Funciones reales de varias variablesº

LEIF MEJLBRO

REAL FUNCTIONS OF SEVERAL VARIABLES LINE INT...

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Leif Mejlbro

Real Functions of Several VariablesExamples of Line Integrales

Calculus 2c-7


Real Functions of Several Variables Examples of Line Integrales Calculus 2c-7 2007 Leif Mejlbro & Ventus Publishing ApsISBN 978-87-7681-258-4


Calculus 2c-7

4

Contents

Contents

Preface

1 Line integrals, rectangular coordinates

2 Line integral, polar coordinates

3 Arc lengths and parametric descriptions by the arc length

4 Tangential line integrals

5

6

25

32

57

whats missing in this equation?

maeRsK inteRnationaL teChnoLogY & sCienCe PRogRamme

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Calculus 2c-7

5

Preface

Preface

In this volume I present some examples of line integrals, cf. also Calculus 2b, Functions of SeveralVariables. Since my aim also has been to demonstrate some solution strategy I have as far as possiblestructured the examples according to the following form

A Awareness, i.e. a short description of what is the problem.

D Decision, i.e. a reection over what should be done with the problem.

I Implementation, i.e. where all the calculations are made.

C Control, i.e. a test of the result.

This is an ideal form of a general procedure of solution. It can be used in any situation and it is notlinked to Mathematics alone. I learned it many years ago in the Theory of Telecommunication in asituation which did not contain Mathematics at all. The student is recommended to use it also inother disciplines.

One is used to from high school immediately to proceed to I. Implementation. However, examplesand problems at university level are often so complicated that it in general will be a good investmentalso to spend some time on the rst two points above in order to be absolutely certain of what to doin a particular case. Note that the rst three points, ADI, can always be performed.

This is unfortunately not the case with C Control, because it from now on may be dicult, if possible,to check ones solution. It is only an extra securing whenever it is possible, but we cannot include italways in our solution form above.

I shall on purpose not use the logical signs. These should in general be avoided in Calculus as ashorthand, because they are often (too often, I would say) misused. Instead of I shall either writeand, or a comma, and instead of I shall write or. The arrows and are in particularmisunderstood by the students, so they should be totally avoided. Instead, write in a plain languagewhat you mean or want to do.

It is my hope that these examples, of which many are treated in more ways to show that the solutionsprocedures are not unique, may be of some inspiration for the students who have just started theirstudies at the universities.

Finally, even if I have tried to write as careful as possible, I doubt that all errors have been removed.I hope that the reader will forgive me the unavoidable errors.

Leif Mejlbro14th October 2007


Calculus 2c-7

6

1 Line integrals, rectangular coordinates

Example 1.1 Calculate in each of the following cases the given line integral, where the curve K isgiven by the parametric description

K = {x Rk | x = r(t), t I}, k = 2 or k = 3.1) The line integral

K ds, where

r(t) = (a(1 cos t), a(t sin t)), t [0, 4].

2) The line integralK

x ds, where

r(t) = (a(1 cos t), a(t sin t)), t [0, 4].

3) The line integralK z ds, where

r(t) =(t, 3t2, 6t3

), t [0, 2].


11 + 6y

ds, where

r(t) =(t, 3t2, 6t3

), t [0, 2].

5) The line integralK (x + e

z), where

r(t) = (cos t, sin t, ln cos t), t [0,

4

].

[Cf. Example 3.3.6.]

6) The line integralK(x

2 + y2 + z2) ds, where

r(t) = (et cos t, et sin t, et), t [0, 2].


x + yz2

ds, where

r(t) =13

(et, et sin t, et

), t [0, u].



2 + y2) ds, where

r(t) =(

1 t21 + t2

,2t

1 + t2

), t R.

9) The line integralK ds, where

r(t) =(2Arcsin t, ln(1 t2), ln 1 + t

1 t)

, t [0,

12

].

Line integrals, rectangular coordinates


Calculus 2c-7

7

10) The line integralK xe

y ds, where

r(t) =(2Arcsin t, ln(1 t2), ln 1 + t

1 t)

, t [0,

12

].


11 + 3x2 + z2

ds, where

r(t) = (cos t, 2 sin t, et), t [1, 1].

A Line integrals.

D First nd r(t) in each case. Then compute the line integral.


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Calculus 2c-7

8

0

2

4

6

8

10

12

y

0.5 11.5 2

x

Figure 1: The plane curve K of Example 1.1.1 and Example 1.1.2 for a = 1.

I 1) Here,

r(t) = a(sin t, 1 cos t),so

r(t) = a

sin2 t + (1 cos t)2 = a2 2 cos t = a

4 sin2t

2= 2a

sin t2 .

Then accordingly,K

ds = 40

2asin t2

dt = 4a 20

| sinu| du = 8a 0

sinu du = 16a.

2) It follows from 1) that

r(t) = a

2(1 cos t),thus

K

x ds =

40

a(1 cos t) a

2(1 cos t) dt

= a

2a 40

|1 cos t| dt = a

2a 40

(1 cos t) dt = 4

2 a

a.

