫

22

02 drr d

d

l 2 2 2 2 20 0

2 ( cos ) sin 2 2 cosa b b d a b ab dπ π= + θ + θ θ = + + θ θ∫ ∫ … (1)

Now, for the lemicon maximum value of r = (a + b) and minimum value of r = (a – b).Therefore, the parametric equation of the ellipse whose semi-axes vectors of the maximum and minimumradii vectors of the lemicon r = a + b cos θ are

( )cos( )sin

x a b ty a b t

= + = −

so that

( )sin

and ( )cos

dx a b tdt

dya b t

dt

= − + = −

… (2)

∴ 22

2 2 2 2( ) sin ( ) cosdydx a b t a b t

dt dt + = + + −

= a2(sin2t + cos2t) + b2(sin2t + cos2t) – 2ab(cos2t – sin2t)

= a2 + b2 – 2ab cos2t … (3)Further, the ellipse is symmetrical about both the axes, and in the first quadrant, ‘t’ varies

for 0 to 2π

.

∴ Whole length of the ellipse, l 22/2

04

dydxdt dt

π = + ∫

/2

2 2

04 2 cos2a b ab t dt

π= + −∫

Take 2t = π – θ so that 2dt = – dθ and when π= θ = π = θ =0, ; , 02

t t … (4)

⇒ l π

π=− θ = + − π − θ + + θ θ ∫ ∫

02 2 2 2

02 2 cos( ) 2 cos

2da b ab a b ab d … (5)

which is clearly equal to the whole length of the lemicon a + bcosθ, (a > b) already obtainedabove in (1).

Further from (1),

12 2

Lemicon 20l 2 1 2 cos ,b ba d

a a

π = + θ + θ ∫ (Putting into binomial format)


22 2

2 20

1 11 2 22 1 2 cos 2 cos2 2!

b b b ba da a a a

π − = + θ + + θ + + … θ

∫

2 2

22 20

1 42 1 cos cos2 8

b b ba da a a

π = + θ + − θ + … θ ∫

2

220

2 1 cos (1 cos )2

b ba da a

π = + θ + − θ θ ∫3

3as is small, neglecting and its higher powersb b

a a

2

220

2 1 cos sin2

b ba da a

π = + θ + θ θ ∫

2

20

1 cos22 1 cos2 2

b ba da a

π − θ= + θ + θ ∫

2 2

2 202 1 cos cos 2

4 4b b ba da a a

π = + θ + − θ θ ∫

2 2

2 20

sin 22 sin4 4 2

b b baa a a

π θ= θ + θ + θ −

2

22 1 approximately

4ba

= π +

Example 32: Find the intrinsic equation of the cardiod r = a(1 – cosθθθθθ).

Solution: Measuring s from the pole where θ = 0,

Step 1: ∴ 2

2 2 2 2 200

(1 cos ) sindrs r d a a dd

θ θ = + θ = − θ + θ θ θ ∫∫ 2 2 2

0(1 cos 2cos sin )a d

θ= + θ − θ + θ θ∫

2

0 02 (1 cos ) 2 sin

2a d a d

θ θ θ= − θ θ = θ∫ ∫

0

cos22 1

2

a

θθ − =


4 cos 12

a θ = − −

28 sin4

a θ= … (1)

Step 2: (1 cos )tan tansin 2

adrrd a

− θ θφ = = =θ θ

imply2θφ = … (2)

Step 3: We know,3

2 2θ θψ = θ + φ = θ + = or 2

3ψθ = or 4 6

ψθ = … (3)

Step 4: Eliminating θ from (3) and (1), we get 28 sin6

s a ψ=

which is the required intrinsic equation of the curve.

π

θ π θ π = = = = =

222

0

: For finding length of cardiod, we know for upper half, goes from 0 to

sin 1i.e. from (i), 8 8 · sin 8 4 implying 82 4 4 2r a a a a s a

Note

Example 33: Find the length of the arc of the parabola ar

= + θ2 1 cos from its vertex and

also obtain the intrinsic equation of the curve. [NIT Kurukshetra, 2007]

Solution: The equation of the given parabola may be written as

2 2

2

2 2 1sec and 2 sec · tan ·1 cos 2 2 2 22cos

2

a a drr a ad

θ θ θ= = = =θ+ θ θ …(1)

Therefore the length s of the arc from the vertex A to any point P(r, θ) is given by

2 210

s r r dθ= + θ∫

2 4 2 4 2

0= sec sec tan

2 2 2a a d

θ θ θ θ+ θ∫ 2 2

0= sec 1 tan

2 2a d

θ θ θ+ θ∫21Put tan sothat sec

2 2 2t d drθ θ= θ =

s tan

22

0= 2 1a t dt

θ

+∫

P

π ψ/2 +

A T

ψ

O(0, 0)θ

T´

φ

Fig. 3.32

( )tan

02 2

0

t 1= 2 1 log 12 2

a t t tθ

+ + + +


= tan sec log tan sec2 2 2 2

a θ θ θ θ + + … (2)

Now 2

2

1tan 2 sec cot tan12 2 2 22 sec · tan ·2 2 2

dr adr a

θ θ θ π θ φ = = = = − θ θ

⇒ 2 2π θφ = − … (3)

Since ψ is the angle made by the tangent to the curve at P(r, θ) with the tangent AT at thevertex A, as shown in the figure 3.32.

Whence 2π + ψ = θ + φ or 2

θψ = (using 3) … (4)

Therefore on using (4), (2) becomes s = a[tan ψ · secψ + log(tanψ + secψ)] … (5)

as the desired intrinsic equation of the curve.

IV. Arc Length of Curves in Pedal Form:

The arc of the curve p = f(r) between the points r = a, r = b is 2 2

,b

a

r drr p−∫ where p = r sinφ

is the length of the perpendicular from the pole on any tangent and φ is the angle betweenthe tangent and the radius vector at that point.

Note: Intrinsic equation of the curve in pedal form is obtained by eliminating r from ·( )

ds dr rrd dp f' r

= ρ = =ψ

and 2 20

r rs drr p

=−∫

Example 34: Find the length of the arc of the parabola p2 = ar for r varying a to 2a.

Solution: Length of the arc of the given parabola p2 = ar is given by

2 2 2

2 2 2 2

(2 )1 2 12 2

a a a

a a a

r a ar rs dr dr drr p r ar r ar

− += = =− − −∫ ∫ ∫

2 2

1 22 2

(2 )12 2 ( )( )

a a

a a

r a a drdrr arr ar

−= +−−

∫ ∫

2 2

1 2 22 22

1 (2 )2

2( )4 4

a a

a a

r a a drdra ar ar r ar

−= +

− − + − ∫ ∫

( ) ( )12 2

2 22 2

( )2

2 2

a a

a a

a drd r ar dra ar

= − +− −

∫ ∫


2

212 12 2( ) cosh2

2

a

a

a

a

arar ar a

−

− = − +

( )1 2 22 2

1using cosh logxdx x x aax a

− = = + − −∫

22 2

2 log2 2 2 2

a

a

a a a aa r r = + − + − −

3 23 22 log 2 log 2 log2 2 2 2

2

a aa a a aa a a a

+ = + + − = +

( )2 log 2 1a = + +

Example 35: Find the intrinsic equation of the curve whose pedal equation is p2 = r2 – a2.

Solution: From the equation, p2 = r2 – a2, we get 2 2dp

p rdr

= or pdr

dp r= … (1)

∴ ·pds drr r p

d dp r= ρ = = =

ψ … (2)

Step 1: Let s be measured from the point where r = a (since for r = 0, p2 = –a2 are imaginaryentity)

Step 2: ∴ 22

2 22 2

1 ( )2 2 2

rr r

a a a

pr r rs dr dr r aa a a ar p

= = = = − =−∫ ∫ … (3)

Step 3: From (2) and (3), eliminating ‘p’, we get

2

2

dsd

sa

ψ

= or 2ds asd

= ψ

or 2ds a ds

= ψ

On integration, 2 2 ,s a= ψ + α where a is a contant of integration. … (4)

Step 4: Let ψ = 0 for s = 0. Therefore from (4), α = 0

Thus, 2 2s a= ψ or 2as = ψ … (5)

which is the desired intrinsic equation of the given curve.


ASSIGNMENT 4

1. Find the length of the arc of the semi-cubical parabola ay2 = x3 from the vertex to theordinate x = 5a.

2. Find the length of the arc of the parabola y2 = 4ax(i) included between the ordinates x = 0 and x = h

(ii) cut of by the line 3y = 8x.

3. Find the parameter of the loop of the curve 3ay2 = x(x – a)2

4. Find the length of the curve y = log(sec x) from x = 0 to π/3.

5. Show that the parameter of the curve 2

211

txt

−=+

, 2

21

tyt

=+

is 2πr.

[The given curve is the circle x2 + y2 = 1] [MDU, 2001]6. Find the length of arc of the curve

x = eθ sinθ, y = eθ cosθ from θ = 0 to π/2.7. Find the length of one full arc of the cycloid

x = a(θ + sinθ), y = a(1 + cosθ)or (x = a(θ – sinθ), y = a(1 – cosθ)) [Jammu Univ, 2001]

8. Find the length of the loop of the curve x = t2, 3

3ty t= − [NIT Kurukshetra, 2008]

9. Find the length of the perimeter of the curve r = 2a cos θ.10. Find the length of spiral of archimedes, r = aθ between the points whose radii vectors

are r1 and r2.

11. In four cuped hypocycloid 2 2 23 3 3x y a+ = , show that

(i)3 cos24as = ψ , s being measured from the vertex.

(ii) s = (3a/2) sin2ψ when (a, 0) is taken as the fixed point.

At the vertex,4π θ =

12. Show that the parabola 2 (1 cos )ar

= + θ , 32

sinds ad

=ψ ψ . Hence show that the arc

intercepted between the vertex and the extremity of the latus ractum is

( )2 log 2 1a + + 13. Find the intrinsic equation of the curve whose equation is p = rsin α.

3.4 VOLUMES OF REVOLUTION

When a plane area made to revolve about a fixed straight line lying in its own plane, generatesa solid body of revolution and its boundary generates a surface of revolution. The fixed lineabout which this plane area is rotated is called the axis of revolution.


e.g.

A B

CD

AB

C

( )i

A BC

( )ii ( )iii

Fig. 3.33

(i) When a rectangle is rotated about one of its base, a right circular cylinder is generated.(ii) When a semicircle is rotated about its bounding diameter, a sphere is generated.

(iii) When a right angle triangle, is rotated about its base, a right circular cone is formedand so on there could be uncountable number of examples.

Whence a plane when revolved about a fixed straight line lying in its own plane, then thebody so generated by the plane area is termed as volume of revolution and the curved surface(i.e. the outer face) generated by the plane area so revolved in called surface of revolution.

I. Volume of Revolution for Cartesian Curves

(i) Revolution about X-axis:The volume of the solid generated by the revolution of the area bounded by the curve

y = f(x), the x-axis and the ordinates x = a, x = b is 2ba y dxπ∫

where y = f(x) is continuous finite and singlevalued function in the interval a ≤ x ≤ b.

Let AB be the curve y = f(x) and CA and BDbe the two ordinates x = a and x = b respectively.

Let P(x, y), Q(x + δx, y + δy) be twoneighbouring points on the curve and LP, NQbe their respective ordinates.

Let V denotes the volume of the solidgenerated by revolution about x-axis of the areaACLP, which is clearly a function of ‘x’. Thenthe volume of the solid generated by therevolution of the area PLNQ is δV. Completethe rectangles PN and LQ.

Clearly, δV lies between the volumes of the right circular cylinders generated bythe revolution of the rectangles PN and LQ.i.e. δV lies between πy2δx and π(y + δy)2δx as PL = y and NQ = (y + δy).

∴Vx

δδ lies between πy2 and π(y + δy)2

Since y is a continuous function of x, therefore, δy → 0 as δx → 0

Taking limits as δx → 0, 2dV ydx

= π

O(0, 0)

A

Q

C L N Dx

B

P

Fig. 3.34


PQ

B

M

AL

NN'

x a = x b = x

y

Fig. 3.35

or 2 [ ]b b

ba

aa

dVy dx dx Vdx

π = =∫ ∫ = (Vol. V when x = b) – (Vol. V when x = a)

= vol. generated by the revolution of the area ACDB – OHence the volume of the solid generated by the revolution of the area ACDB about

x-axis is ba y dxπ∫ 2

(ii) Revolution about Y-axis: The volume of the solid generated by revolution about y-axis

of the area bounded by the curve x = f(y), y-axis and abscissae y = a, y = b is 2ba x dyπ∫

where x= f(x) is continuous finite and single valuedfunction in the interval a ≤ x ≤ b.

(iii) Revolution about any axis:Let the curve AB, y = f(x), revolves about any linesay LM other than x-axis or y-axis and take one fixedpoint on this line. If P and Q, be two adjusant pointson the curve with perpendicular meeting the line LMin N and N’ respectively, then NN’ = d(LN) and thevolume of revolution of the elementary disc ofrevolution about LM is π(PN)2 d(LN) and hence the

total volume is given by 2( ) ( )PN d LNπ∫ with properlimits of integration.

Example 36: The hyperbola yxa b

− =22

2 21 revolves

about the axis of X. Show that the volume cutoff from one of the two solids thus obtained bya plane perpendicular to X-axis, and distant h

from the vertex, is b h a ha

π +2 2

2(3 )

3

Solution: Clear from the figure 3.36, volume ofthe solid cut off by a plane ⊥ to X-axis and at adistance h from the vertex is the same as volumeobtained by rotating the portion of the curve fromx = a to x = a + h (i.e. portion of the curve to theright hand side of Y-axis).