3) It follows from r(t) = (1, 6t, 18t2) that

r(t) =

1 + 36t2 + 324t4 =

(1 + 18t2)2 = 1 + 18t2,

henceK

z ds = 20

6t3(1 + 18t2) dt =[64t4 +

6 186

t6]20

=32 16 + 18 64 = 24 + 1152 = 1176.



Calculus 2c-7

9

0

0.5

1

1.5

2

0.20.4

0.60.8

11.2

1.4

0.20.4

0.6

Figure 2: The curve K for t [0, 710 ]. It is used in Example 1.1.3 and Example 1.1.4 for t [0, 2].

4) It follows from 3 above that r(t) = 1 + 18t2, soK

11 + 6y

ds = 20

1 + 18t2

1 + 18t2dt = 2.

0.350.30.250.20.150.10.05

00.1

0.20.3

0.40.5

0.60.7

0.750.8

0.850.9

0.951

Figure 3: The curve K of Example 1.1.5.

5) It follows from

r(t) =( sin t, cos t, sin t

cos t

),

that

r(t) =

sin2 t + cos2 t +sin2 tcos2 t

=

1 +

sin2 tcos2 t

=1

| cos t| ,

henceK

(x + ez) ds =

4

0

(cos t + eln cos t

) 1cos t

dt =

4

0

2 dt =

2.



Calculus 2c-7

10

1

2

3

4

5

6

7

01

23

45

6

32

1

1

Figure 4: The curve K of Example 1.1.6 and apart from e factor 1/3 of Example 1.1.7.

6) It follows from

r(t) =(et(cos t sin t), et(sin t + cos t), et) ,

that

r(t) = et

(cos t sin t)2 + (sin t + cos t)2 + 1 =

3 et,

thusK(x2 + y2 + z2) ds =

20

e2t(cos2 t + sin2 t + 1

) 3 et dt= 2

3 20

e3t dt =2

33

(e6 1) .

7) If we rst divide by

3, we get by Example 1.1.6 the more nice expression r(t) = et. Thenthe line integral becomes

Kds =

u0

et dt = eu 1.

8) We get by just computing

r(t) =(2t(1 + t2) 2t(1 t2)

(1 + t2)2,2(1 + t2) 2t 2t

(1 + t2)2

)

=( 4t

(1 + t2)2,2(1 t2)(1 + t2)2

)=

1(1 + t2)2

(2t, 1 t2),

hence

r(t) = 1(1 + t2)2

4t2 + (1 t2)2 = 1

(1 + t2)2

(1 + t2)2 =1

1 + t2.

Then nally,K(x2 + y2) ds =

+

1(1 + t2)2

{(1 t2)2 + 4t2} 2

1 + t2dt

= +

(1 + t2)2

(1 + t2)2 21 + t2

dt = [2Arctan t]+ = 2.



Calculus 2c-7

11

1

0.5

0

0.5

1

y

1 0.5 0.5 1

x

Figure 5: The curve K of Example 1.1.8, i.e. a circle except for the point (1, 0).

Alternative. The computation above was a little elaborated. However, the line integralis independent of the chosen parametric description, and K is a circle with the exception ofthe point (1, 0), which is of no importance for the integration. Therefore, we can apply thesimpler parametric description

r(t) = (cos t, sin t), t ] , [,where

r(t) = ( sin t, cos t) og r(t) =

sin2 t + cos2 t = 1.

Then the line integral becomes almost trivial,K(x2 + y2) ds =

12 dt = 2.

00.20.40.60.811.21.41.6

0.70.6

0.50.4

0.30.2

0.1

0.20.4

0.60.8

11.2

1.41.6

Figure 6: The curve K of Example 1.1.9 and Example 1.1.10.

9) Here

r(t) =(

21 t2 ,

2t1 t2 ,

11 + t

+1

1 t)

=2

1 t2(

1 t2,t, 1),



Calculus 2c-7

12

hence

r(t) = 21 t2

1 t2 + t2 + 1 = 2

2

1 t2 =

2(

11 + t

1t 1

).

The line integral isK

ds = 1

2

0

2(

11 + t

1t 1

)dt =

2[ln

1 t1 t

] 12

0

=

2 ln

(1 + 1

2

1 12

)=

2 ln

(2 + 12 1

)= 2

2 ln(

2 + 1).

10) We consider the same curve as in Example 1.1.9, so we can reuse that

r(t) = 2

21 t2 =

2(

11 + t

1t 1

), t

[0,

12

],


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Calculus 2c-7

13

and the line integral becomesK

x ey ds = 1

2

0

2Arcsin t (1 t2) 2

21 t2 dt = 4

2 1

2

0

Arcsin t dt

= 4

2

4

0

u cosu du = 4

2[u sinu + cosu]40 = 4

2{

4 1

2+

12 1

}= 4 + 4

2 = 4(

2 1).

0.5

1

1.5

2

2.5

1.51

0.5

0.51

1.50.6

0.7

0.8

0.9

1


11) Here

r(t) =( sin t, 2 cos t, et) ,

so

r(t) =

sin2 t + 4 cos2 +e2t =

1 + 3 cos2 +2 e2t.

The parametric description of the integrand restricted to the curve is1 + 3x2 + z2 =

1 + 3 cos2 t + e2t,

so the line integral becomes easy to computeK

11 + 3x2 + z2

ds = 11

11 + 3 cos2 t + e2t

1 + 3 cos2 t + e2t dt =

11

1 dt = 2.