∴ 2

2 2 22

The required volume ( )b a h

a a

by dx x a dxa

+= π = π −∫ ∫

22

2 2using 1

yxa b

− =

2 3

22 3

a h

a

b x a xa

+π = −

h

x a = x a h = +

M

( + , 0)a hA ( , 0)a– xA'

y

P

xO(0, 0)

Fig. 3.36


3 32 2 2 3

22 2

( ) 3( )3 3

a h ab b ah ha a h aa a

+ − π π + = − + − =

2 2

2

(3 )3

b h a ha

π +=

Example 37: Obtain the volume of the solid of revolution of the loop of the curvey2(a – x) = x2(a + x) about Y-axis.

Solution: As shown in the geometry, the curve2

2 ( )( )

x a xya x

+=−

has one loop lying between –a ≤ x ≤ a.

The volume of the solid of revolution of the loopabout X-axis is equal to

0 20

2 ( )( )aa

x a xV y dx dxa x−−

+= π = π−∫ ∫

Put (a – x) = t so that dx = – dt,

2

2

( ) (2 )a

a

a t a tdt

t− −= −π∫

2 3 2 2 32 5 4a

a

a a t at t dtt

− + − = π ∫

2 3

2 22 5 4a

a

a a at t dtt

= π − + − ∫

23

3 2 22 log 5 23

a

a

ta t a t at = π − + −

3 3

3 2 2 3 2 2(2 )2 log 2 5 (2 ) 2 (2 ) 2 log 5 2 ( )3 3a aa a a a a a a a a a a a

= π − + − − − − −

3 22 log 23

a = π −

Example 38: Find the volume of spherical cap of height h cut of from a sphere of radius a.

Solution: Let the equation of the circle be x2 + y2 = 1 … (1)Let the plane PA’A cut the circle at a distance h from A so that AA’ = h and OA’ = a – h, sinceOA = a.

Fig. 3.37

y

xA(– , 0)a

O

x a =

x


The required volume of the spherical cap (shown shaded in fig 3.38) is the volume of thesolid generated by the revolution of the Arc AP (of the circle x2 + y2 = a2) about the X-axisbounded between the ordinates x = a – h to x = a.

O

P

P´

a h –

A' Nh A a( , 0) x

y

Fig. 3.38

∴ 3

2 2 2 2( )3

aaa

a h a h a h

xV y dx a x dx a x− − −

= π = π − = π − ∫ ∫

{ } { }2 3 31( ) ( )3

a a a h a a h = π − − − − −

{ }2 2 2 33 33

a h a h ah hπ= π − − +

3

2 2

3 3h hah h a = π − = π −

Cor. If the segment is cut off by a plane at a distance a/2 either from ‘a’ or from the centre,

then a – h becomes a/2 and V becomes 2

35 units4 6 24h aa a π − = π , whereas the volume of

the sphere of radius ‘a’ is 343

aπ (i.e. put limits x = 0 to a). Whence, the volume of the segment

is 3

3

5524 times4 32

3

a

a

π=

π that of the volume of the sphere.

Example 39: A basin is formed by revolution of the curve x3 = 64y (y > 0) about the Y-axis.If the depth of the basin is 8 inches, how many cubic metres of water will it hold?

Solution: The equation of the generating curve is x3 = 64y


This curve is symmetrical in opposite quadrants i.e. by rotating through 180°, the geometryremains the same.

The height of the basin is given 8 inches, so that y = 8Therefore, x3 = 64 × 8 i.e. x = 8.Hence, the portion OA of the curve with point A(8, 8) only

is considered for generating volume by revolution aboutY-axis.

∴ 8

2

0The required volume x dy= π∫

283

0(64 )y dy= π∫

2

8 3016 y dy= π ∫

853

0

16 53

y = π

5348 · (8) 0

5

= π −

5

3 348 48 1536(2 ) 36 cubic inches5 5 5

= π × = π × = π

Example 40: Show that the volume of the solid generated by the revolution of the curve

(a – x)y2 = a2x, about its asymptote is aπ2 312

[NIT Kurukshetra, 2007]

Solution: As this problem lies under the category of revolutionabout any axis.

Let P(x, y) be any point on the curve, shown in Fig. 3.40.The desired volume is obtained by revolution of the curve

about its asymptote x = a,

2

02 ( ) ( )V PM d PN

∞= π∫ ,

where PM = OA – ON = a – x

2

02 ( )V a x dy

∞= π −∫

22

2 202 ,

aya dy

a y

∞ = π − + ∫ using given curve

( )6

22 202 a dy

a y

∞= π

+∫

Fig. 3.39

Fig. 3.40

y = 8

N P

O

y

A(8, 8)

xy = 0

8´´

( – ) = a x y a x2 2

P( , )x y

M

NO

A a( , 0)

Asymptote

x a =

(0, 0)


Put y = a tan θ so that dy = a sec2θdθ

implying /2 2

62 2 2 20

sec2( tan )

a dV aa a

π θ= π+ θ∫ /2 2 3

3 2 3

0

12 cos 2 ·2 2 2

aa d aπ π π= π θ θ = π =∫

Alternately: In 2

02 ( )PM d PN

∞π∫ , we may keep it unchanged and simply find

2

32

( ) ( )2 ( )

a x ad PN d y d dxa x

x a x

= = = − −

Thus, implying2

23

0 02

( ) · 22 ( )

a aa a xV a x dx dxxx a x

−= π − = π−

∫ ∫

Put x = a sin2θ, dx = 2a sin θcosθdθ, limits are θ = 0 to .2π

implying /2 2 3

3 2

02 cos

2aV a d

π π= π θ θ =∫Example 41: Find the volume of the solid in the form of a Torus formed by the revolutionof the circle x2 + (y – b)2 = a2 (b > a) about x-axis.

Solution: For any general point P(x, y) on the circle x2 + (y – b)2 = a2,

(y – b)2 = (a2 – x2) or 2 2y b a x− = ± − or 2 2y b a x= ± −

i.e. P takes two positions P1 and P2 above and below the origin (0, b) such that

2 21y b a x= − − and 2 2

2y b a x= + −Therefore the required volume of the Torus so formed

by revolving the shaded area (shown in the figure) aboutY-axis is given by

( )2 2 2 22 1 2 1

0 002 2

aa aV y dx y dx y y dx = π − π = π − ∫ ∫ ∫

( )2 1 2 10

2 ( )a

y y y y dx= π + −∫ ;

2 2

02 2 · 2

ab a x dx= π −∫

2 2 2

1

0

8 sin2 2

ax a x a xb

a− −= π +

2

2 28 · 22 2ab a b π = π = π

1Since only sin is non-zero at x x aa

− =

Fig. 3.41

x = 0

A(0, – )b a

( , )x y1 1

x a =

( , )x y2 2

B

O´ (0, )b

(0, + )b a

O

P2

P1


Example 42: Show that the volume of solid formed by revolving the ellipse yx

a b+ =

22

2 21

(or x = a cosθθθθθ, y = b sinθθθθθ) about the line x = 2a is 4πππππ2a2b. [NIT Kurukshetra, 2004]

Solution: For any point P(x, y) on the ellipse 22

2 21

yxa b

+ = ,

we get, 2 2 22

2 2 21

y b yxa b b

−= − = or2 2b yx

a b

± −= or 2 2ax b y

b= −

Means P(x, y) corresponds two values of x against points P1 and P2 as (Taking +ve sign)x1 = 2a – x and x2 = 2a + x from the line x = 2a.

Now volume of the solid formed by revolving the given ellipse about the x = 2a will betwo times the volume generated by revolving the upper half of the curve about x = 2a

C(0, – )a

P a x y2(2 + , )

D b(0, – )

A a( , 0)y = 0

P a x y1(2 – , )

B b(0, ) y b =

x a = 2

Fig. 3.42

i.e. 20

2b

V x dy= π∫

2 22 100

2b b

x dy x dy = π − π ∫ ∫

2 22 1 2 1 2 10

2 ( ) 2 ( )( )b b

ax x dy x x x x dy = π − = π + − ∫ ∫

0

2 4 ·2b

a x dy= ∫ as x2 + x1 = 4a and x2 – x1 = 2x

2

2 2 2 20 00

1616 16 ·b b ba aa xdy a b y dy b y dy

b bπ= π = π − = −∫ ∫ ∫

2 22 2

1

0

16 sin2 2

by b y ya b

b b−

−π= +

2 2

2 216 · 42 2

a b a bbπ π = = π

(As only 2nd term is non-zero at limit x = b)


Example 43: Find the volume (generated) of the frustrum of a right circular cone whoselower base has radius R, upper base is radius r and altitude is h. [NIT Kurukshetra, 2003]

Solution: From the geometry of the frustrumit is clear that if the cone AB is revolved aboutY-axis (any axis PN) it will generate volumegiving shape of the frustrum. So the equation ofAB can be written with the help of two point formwith coordinates of A(R, 0), B(r, h) as

( ) ( )1

2 11 1

2

( )( )y y

y y x xx x

−− = −−

or 00 ( )hy x R

r R−− = −−

so that( )r Rx y R

h−= +

Now volume of the frustrum is the volumeobtained by revolving AB about Y-axis, i.e.

2 2

0 0( ) ( )

h hV PN d ON x dy= π =∫ ∫

2

2

00,

h hr R y R dy ay R dyh

− = π + = π + ∫ ∫ when − = (const.)r R ah

3

3 3

0

( ) 1· ( ) (0 )3 3

hay R

ah R Ra a

+ π= π = + − +

3

3· ·3

h r R h R Rr R h

π − = + − −

3 3 2 2( )

3( ) 3h hr R r R rR

r Rπ π= − = + + −

Alternately: NP = NN‘ + N’P = OA + N’P = R + AN’tanθ tan r RR y R yh−= + θ = +

'Since in , tan or tan tan ,'

where in the , tan or tan

N PAN'P N'P AN' yAN

N'B r RABN'N''A h

∆ = θ = θ = θ −∆ = θ = θ

∴ 2 2

0 0( ) ( ) ( tan )

hh

yV x PN d ON y R dy

== π = = π θ +∫ ∫

3

0

( tan ) 13 tan

hy R θ + = π θ

Fig. 3.43

O

Qy

OO' h= O'B r =

B r h( , )O´ N´´( , )r h B´

A´ R( , 0)

A R( , 0)

P x y( , )

x-axis

y-axis

h

NN


3 3

0 03 tan 3tan

h h

r R r Ry R h Rh h

π − π − = + = + θ θ

[ ] ( )3 3 2 2

3( ) 3h hR r r R rh

r Rπ π= − = + +−

Note: In case of the revolution about X-axis, limits will be R to r and the frustrum will be along horizontal line.

Example 44: A quadrant of a circle of radius a, revolves about its chord. Show that the

volume of the spindle generated is aπ − π 2(10 3 )6 2

Solution: The equation of the generating circle ofradius ‘a’ be x2 + y2 = a2

From the geometry, if A(a, 0) and B(0, a) be theextremities of the arc AB (in the positive quadrant),then the equation of the line AB, using two pointform, is given by

2 1

1 12 1

( )y y

y y x xx x

−− = −−

or 00 ( )

0ay x a

a−− = −− or x + y – a = 0

Let P(x, y) be any point on the arc AB. Draw PN ⊥ on the Chord AB and form AP.PN is perpendicular distance of P(x, y) from the chord AB or the line x + y – a = 0

∴ PN 2 2( ) ( )

2 2 2x y a x a y x a a x+ − − + − + −= = =

implying ( ) ( )

22 2 2 2 2

2 2 2( ) 2( )

( )2

x a a x x a a xPN a x a a x

− + − + − −= = − − −

And ( )( )2

22 2 2 2 ( )( 0)

2x y a

AN AP PN x a y+ − = − = − + − −

2 2 2 21 2( ) 2 ( ) 2 ( )2

x a y x a y y x a= − + − − − − −

2 21 ( ) 2 ( )2

x a y y x a= − + − −

21 ( )2

x a y= − −

implying ( )2 21 1( ) ( )2 2

AN x a y x a a x= − − = − − −

∴ 2 2

2 2 2 2

1( ) 12 2

x a x xd AN dx dxa x a x

− += + = − −

O

B a(0, )

y

A a( , 0)x

P x y( , )

N

Fig. 3.44


As for the arc AB, x varies from x = 0 to x = a, therefore the required volume,

( ) 2 22 2 2

2 20 0( ) ( ) ( )

2

aa a x xV PN d AN a x a a x dxa x

− += π = π − − −−∫ ∫

Put x = a sinθ, so that dx = a cosθ dθ and limits are θ = 0, θ = π/2.

/2

0

cos sin· (1 sin )( (1 cos )) · cos2 · cos

a aa a a da

π θ + θ= π − θ − θ θ θ θ ∫

3 /2

0(1 sin )(1 cos )(cos sin )

2a d

ππ= − θ − θ θ + θ θ∫

2 2/23

2 20

sin sin sin cos cos sin coscos sin cos cos sin2

a dπ θ − θ − θ θ + θ θ + θπ= θ − θ − θ θ − θ θ ∫

2 2/23

2 20

(sin cos ) 2sin cos (sin cos )(sin cos cos sin )2

a dπ θ + θ − θ θ − θ + θ π= θ + θ θ + θ θ ∫

/23

3 3

0

1 1(cos ) (sin ) 2sin · (sin ) 1 (sin ) (cos )2 3 3a d d d d d d

ππ = − θ + θ − θ θ − + θ − θ θ ∫

/23 2

3 3

0

sin 1 1cos sin 2 sin cos2 2 3 3a

ππ θ = − θ + θ − − θ + θ − θ

3 35 (10 3 )2 3 2 6 2a aπ π π = − = − π

II. Volume of Revolution for Parametric Curves

If x = f(t) and y = φ(t) are parametric equations of a curve, then the volume of the solidgenerated by revolving the area about X-axis is

2 dxy dtdt

π∫with proper limits of integration.