Calculus 2c-7

14

Example 1.2 Calculate in each of the following cases the given line integral along the given planecurve K of the equation y = Y (x), x I.1) The line integral

K x

2 ds along the curve

y = Y (x) = lnx, x [1, 2

2].


11 + 4y

ds along the curve

y = Y (x) = x2, x [0, 1].3) The line integral

K y

2 ds along the curve

y = Y (x) = x, x [1, 2].


12 y2 ds along the curve

y = Y (x) = sinx, x [0, ].


12 + y2

ds along the curve

y = Y (x) = sinhx, x [0, 2].6) The line integral

K y e

x ds along the curve

y = Y (x) = ex, x [0, 1].A Line integrals along plane curves.

D Sketch if possible the den plane curve. Compute the weight function

1 + Y (x)2 and nallyreduce the line integral to an ordinary integral.

0.2

0

0.2

0.4

0.6

0.8

1

1.2

1.4

y

0.5 1 1.5 2 2.5 3

x


I 1) It follows from Y (x) =1x

that

1 + Y (x)2 =

1x

1 + x2, x [1, 2

2].



Calculus 2c-7

15

Thus we get the line integralK

x2 ds = 221

x2 1x2

1 + x2 dx =

220

1 + x2 x dx

=[12 23(1 + x2)

32

]221

=13

{932 2 32

}=

13(27 2

2) = 9 2

3

2.


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Calculus 2c-7

16

0.2

0

0.2

0.4

0.6

0.8

1

1.2

y

0.2 0.2 0.4 0.6 0.8 1 1.2

x


2) From Y (x) = 2x follows that1 + Y (x)2 =

1 + 4x2,

and thusK

11 + 4y

ds = 10

1 + 4x2

1 + 4x2dx =

10

11 + (2x)2

dx =12

20

11 + t2

dt

=12

[ln(t +

1 + t2

)]20=

12

ln

(2 +

5

1

)=

12

ln(2 +

5).

0.2

0

0.2

0.4

0.6

0.8

1

1.2

y

0.2 0.2 0.4 0.6 0.8 1 1.2

x


3) Here clearly

1 + Y (x)2 =

1 + 12 =

2, soK

y2 ds = 21

x2

2, ds =

23[x3]21=

73

2.

4) We get by dierentiation of Y (x) = sinx that Y (x) = cosx, hence the weight function is1 + Y (x)2 =

1 + cos2 x =

2 sin2 x.



Calculus 2c-7

17

0.2

0

0.2

0.4

0.6

0.8

1

1.2

y

0.5 1 1.5 2 2.5 3

x


We nally get the line integral by insertionK

12 y2 ds =

0

12 sin2 x

2 sin2 x dx =

0

dx = .

0

1

2

3

4

y

0.5 1 1.5 2

x


5) When Y (x) = sinhx, then Y (x) = coshx, and the weight becomes

1 + Y (x)2 =

1 + cosh2 x =

2 + sinh2 x.

We get nally the line integral by insertionK

12 + y2

ds = 20

12 + sinh2 x

2 + sinh2 x dx =

20

dx = 2.

6) When Y (x) = ex then also Y (x) = ex, so the weight function becomes

1 + Y (x)2 =

1 + e2x.



Calculus 2c-7

18

0

0.5

1

1.5

2

2.5

3

y

0.2 0.4 0.6 0.8 1 1.2

x


We get the line integral by insertionK

yex ds = 10

ex ex

1 + e2x dx = 10

1 + e2x e2x dx

=12

1x=0

1 + e2x d(1 + e2x) =

12 23

[(1 + e2x

) 32]10=

13

{(1 + e2)

32 1

}.


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Calculus 2c-7

19

Example 1.3 A space curve K is given by the parametric description

r(t) =(et, t

2, et

), t [0, ln 3].

Prove that the curve element ds is given by (et + et) dt, and then nd the line integralK

x3z ds.

A Curve element and line integral.

D Follow the guidelines.

0.40.50.60.70.80.91

00.2

0.40.6

0.81

1.21.4

1

1.5

2

2.5

3

Figure 14: The curve K.

I The curve is clearly of class C. Furthermore,

r(t)2 = (et)2 + (2)2 + (et) = e2t + 2 + e2t = (et + et)2 ,and we get the curve element

ds = r(t) dt = (et + et) dtwith respect to the given parametric description.

Then compute the line integral,

K

x3z ds = ln 30

e3tet(et + et

)dt =

ln 30

(e3t + et

)dt

=[13e3t + et

]ln 30

=13 33 + 3 1

3 1 = 3

2,

where we alternatively rst can apply the change of variables u = et, from which

K

x3z ds = 31

(u2 + 1) du =[13u3 + u

]31

= 9 + 3 13 1 = 32

3.



Calculus 2c-7

20


r(t) =(t + 4 cos t,

43t 3 cos t, 5 sin t

), t

[0,

2

].

Find the value of the line integralK

x ds.

A Line integral.

D First nd the curve element ds = r(t) dt.

0

1

2

3

4

5

3

2

1

01

2

2

3

4

Figure 15: The space curve K.

I From

r(t) =(1 4 sin t, 4

3+ 3 sin t, 5 cos t

),

follows that

r(t)2 = (14 sin t)2+(

43+3 sin t

)2+25 cos2 t

= 18 sin t+16 sin2 t+169

+8 sin t+9 sin2 t+25 cos2 t =259

+ 25 =259 10,

thus

ds = r(t) dt = 53

10 dt.

The line integral is

K

x ds =

2

0

(t + 4 cos t) 53

10 dt =

53

10[t2

2+ 4 sin t

]2

0

=53

10(

2

8+ 4

)

=52

10

24+

20

103

.



Calculus 2c-7

21


r(t) =(t2, e2t,

12t4)

, t [1, 2].

Find the value of the line integralK

1x + 2xz + y2

ds.

A Line integral.

D First calculate the weight function r(t).

1

2

3

4

5

6

7

8

1020

3040

50

1.52

2.53

3.54

Figure 16: The curve K. Notice the dierent scales on the axes.

I We get from

r(t) =(2t, 2e2t, 2t3

)= 2

(t, e2t, t3

)that

r(t) = 2

e4t + t2 + t6.

Then by insertion and reduction,K

1x+2xz+y2

ds = 21

1t2+2t2 12 t4+e4t

2

e4t+t2+t6 dt = 2.



Calculus 2c-7

22


r(t) =(ln t, t2,

12t4)

, t [1, 2].

Prove that the curve element ds is given by(

1t+ 2t3

)dt, and then compute the line integral

K

y ex

zds.

A Line integral.

D Follow the guidelines.

1

2

3

4

5

6

7

8

1 1.5 2 2.5 3 3.54

00.1

0.20.3

0.40.5

0.60.7


I Clearly, r(t) is of class C for t ]1, 2[. Then

r(t) =(

1t, 2t, 2t3

), t ]1, 2[,

implies that

r(t)2 = 1t2

+ 4t2 + 4t6 =(

1t+ 2t3

)2,

so

ds = r(t) dt =1t + 2t3

dt =(

1t+ 2t3

)dt for t ]1, 2[.

We get by insertion of the parametric description,

K

y ex

zds =

21

t2 t12 t

4(

1t+ 2t3

)dt = 2

21

(1t2

+ 2t2)

dt

= 2[1

t+

23t3]21

= 2(1

2+

163

+ 1 23

)=

313

.



Calculus 2c-7

23


r(t) =(ln t, t2, 2t

), t

[12,32

].

1) Find a parametric description of the tangent to K at the point r(1).

2) Prove that the curve element ds is given by(

1t+ 2t

)dt.

3) Compute the value of the line integralK

(ex +

y + 2z) ds.

A Space curve; tangent; curve element; line integral.

D Find r(t), and apply that ds = r(t) dt.

1

2

3

4

1

1

2

3

1

0.5

0.5

1

Figure 18: The curve K and its tangent at (0, 1, 2).

I 1) Since r(1) =(ln 1, 12, 2 1) = (0, 1, 2), and

r(t) =(

1t, 2t, 2

), r(1) = (1, 2, 2),

a parametric description of the tangent is given by

(x(u), y(u), z(u)) = (0, 1, 2) + u (1, 2, 2) = (u, 2u + 1, 2u + 2), u R.2) Since

r(t)2 = 1t2

+ 4t2 + 4 =(

1t+ 2t

)2,

we get for t [12,32

]that

ds = r(t) dt =1t + 2t

dt =(

1t+ 2t

)dt.



Calculus 2c-7

24

3) Then by insertion and computation we get the line integralK

(ex +

y + 2z) ds = 3

2

12

(eln t +

t2 + 2 2t

)(

1t+ 2t

)dt

= 3

2

12

(t + t + 4t) 1t(1 + 2t2) dt = 6

32

12

(1 + 2t2) dt

=[6t + 4t3

] 3212= 6 + 4

{(32

)3(

12

)3}= 6 +

12(27 1) = 19.


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Calculus 2c-7

25

2 Line integral, polar coordinates

Example 2.1 Compute in each of the following cases the given line integral along the plane curve Kwhich is given by an equation in polar coordinates.


2 + y2) ds along the curve given by

= e, [0, 4].

2) The line integralK y ds along the curve given by

= a(1 cos), [0, ].


y ds along the curve given by

= Arcsin , [0, 1].


y4a 3 ds along the curve given by

= a cos2 , [0,

2

].


1x2 + y2

ds along the curve given by

=a

cos,

[0,

4

].

6) The line integralK(

x2 + y2 1)ds along the curve given by

= ln , [1, 2].


2 + y2) ds along the curve given by

= , [1, 2].

A Line integral in polar coordinates.

D First compute the weight function

2 +

(d

d

) possibly

1 +(

d

d

)2, and then the lineintegral.

I 1) Fromd

d= e = follows that

2 +

(d

d

)2=

2 =

2 e,

and thusK(x2 + y2) ds =

40

2

2 e d =

2 40

e3 d =

23(e12 1) .