Likewise, the volume of the solid formed in revolving the same curve about Y-axis is

2 dyx dt

dtπ∫

with proper limits of integration.

Example 45: Find the volume of the real formed bythe revolution of the cycloid x = a(θθθθθ + sinθθθθθ),y = a(1 – cosθθθθθ) about the tangent at the vertex (orabout the X-axis).

Solution: Clearly in the geometry of the given cycloid,the point O (the farthest point on it) is its vertex andthe axis OX is the tangent to it at the vertex O. Fig. 3.45

x´

B( = – )θ π A( = )θ π

xLθ = 0O

θ = 0

θ


Thus, the volume of the reel formed = 2 [volume generated by the revolution of areaOQALO about the tangent of O viz about X-axis].

Or in otherwords, the volume of the reel is the volume generated by revolution of thearea formed below by one full branch BOA of the cycloid about X-axis.

∴ 2 2

0 0The required vol. 2 2

a dxy dx y dd

π θ=π

θ== π = π θ

θ∫ ∫ , where θ is the parameter

22 2 3 2 2

0 02 (1 cos ) · (1 cos ) 2 2 sin 2 cos

2 2a a d a d

π π θ θ = π − θ + θ θ = π θ ∫ ∫3 4 2

016 sin · cos

2 2a d

π θ θ= π θ∫/2

3 4 2

032 sin cosa t t dt

π= π ∫ , when 2

tθ =

3 2 33 · 1 · 1

32 ,6 · 4 · 2 2

a aπ= π = π π − … − … π = + + − …

∫/2

0

( 1) 1( 1) 1using sin cos

( )( 2) 1 2m n m n

t t dtm n m n

Example 46: Find volume generated by revolving one arch of the cycloid x = a(θθθθθ – sinθθθθθ), y = a(1 – cosθθθθθ)

(i) about its base (ii) about Y-axis. (NIT Kurukshetra, 2002)

Solution:(i) The equation of the cycloid are

x = a(θ – sinθ), y = a(1 – cosθ)See the geometry, for the first half of the cycloid in the first quadrant, θ varies from 0to π.

∴ 2 2

0 02 2 dxV y dx y d

d

π π= π = π θ

θ∫ ∫ 2 2

02 (1 cos ) · (1 cos )a a d

π= π − θ − θ θ∫

3

3 2

02 2 sin

2a d

π θ = π θ ∫Take 2

tθ = , dθ = 2dt

and for θ = 0; t = 0 for θ = π, 2

t π=

Thus /2

3 2 3

02 (2sin ) 2V a t dt

π= π ∫

/2

3 6

032 sina t dt

π= π ∫

3 2 35 · 3 · 1

32 · 56 · 4 · 2 2

a aπ= π = π

Fig. 3.46

B

2a

Aθ = 0

θπ

= /2

θ π =

O θ π = 2 xC

θπ

= /2

awaπ


(ii) For volume of revolution about Y-axis, seethe fig. 3.47 Here, we first obtain the volumeof the reel generated by the revolution ofthe cycloid about Y-axis, i.e. it is the volumeof revolution of the area ABLA about y-axis,

V 2

2

0

y a

yx dy

=

== π∫

2 2 2

0 0( sin ) · sin

dyx d a a d

d

π π= π θ = π θ − θ θ θ

θ∫ ∫ ( )3 2 2 3

0· sin 2 sin sina d

π= π θ θ − θ θ + θ θ∫

I1 I2 I3

Now integral 2 21 · sin ( cos ) 2 ( cos )I d d = θ θ θ = θ − θ − θ − θ θ ∫ ∫

2 2= · cos 2 sin 2sin · cos 2 sin 2cosd −θ θ + θ θ − θ θ = −θ θ + θ θ + θ ∫ 2

2 2 sin (1 cos 2 ) ( cos 2 )I d d d= θ θ θ = θ − θ θ = θ − θ θ θ∫ ∫ ∫ { }2 2 · sin 2sin 2 sin 2 cos 2· 1

2 2 2 2 2 4d

θ θ θ θ θ θ θ = − θ − θ = − − ∫ 3 2 2

3 sin sin sin (1 cos )sinI d d d= θ θ = θ θ θ = − θ θ θ∫ ∫ ∫ 2 2sin sin · cos sin ( sin )cosd d d= θ − θ θ θ = θ θ + − θ θ θ ∫ ∫ ∫

3coscos3

θ= − θ + , using 1( ( ))

( )( ( ))1

nn f

f' f dn

+θθ θ θ =

+∫

∴ 2

3 2 · sin 2 cos2· cos 2 · sin 2cos2 2 4

V a

θ θθ θ= π −θ θ + θ θ + θ− + +144444424444443

14444244443

3

0

coscos3

π

θ− θ +

1442443

2

3 2 1 1 1 11 12 4 3 4 3

a π = π π − − + − − + +

,

(All sine terms are zero for both the limits.)

2

3 82 3

a π = π − Now volume generated by the revolution of the area ALBCA about Y-axis,

22

2 3 2( sin ) · sindy

V' x d a dd

ππ

θ=π π= π θ = π θ − θ θ θ

θ∫ ∫

223 2 2 31 1cos 2 sin cos sin 2 cos cos

2 2 4 3a

π

π

θ θ = π −θ θ + θ θ + θ − + θ + θ + θ

B´L

B θ π =

A C

P x y( , )

θ = 0 θ π = 2

Q

Fig. 3.47


2 2

3 2 24 1 1 1 14 1 12 4 3 2 4 3

a π π = π − π + − + + − π − − + +

3 213 82 3

a = π π − Thus the desired volume of the solid generated by revolution of the cycloid about Y-axis

2

3 2 3 313 8 8 62 3 2 3

V' V a a π − = π π − − − = π

III. Volumes of Revolution for Polar Curves

The volume of the solid generated by the revolution of the area bounded by the curver = f(θ) and the radii vector θ = α, θ = β:

(i) about the initial line OX (θθθθθ = 0) 323

r dβ

α= π θ θ∫ sin

(ii) about the line 323

dβ

α

= π θ θ ∫ cosðOY è =

2

(The proof of these formulae depend on Pappus theorem).