Line integrals, polar coordinates


Calculus 2c-7

26

40

30

20

10

35 30 25 20 15 10 5


0

0.2

0.4

0.6

0.8

1

1.2

2 1.5 1 0.5




Calculus 2c-7

27

0

0.2

0.4

0.6

0.8

1

0.1 0.2 0.3 0.4 0.5


2) Fromd

d= a sin, follows that

2 +(

d

d

)2= a2

{(1 cos)2 + sin2 } = a2 2(1 cos),


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Calculus 2c-7

28

0

0.1

0.2

0.3

0.2 0.4 0.6 0.8 1


henceK

y ds = 0

() sin a

2

1 cosd = a2

2 0

(1 cos) 32 sind

= a2

2[25(1 cos) 52

]0

= a2

2 25 2 52 = 16a

2

5.

3) It follows from

1 +

{

d

d

}2=

1 +{

11 2

}2=

11 2 ,

thatK

y ds =

10

sin() 1

1 2 d = 10

2 1

1 2 d

= 10

1 2 d =

[

1 2]10= 1.

Alternatively, = sin, [0,

2

]and

2 +

(d

d

)2=

sin2 + cos2 = 1,

thusK

y ds =

2

0

sin2 1 d =

2

0

sind = 1.

4) It follows from

d

d= 2a sin cos,



Calculus 2c-7

29

0

0.2

0.4

0.6

0.8

1

0.5 1 1.5 2

Figure 23: The curve K of Example 2.1.5. (In fact, a segment of the line x = a.

that

2 +(

d

d

)2= a2 cos4 + 4a2 sin2 cos2 = a2 cos2 {cos2 + 4 sin2 }= a2 cos2

{4 3 cos2 } .

Now, cos 0 for [0,

2

], so

2 +(

d

d

)2= a cos

4 3 cos2 ,

and the line integral becomesK

y4a 3 ds =

K

sin4a 3 ds =

2

0

a cos2 sin4a 3a cos2 a cos

4 3 cos2 d

= a

a

2

0

cos3 sind = aa[cos

4

4

]2

0

=a

a

4.

5) If =a

cos, then

d

d=

a sincos2

,

hence

2 +(

d

d

)2= a2

{1

cos2 +

sin2 cos4

}=

a2

cos4 ,

where2 +

(d

d

)2=

a

cos2 .

The line integral is obtained by insertion,K

1x2 + y2

ds =K

12

ds = 0

cos2 a2

acos2

d =

4a.



Calculus 2c-7

30

0

0.5

1

1.5

2

0.2 0.4 0.6 0.8 1


0.84

0.86

0.88

0.9

0.92

0.94

0.96

0.6 0.8 1 1.2 1.4 1.6


6) If = ln , thend

d= 1 1

=

1

,

hence1 +

(d

d

)2=

1 + ( 1)2.

Finally, we get the line integral by insertion,K

(x2 + y2 1

)ds =

K( 1) ds =

21

1 + ( 1)2 ( 1) d

=13

[{1 + ( 1)2} 32 ]2

1=

13(2

2 1).



Calculus 2c-7

31

7) If =1, then

d

d= 1

2, sa

1 +

(d

d

)2=

1 +12

=

1 + 2

.

We get the line integral by insertion

K(x2 + y2) ds =

21

2

1 + 2

d =

13

[{1 + 2

} 32]21=

13{5

5 2

2}.


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Calculus 2c-7

32

3 Arc lengths and parametric descriptions by the arc length

Example 3.1 Compute in each of the following cases the arc length of the plane curve K given by anequation of the form y = Y (x), x I.1) The arc length

K ds of the curve

y = Y (x) =x4 + 48

24x, x [2, 4].

2) The arc lengthK ds of the curve

y = Y (x) = a coshx

a, x [a, a].

[Cf. Example 3.4.1 and Example 4.1.8.]


y = Y (x) = lnex 1ex + 1

, x [2, 4].


y = Y (x) = x32 , x [0, 1].


y = Y (x) = x23 , x [0, 1].

A Arc lengths of plane curves.

D Sketch the plane curve. Calculate the weight function

1 + Y (x)2 and reduce the line integral ofintegrand 1 to an ordinary integral.

1.4

1.6

1.8

2

2.2

2.4

2.6

2.8

3

3.2

2 2.2 2.4 2.6 2.8 3 3.2 3.4 3.6 3.8 4

x


Arc lengths and parametric descriptions by the arc length


Calculus 2c-7

33

I 1) It follows from

Y (x) =x3

24+

2x, thus Y (x) =

x2

8 2

x2=

x4 168x2

,

that

1 + Y (x)2 =

18x2

64x4 + (x4 16)2 = 1

8x2(x4 + 16) =

x2

8+

2x2

.

We get the arc length by insertion,K

ds = 42

1 + Y (x)2 dx =

42

{x2

8+

2x2

}dx =

[x3

24 2

x

]42

=64 8

24 2

4+

22

=5624 1

2+ 1 =

73

+12

=176

.

0.2

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

y

1 0.5 0.5 1

x

Figure 27: The curve K of Example 3.1.2 for a = 1.

2) From Y (x) = sinhx

afollows that

1 + Y (x)2 =

1 + sinh2

(xa

)= cosh

h

a.

The arc length isK

ds = aa

coshx

adx = a

[sinh

x

a

]aa

= 2a sinh 1 =a

e(e2 1).

3) It follows from

Y (x) =ex

ex 1 ex

ex + 1=

2ex

e2x 1 ,

that

1 + Y (x)2 =

1e2x 1

(e2x 1)2 + 4e2x = e

2x + 1e2x 1 =

coshxsinhx

,



Calculus 2c-7

34

0.3

0.25

0.2

0.15

0.1

0.05

0

y

1 2 3 4x


so the arc length becomesK

ds = 42

coshxsinhx

dx = [ln sinhx]42 = ln(

sinh 4sinh 2

)= ln

(2 sinh 2 cosh 2

sinh 2

)= ln(2 cosh 2) = ln

(e2 + e2

)= ln(e4 + 1) 2.

4) Here, Y (x) =32

x, so

1 + Y (x)2 =

1 +

94x.

The arc length is

K

ds = 10

1 +

94x dx =

49 23

[(1 +

94x

) 32]10

=827

{(134

) 32

1}

=127{13

13 8}.

0.2

0

0.2

0.4

0.6

0.8

1

1.2

y

0.2 0.2 0.4 0.6 0.8 1 1.2

x




Calculus 2c-7

35

5) Since the arc length of y = x23 , x [0, 1], is equal to the arc length of x = y 32 , it follows from

Example 3.1.4 thatK

ds =127{13

13 9}.

Alternatively, Y (x) =23x

13 , thus

K

ds = 10

1 +

49x

23 dx =

10

x13

x

23 +

49dx =

32

10

t +

49dt

=

[(t +

49

) 32]10

=(

139

) 32

(

49

) 32

=13

1327

827

=127{13

13 8}.


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Calculus 2c-7

36

I

1

0.5

0

0.5

1

0.2 0.2 0.4 0.6 0.8 1


Example 3.2 Compute in each of the following cases the arc length of the given plane curve K by anequation in polar coordinates.

1) The arc lengthK ds of the curve given by

= a cos4

4, [0, 4].


= a(1 + cos), [0, 2].


= ln , [1, e].


= a sin3

3, [0, 3].

A Arc lengths in polar coordinates.

D First calculate the weight function

2 +

(d

d

) possibly

1 +(

d

d

)2, and then the lineintegral.

1) Since

d

d= a cos3

4 sin

4,

the weight is given by

2 +(

d

d

)2= a2 cos8

4+ a2 cos6

4 sin2

4= a2 cos6

4,



Calculus 2c-7

37

0

0.2

0.4

0.6

0.8

1

1.2

0.5 1 1.5 2


0

0.1

0.2

0.3

0.4

0.5

0.6

0.8 0.6 0.4 0.2

Figure 32: The curve K of Example 3.2.3. (Part of the curve of Example 2.1.2).

henceK

ds = 40

2 +

(d

d

)2d =

40

acos3

4

d = 4a 0

| cos3 t| dt

= 8a

2

0

cos3 t dt = 8a

2

0

(1 sin2 t) cos t dt = 8a[sin t 1

3sin3 t

]2

0

=16a3

.

2) In this case,2 +

(d

d

)2= a

(1 + cos)2 + sin2 = a

2(1 + cos) = a

4 cos2

2= 2a

cos 2

,so

Kds =

20

2acos

2

d = 4a 0

| cos t| dt = 8a

2

0

cos t dt = 8a.



Calculus 2c-7

38

1

0.8

0.6

0.4

0.2

0.2

0.6 0.4 0.2 0.2 0.4 0.6


3) From1 +

{

d

d

}2=

1 +

{ 1

}2=

2,

follows thatK

ds =1

2 d =

2 (e 1).

Alternatively, = e, [0, 1], so (cf. Example 2.1.1)2 +

(d

d

)2=

2 e,

henceK

ds = 10

2 e d =

2(e 1).

4) Hered

d= a sin2

3 cos

3, so

2 +

(d

d

)2= a

sin6

3+ sin4

3 cos2

3= a sin2

3,

thusK

ds = 30

a sin2

3d = 3a

0

sin2 t dt =3a2

0

(1 cos 2t) dt = 3a2

.



Calculus 2c-7

39

3.532.521.510.5

00.2

0.40.6

0.81

0.20.4

0.60.8

1

Figure 34: The space curve of Example 3.3.1.

Example 3.3 Below are given some space curves by their parametric descriptions x = r(t), t I.Express for each of the curves there parametric description with respect to arc length from the pointof the parametric value t0.

1) The curve r(t) = (cos t, sin t, ln cos t), from t0 = 0 in the interval I =[0,

2

[.

2) The curve r(t) =13(et cos t, et sin t, et) from t0 = 0 in the interval I = R.


3) The curve r(t) = (ln cos t, ln sin t,

2 t) from t0 =

4in the interval I =

]0,

2

[.

4) The curve r(t) = (7t + cos t, 7t cos t,2 sin t) from t0 = 2 in the interval I = R.

5) The curve r(t) = (cos(2t), sin(2t), 2 cosh t) from t0 = 0 in the interval I = R.

6) The curve r(t) = (cos t, sin t, ln cos t) from t0 = 0 in the interval I =]

2,

2

[.


A Parametric description by the arc length.