Example 47: Find the volume of the solid generated by the revolution of the cardiodr = a(1 + cosθθθθθ) about the intial lines. [KUK, 2000]

Solution: The Cardiod r = a(1 + cosθ) is symmetrical about the initial line and for its upperhalf, θ varies from 0 to π (see fig. 3.30)

∴ 3 3 3

0 0

2 2Required volume sin (1 cos ) sin3 3

r d a dπ ππ= π θ θ = + θ θ θ∫ ∫

3

3

0

2 (1 cos ) ( sin )3a d

ππ= − + θ − θ θ∫ ,1( )

using ( ) ( )1

nn f

f f' dn

+ θθ θ θ =

+∫

43 3 3

0

(1 cos )2 8(0 16)3 4 6 3a a a

π+ θπ π π= − = − − =

Example 48: Find the volume of the solid obtained byrevolving the lemniscate r2 = a2cos2θθθθθ about the initial line.

[NIT Kurukshetra, 2009]

Solution: The curve r2 = a2cos2θ is symmetrical about the initialline and consists of two equal loops.

The required volume is therefore twice the volumegenerated by the revolution of the portion OA of the curveabout the initial line.

B

O

θ π = /4

A x

y

Fig. 3.48


Implying,/4 /4

3 2

00

2 4 32 sin ( cos 2 ) sin3 3 2

V r d a dπ ππ π= θ θ = θ θ θ∫ ∫

3/43

2 20

4 (2cos 1) sin3a d

ππ= θ − θ θ∫ , using cos2θ = (2 cos2θ – 1)

Take 2 cos secθ = φ so that 2 sin sec tand d− θ = φ φ φ

and limits,

for 0,4

, 04

πθ = φ = πθ = φ =

Implying 03 32 2/4

sec tan4 (sec 1) ·3 2aV d

π

φ φπ= φ − − φ∫ … (1)

3 /4

3

0

4 tan · (sec )3 2

a d dππ= φ φ φ∫ (Integration by parts)

Ist fun. IInd fun.

( )3 /4/43 2 20 0

4 tan sec (3 tan · sec ) · sec ,3 2

a dπππ = φ φ − φ φ φ φ ∫

( )3 /4 2 20

4 2 0 3 tan (1 tan )sec3 2

a dππ = − − φ + φ φ φ ∫

/4 /43 3 3

2 3

0 0

4 4 4tan sec tan (sec tan )3 2 2a a aV d d

π ππ π π= − φ φ φ − φ φ φ φ∫ ∫⇒

3 34

0

4 4 tan · (sec ) 33 2a aV d d V

ππ π= − φ φ φ −∫ (using (1))

Ist fun. IInd fun.

or ( )3 3 /4 4 2

0 0

4 44 tan sec (sec ) · sec3 2a aV d

ππ π π= − φ φ − φ φ φ

∫

( )13 3 /4

2 220

4 4 2 0 (sec ) sec3 2a a d

π π π= − − − φ φ φ ∫

Take tanφ = t so that sec2φdφ = dt and limits are 0 to 1.

∴ 3 3 1 2

0

4 44 2 13 2a aV t dtπ π = − − + ∫

( )1

3 3 3 22

0

1 12 log 13 2 2 2 2a a a t tV t t

π π π += − × + + + +

( ) ++ = + + + ∫2 2 2

2 2 2 2using log2 2

x x a ax a dx x x a


( )3 32 2 1 log 1 23 2 2 2

a aV − π π= + + +

( ) ( )3 3 3 1 1log 2 1 log 2 16 2 2 2 2 3a a a −π π π= + + = + −

( )( )3 2 3 log 2 1 212aπ= + −

Alternately:

If we take }cos ,sin

x ry r

= θ= θ , the given equation r2 = a2cos2θ becomes r2 = a2(cos2θ – sin2θ)

or r4 = a2(r2cos2θ – r2sin2θ) becomes (x2 + y2)2 = a2(x2 – y2) or y4 + (2x2 + a2)y2 + (x4 – a2x2) = 0which is a quadratic in y2 and on solving it for y2, we get

2 2 2 2 2 4 2 2

2 2 2 2 2 4(2 ) (2 ) 4( ) 1 (2 ) 82 2

x a x a x a xy x a a x a

− + ± + − − = = − + ± +

2

2 2 21 (2 ) 2 22 8

ax a a x

= − + ± +

(Negative sign before under root is left as with negative sign, y2 becomes negative and ybecomes imaginary)

Hence the required volume,

2

02 ·

aV y dx= π∫

22 2 2

0

12 (2 ) 2 22 2 2

a ax a a x dx = π − + + +

∫

2 22

3 2 1

0

2 8 82 2 sinh3 2 2

2 2

aa ax x

xx a x a a−

+ = π − − + +

2

2 2 2 2 1using sinh2 2x a xa x dx a x

a− + = + + ∫

⇒ 2

3 3 12 32 2 · sinh 2 23 2 2 2 16

a a aa a a − −= π − + +

⇒ ( )3 33 32 3 log 2 2 3 ,

3 2 4 2a aa a −= π − + + +

( )( )1 2sinh log 1z z z− = + +

⇒ ( )3 1 1 log 2 1 2 22 3 2 2a π= − + + +

( )3 21 1 log 2 12 3 2 2a π= − + +


⇒ ( )3 1 1log 2 12 2 3a π= + −

ASSIGNMENT 5

1. Find the volume formed by the revolution of the loop of the curve y2(a + x) = x2(3a – x),about x-axis.

2. Show that the volume of the spindle solid generated by the revolution of the asteriod

x2/3 + y2/3 = a2/3 about x-axis is 332 .105

aπ

3. Show that the volume of the solid formed by the revolution of the loop of the curve

y2(a + x) = x2(a – y) about x-axis is 3 22 log 23

a π − .

4. A sphere of radius ‘a’ is divided into two parts by a plane distant 2a

from the centre.

Show that the ratio of the volumes of the two parts is 5 : 24.5. Find the volume of the solid formed by the revolution about y-axis, of the area enclosed

by the curve xy2 = a2(a – x) and its asymptote. [NIT Kurukshetra, 2007]6. Find the volume generated by the revolution of the one full arc of the cycloid

x = a(θ + sinθ), y = a(1 + cosθ) about its base (or about x-axis).[Here for one full arc, θ varies from π to 2π].

7. Find the volume of the solid generated by the revolution of the cissoid x = 2a sin2t,3sin2

costy at

= , 2 2t−π π ≤ ≤ , about its asymptote.

8. Find the volume of the solid generated by revolution of the tractrix

21cos log tan2 2

tx a t a= + , y = a sint about its asymptote.