D Find s(t) = r(t) and then s = s(t) and t = (s), where we integrate from t0. Finally, insert inx = r(t) = r((s)).

I 1) From

r(t) =( sin t, cos t, sin t

cos t

), t

[0,

2

[,

follows that

s(t) = r(t) =

sin2 t + cos2 t +sin2 tcos2 t

=1

cos t,



Calculus 2c-7

40

hence

s(t) = t0

1cosu

du = t0

cosu1 sin2 u du =

t0

12

(1

1 + sinu+

11 sinu

)cosu du

=12

[ln(

1 + sinu1 sinu

)]t0

=12

ln(

1 + sin t1 sin t

).


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Calculus 2c-7

41

1

2

3

4

1

2

3

1.51

0.5

0.5


Then

1 + sin t1 sin t = e

2s, dvs. sin t =e2s 1e2s + 1

= tanh s, s 0.

Notice that it follows from t [0,

2

[that

cos t =2es

e2s + 1=

1cosh s

.

Thus

t = Arcsin(

e2s 1e2s + 1

)= Arcsin(tanh s), s 0,

and the parametric description by the arc length is

r(s) = (cos t, sin t, ln cos t) =(

2es

e2s + 1,e2s 1e2s + 1

, ln(

2es

e2s + 1

))

=(

1cosh s

, tanh s, ln cosh s)

, s 0.

2) Here

r(t) =13et (cos t sin t, cos t + sin t, 1),

so

s(t) = r(t) = 13et

(cos t sin t)2 + (cos t + sin t)2 + 1 = et,

hence

s(t) = t0

e du = et 1, and t = ln(s + 1), s > 1.



Calculus 2c-7

42

0.5

1

1.5

2

3.53

2.52

1.51

0.50

3.53

2.52

1.51

0.50


Finally, we get the parametric description by the arc length,

r(s) =13

((s + 1) cos(ln(s + 1)), (s + 1) sin(ln(s + 1)), s + 1)

=s + 1

3(cos(ln(s + 1)), sin(ln(s + 1)), 1) , s > 1.

3) From

r(t) =( sin t

cos t,cos tsin t

,

2)

, t ]0,

2

[,

follows that

s(t) = r(t) =

sin2 tcos2 t

+ 2 +cos2 tsin2 t

=sin tcos t

+cos tsin t

=1

cos t sin t,

as t ]0,

2

[. Then

s(t) = t

4

1cosu sinu

du = t

4

(sinucosu

+cosusinu

)du = ln

sin tcos t

= ln tan t,

and thus s R and tan t = es, and

cos t =+1

1 + tan2 t=

11 + e2s

, and sin t =es

1 + e2s.

The parametric description by the arc length is

r(s) =(1

2ln(1 + e2s), s 1

2ln(1 + e2s),

2Arctan(es)

), s R.

4) Here,

r(t) = (7 sin t, 7 + sin t,

2 cos t),



Calculus 2c-7

43

0.2

0.4

0.6

0.8

1

1.2

1.4

0 510

1520

5

10

15

20


2.2

2.4

2.6

2.8

3

1

0.5

0

0.5

1

0.40.2

00.2

0.40.6

0.81


so

s(t) = r(t) =

(7 sin t)2 + (7 + sin t)2 + 2 cos2 t

=

2 49 + 2 sin2 t + 2 cos2 t = 98 + 2 = 10,and thus

s(t) = t

2

10 du = 10(t

2

), sa t =

s

10+

2, s R,

and the parametric description with the arc length as parameter from the point(

72

,72

,

2)

is

r(s) =(

7s + 3510

sin( s10

),7s + 35

10+ sin

( s10

),

2 cos( s10

)),

for s R.



Calculus 2c-7

44

5) It follows from

r(t) = (2 sin 2t, 2 cos 2t, 2 sinh t),


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Calculus 2c-7

45

0.7

0.6

0.5

0.4

0.3

0.2

0

0.80.6

0.40.2

0.20.4

0.60.8

0.60.7

0.80.9

1

Figure 39: The space curve of Example 3.3.6; cf. Example 3.3.1.

that

s(t) = r(t) = 2

sin2(2t) + cos2(2t) + sinh2 t = 2 cosh t,

hence

s(t) t0

2 coshu du = 2 sinh t,

so s R and

t = Arsinh(s2

)= ln

(12

(s +

s2 + 4

)), s R.

The parametric description with the arc length as parameter is

r(s) =

(cos

(2Arsinh

(s2

)), sin

(2Arsinh

(s2

)), 2

1 +

s2

4

)

=(cos

(2Arsinh

(s2

)), sin

(2Arsinh

(s2

)),

4 + s2),

for s R.6) This is an extension of the curve of Example 3.3.1, with the same parametric description

evaluated from the same point t0 = 0. We can therefore reuse this example, since the onlychange is that s R,

r(s) =(

1cosh s

, tanh s, ln cosh s)

, for s R.



Calculus 2c-7

46

0.2

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

y

1 0.5 0.5 1

x

Figure 40: The chain curve for a = 1, cf. Example 3.4.1.

Example 3.4 Find for every one of the given plane curves below an equation of the form

(1) = (s),

where the signed arc length s is computed from a xed point P0 on the curve, while is the anglebetween the oriented tangents at P0 and at the point P on the curve given by s.