0 22 cosHINT : 2 ,

sindx dx a ty dtdt dt t−∞

π = ∫ 9. Find the volume of the solid generated by revolution of the cardiod r = a(1 – cosθ) about

initial axis.10. Find the volume of the solid generated by revolving the lemniscate r2 = a2cos2θ about

the line 2πθ = . HINT : 2 sin sin θ = φ

11. Show that the volume of the solid formed by the revolution of the lemicon r = a + b cosθ,

a > b about the initial axis is 2 24 ( ).3

a a bπ +


12. Find the volume generated by revolving the area bounded by the curve ,x y a+ =x = 0, y = 0 about the x-axis [NIT Kurukshetra, 2006]

3.5 SURFACES OF REVOLUTION

The area of the surface of the solid generated by revolution about x-axis of the area boundedby the curve y = f(x), the x-axis and the ordinates x = a, x = b is

ba ydsπ∫ 2

where s is the length of the arc of the curve measured from a fixed point on it to any point (x, y).Let AB be the curve y = f(x), and LA and MB be the ordinates

x = a and x = b respectively. Let P(x, y) and Q(x + δx, y + δy) bethe neighbouring points on the curve and NP and N’Q betheir respective ordinates (Fig 3.49)

Let the arc AP = s so that PQ = δs. If S denotes the curvedsurface of the solid generated by the revolution about axis ofx of the area ALNP, then S is clearly a function of x and δS thecurved surface area of the solid generated by the revolutionabout the axis of x, of the area PN N’Q.

Since δS is small, then it may be taken as axiom that thecurved surface δS may be taken between the curved surfacesobtained by revolution of the right circular cylinders whosebase radii are NP and N’Q and of the same thickness δs.

Thus δS lies between 2πyδs and 2π(y + δy)δs

∴ Ss

δδ lies between 2πy and 2π(y + δy)

Since y = f(x) is continuous, therefore, as δx → 0, also δy → 0

Taking limits as Q → P, 2dS yds

= π

∴ 2x b

x bx a

x ayds S

===

=π =∫ = (value of S when x = b) – (value of S when x = a) = curved surface of the

solid generated by the revolution of the area ALMB

Hence the required surface area of revolution 2x bx a y ds=

== π∫Note:

1. Here we have assumed that the curve does not cross the axis of revolution (i.e. X-axis) and the ordinatesare continuously increasing. However, it is also true when the ordinates are continously decreasing.

2. If we interchange x and y in the above formula, then the curved surface of the solid generated by therevolution about Y-axis of the area bounded by the curve x = f(y), Y-axis and the abscissae y = a, y = b is

2y by a xds=

= π∫3. Surface area of the solid generated by the revolution about the line LM (i.e. about any axis other than

X-axis or Y-axis) of the arc AB is 0

2x Mx OL

PN ds==

π∫ , PN is the perpendicular from any point P(x, y) of the

arc AB on LM such that x = ON and y = NP, where O is the fixed point on the line LM.

Fig. 3.49

yB

A

L N N´ MxO (0, O)

s P( )x, y

δs

θ


For Practical purposes, Surface of revolution is as follows:

(i) Cartisian Form (for the curve y = f(x))

2 dsS y dxdx

= π∫ ,2

1dyds

dx dx = +

(ii) Parametric Form (for the curve x = φ(t), y = ψ(t))

2 dsS y dtdt

= π∫ ,22 dyds dx

dt dt dt = +

(iii) Polar Form (for the curve r = f(θ))

2 dsS y dd

= π θθ∫ ,

22ds drr

d d = + θ θ

Example 49: Find the area of the surface formed by the revolution of y2 = 4ax, about the X-axis by the arc from the vertex to one end of the latus rectum.

Solution: For the given parabola y2 = 4ax, 2dy a

dx y=

and 2 2 2

24 41 1 1

4dyds a a x a

dx dx y ax x+ = + = + = + =

For the arc from the vertex, O to L, (the end of the latus rectum),x varies from o to a (Fig. 3.50).

∴ Required surface of revolution 0

2a dsy dx

dx= π∫

0

2 4 ·a x aax dx

x= π +∫

31 22

0

0

4( )( ) 4 3

2

a

a a xa a x dx a= π ++ = π∫

( )28 2 2 13

a= π −

Example 50: Find the sphere of the solid generated by the revolution of the asteroidx = a cos3t, y = a sin3t, about the Y-axis.

Solution: The asteriod x = a cos3t, y = a sin3t (or x2/3 + y2/3 = a2/3) is symmetric about both the axisand in the first quadrant, θ varies from 0 to π/2 (see details, under tracing of curves).

Here 23 cos sindx a t tdt

= − and 23 sin cosdy

a t tdt

=

Fig. 3.50

O

L

S

L

( , 0)a


so that 22

2 4 2 2 4 29 cos sin 9 sin cos 3 sin cosdyds dx a t t a t t a t t

dt dt dt = + = + =

∴ The required surface area = 2 times the surface area generated by revolution of the areaunder the curve in the first quadrant

/2 /4 /2

3 2 4

0 0 02 2 4 cos · 3 sin cos 12 cos sindsx dt a t a t t dt a t t dt

dt

π π π= π = π = π∫ ∫ ∫

/2

2 4

012 cos ( sin )a t t dt

π= − π −∫

/25

2 2

0

cos 1212 (0 1),5 5

ta aπ −= − π = π −

θ θ θ θ = +∫ ( )

using ( ) ( )1

n ff f' d

n

2125

a= π

Example 51: Prove that the surface generated by the revolution of the tractrix

2cos log tan2 2a tx a t = + , y = a sint about its asymptote is equal to the surface of a sphere

of radius ‘a’.

Solution: For the given equation 2cos log tan2 2a tx a t= + , y = a sint

We get

21 1sin · · sec · sin

2 2tan 2sin cos2 2 2

dx t aa t a a tt t tdt= − + = − +

sinsin

aa tt

= − + 2 2(1 sin ) cos

sin sint ta a

t t−= =

and cosdy

a tdt

= ,

So that 22 2 2 2 24

2 2 22 2

cos (cos sin )cos coscossin sin sin

a t t tdyds dx t ta a t adt dt dt t t t

+ = + = + = =

In the Ist quadrant, above the initial axis, θ varies from 0 to π/2 (see the details undertracing … ). Therefore, the required surface generated by revolution of the curve about x-axis,

S /2 /2 /2

2 2

00 0

cos2 2 4 sin 4 cos 4sin

ds a ty dx a t dt a tdt adx t

π π π= π = π = π = π∫ ∫ ∫

which is equal to the surface of a sphere of radius ‘a’.

Example 52: Find the surface of the solid of revolution formed by revolving the cardiodr = a(1 + cosθθθθθ) about the initial axis. [Jammu Univ, 2001]

Solution: The cardiod r = a(1 + cosθ) is symmetrical about the initial axis and for the upperhalf θ, varies from 0 to π (Fig. 3.51).


Here sindr ad

= − θθ

so that

2 2 2 2 2 21 (1 cos ) sinds r r a a

d= + = + θ + θ

θ 2 2(1 cos ) 2 cos2

a θ= + θ =

∴ Required Surface area of revolution,

0 0

2 2 sin 2 cos2

dsS y d r a dd

π π θ= π θ = π θ θθ∫ ∫

0

4 (1 cos )sin · 2 cos2

aa a a dθ= π + θ θ θ∫

2 2

04 2cos · 2sin cos · 2cos

2 2 2 2a d

π θ θ θ θ= π θ∫ 2 4

016 cos sin

2 2a d

π θ θ= π θ∫ 2 5

0

132 cos · sin · ,2 2 2

a dπ θ θ = − π − θ ∫

+ θ θ θ θ = +∫1( )

using ( ) ( )1

nn f

f f' dn

5

2 22

0

cos 32 32232 (0 1)5 5 5

a aa

πθ − π π= − π = − =

Example 53: The arc of the cardiod r = a(1 + cosθθθθθ) included between −π π≤ θ ≤2 2

is rotated

about the line πθ =2

. Find the area of the surface generated.

Solution: The given cardiod r = a(1 + cosθ) is symmetricalabout the initial axis as shown in the figure. The requiredsurface area is obtained by revolving the shaded area

OCAB about the axis 2πθ = , in other words the required

surface is two times the surface area generated byrevolution of the area under the portion AB of the curves.

i.e. /2

02 2 dsS x d

d

π= π θ

θ∫Here sindr a

d= − θ

θ so that

2

2 2 2 2 2(1 cos ) sin 2(1 cos ) 2 cos2

ds drr a a a ad d

θ = + = + θ + θ = + θ = θ θ

Fig. 3.51

Oθ = πθ = 0A

C

B

θ = /2π


∴ The required surface area of revolution,

/2 /2

0 04 cos · 2 cos · 8 (1 cos )cos cos

2 2S r a d a a d

π πθ θ= π θ θ = π + θ θ θ∫ ∫

/22 2 2

08 2cos · 1 2sin · cos

2 2 2a d

π θ θ θ = π − θ ∫

/22 2 2

016 1 sin 1 2sin cos

2 2 2a d

π θ θ θ = π − − θ ∫

/22 2 4

016 1 3sin 2sin cos

2 2 2a d

π θ θ θ = π − + θ ∫Put sin

2tθ = so that 1 cos

2 2d dtθ θ = and limits are 0 to 1 .

2

11 5 222 2 4 2 3

0 0

232 (1 3 2 ) 325tS a t t dt a t t = π − + = π − + ∫

2

2 2 21 1 2 1 20 10 2 12 993 3 3 ·2 2 2 5 4 2 20 2 20 2 5 2

aa a a − + π= π − + = π = π =

Example 54: A quadrant of a circle of radius ‘a’, revolves around its chord. Show that the

surface of spindle formed is aπ − π2

(4 )2

. [NIT Kurukshetra, 2005]

Solution: Let us take a circle of radius ‘a’ with centre as Oand point C as the mid point of its quadrantal arc AB (Fig.3.52).

Let P be any point on the arc AB, so that in parametricform, x = a cosθ, y = a sinθ.

Draw PM Perpendicular to OX (as X-axis) and PNPerpendicular to AB so that

cos4

PN LM OM OL x a π= = − = −

1cos cos cos

4 2a a a θ= θ − = θ −

and 22

2 2( sin ) ( cos )dyds dx a a a

d d d = + = − θ + θ = θ θ θ

It is a surface of revolution about any axis, therefore here the required surface area ofrevolution is the surface area generated by revolution the arc AB about its chord AB. Or it is2 times the surface area generated by revolution of the area formed by revolution of the arcAC about the line AL.

Implying, /4 /4

0 0

12 2 4 cos ·2

dsS PN d a a dd

π π = π θ = π θ − θ θ∫ ∫

O

Na P x y( , )

L MC

A

B

x

y

π4

Fig. 3.52


/42 2

0

14 sin 4 (0 0)2 2 4 2

a aπ θ π= π θ − = π − − −

2

(4 )2aπ= − π

ASSIGNMENT 6

1. Find the area of the surface generated by revolution of the arc of the catenary cosh xy cc

=

from x = 0 to x = c about X-axis.

2. Find the surface of the solid generated by the revolution of the ellipse 22

2 21

yxa b

+ = about

X-axis.3. Find the surface of the right circular cone formed by the right angled triangle about a

side which contains the right angle.4. Obtain the surface area of a sphere of radius a.5. Show that the surface area of the spherical zone contained between two parallel planes

is 2πah, where a is the radius and h the distance between the planes.6. Find the area of the surface formed by rotating about y-axis, the curve y2 = x3 from x = 0

to x = 4.7. The part of the parabola y2 = 4ax cut off from by the latus-ractum revolves about the

tangent at the vertex. Find the curved surface of the reel thus formed.8. Find the surface area of the solid generated by revolving the cycloid x = a(t – sint),

y = a(1 – cost) about the base.9. Find the surface of the reel formed by the revolution round the tangent at the vertex of

an arch of the cycloid x = a(θ + sinθ), y = a(1 – cosθ)10. Find the surface of the solid formed generated by the revolution of the cardiod

r = a(1 – cosθ) about the initial line.11. Find the surface of the solid generated by the revolution of the lemniscate r2 = a2cos2θ

about the initial line.12. Find the area of the surface generated if an arch of the cycloid x = a(θ – sin θ),

y = a(1 – cosθ) revolves about the line y = 2a.

ANSWERS

Assignment 1

1. πab 2. πa2 3. 2815

a


4.2

8aπ

5. (π – 2)a2 6. 3πa2

7. 4πa2 8.17 units6 9.

9 units8

10. 36 units 11. 2643

a

Assignment 2

3. πa2 4. 3πa2 5. 3πa2

Assignment 3

1.23

2aπ

2. 11π 3.2

8aπ

4.2

4aπ

5. a2(πππππ – 1) 8. ( ) 215 3 8

16a− π

Assignment 4

1.111227

a

2. 2(i) log ,h a hah h aa

+ ++ + (ii) 15 log 216

a +

3.43a

4. ( )log 2 3+

6. ( )22 1eπ

− 7. 8a

8. 4 3 9. 2aπ

10. 2 2 2 12 1

1 ( ) ( ) , where ( ) sinh2

rf r f r f r r r a aa a

−− = + +

12. s = a eψ · cotα

Assignment 5

1. πa3(8log2 – 3) 5.2 3

2aπ

6. 5π2a3

7. 2πa3 9. 2325

aπ 10.2 32

8aπ


Assignment 6

1.2

(2 sinh 2)2aπ + 2.

2 21

2 22 sinb a a bab

a aa b−

−π + −

3. 2 2 ,r r hπ + where r is the base radius and h is the height of the cone.

4. 4πa2 6. ( )128 1 125 101215

π +

7. ( )2 3 2 log 2 1a π − + 8. 2643

aπ 9. 2323

aπ

10. 2325

aπ 11. 2 14 12

a π − 12.3

12aπ

Documents

nitkkr.ac.innitkkr.ac.in/docs/3-Geometrical Applications of Integration.pdfnitkkr.ac.in