1) The chain curve given by y = a coshx

a, from P0 given by x = 0.


2) The asteroid given by

r(t) = a( cos3 t, sin3 t) , t [0,

2

],

from P0 given by t = 0.

3) The winding of a circle given by

r(t) = a(cos t + t sin t, sin t t cos t), t R+,

from P0 given by t = 0.

4) The cycloid given by

r(t) = a(t sin t, 1 cos t), t [0, 2],

from P0 given by t = .

It can by proved that (1) determines the curve uniquely with exception of its placement in the plane.Therefore, (1) is also called the natural equation of the curve.

A Natural equation.

D Find the arc length s, and then by a geometrical analysis.



Calculus 2c-7

47

1

0.5

0

0.5

1

y

1 0.5 0.5 1

x

Figure 41: The asteroid of Example 3.4.2.

1) The point P0 has the coordinates (0, a). A parametric description of the chain curve is e.g.

r(t) =(t, a cosh

t

a

),

hence

r(t) =(1, sinh

t

a

), where r(0) = (1, 0),

and thus = Arctan(sinh

t

a

).

From

s(t) = r(t) =

1 + sinh2t

a= cosh

t

a,

follows that

s(t) = t0

coshu

ady = a

[sinh

u

a

]t0= a sinh

(t

a

).

The natural equation is

= (s) = Arctan( sa

).

2) The point P0 has the coordinates (a, 0), andr(t) = a

(3 cos2 t sin t, 3 sin2 t cos t) = 3a cos t sin t(cos t, sin t).

For t 0+ we get r(0) = 0, and by considering a gure we may conclude that we have a horisontalhalf tangent. Then it follows that = t.

It follows from

s(t) = r(t) = 3a cos t sin t,



Calculus 2c-7

48

10

8

6

4

2

0

2

4

6

8

10

6 4 2 2 4 6 8 10

Figure 42: The winding of the circle of Example 3.4.3.

that

s(t) = t0

3a cosu sinu du = 3a2

t0

sin 2u du =3a4{1 cos 2t},

hence

cos 2t = 1 4s3a

,

and

= (s) = t =12Arccos

(1 4s

3a

).

3) The point P0 has the coordinates (a, 0), and

r(t) = a( sin t + sin t + t cos t, cos t cos t + t sin t) = at(cos t, sin t).It follows that = t.

As t > 0 we have

s(t) = r(t) = at,thus

s(t) = a t0

u du =a

2t2, sat =

2sa

,

and hence

= (s) =

2sa

.

4) The point P0 is described by r() = a(, 0). The curve has a vertical half tangent at P0. From

r(t) = a(1 cos t, sin t) = a(2 sin2

t

2, 2 sin

t

2cos

t

2

)= 2a sin

t

2

(sin

t

2, cos

t

2

),



Calculus 2c-7

49

0.5

0.51

1.52

2.5

y

2 2 4 6 8 10

x

Figure 43: The cycloid of Example 3.4.4.

follows that

s(t) = r(t) = 2a sin t2,


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Calculus 2c-7

50

so

s(t) = t

2a sinu

2du = 4a

[ cos u

2

]t

= 4a{ cos t

2

},

and hence

cost

2= s

4a, dvs. t = 2Arccos

( s

4a

).

Since must have the form at + b, it is easy to derive that

= t = 2Arccos( s

4a

).

Example 3.5 A plane curve K is given by the parametric description

r(t) =(a

t0

sin(u2)du , a

t0

cos(u2)du

), t R.

The signed arc length from the point (0, 0) is called s.

1. Find s, and nd the parametric description of the curve given by the arc length.

It is proved in Dierential Geometry that any plane curve has a curvature

(t) ={ez r(t)} r(t)

r(t)3 ,

where we let the plane of the curve be the (X,Y )-plane in the space.

2. Prove that is proportional to s for K.The curve under consideration has many names: the clothoid, Eulers spiral, Cornus spiral.

Remark. Clothoid means in koine, i.e. Ancient Greek: = I spin.

A Parametric description with respect to arc length, curvature.

D Find ds and then compute.

I 1) As s(t) = r(t) and r(t) = a (sin (t2) , cos (t2)), fass(t) = a

sin2 (t2) + cos2 (t2) = a,

we get

s(t) = at and t(s) =1a

s.

The parametric description with the arc length is

x = r(t) = a( t

0

sin(u2)du ,

t0

cos(u2)du

)= a

( sa

0

sin(u2)du ,

sa

0

cos(u2)du

).



Calculus 2c-7

51

1

0.5

0.5

1

0.8 0.6 0.4 0.2 0.2 0.4 0.6 0.8

Figure 44: The clothoid for a = 1 and s [4, 4].

2) From

r(t) = a(sin(t2), cos

(t2)) a (sin (t2) , cos (t2) , 0) ,

and

r(t) = 2ta(cos

(t2), sin (t2)) 2ta (cos (t2) , sin (t2) , 0) ,

follows that

{ez r(t)} r(t) =

2ta cos(t2) 2ta sin (t2) 0

0 0 1

a sin(t2)

a cos(t2)

0

=

2ta cos

(t2) 2ta sin (t2)

a sin(t2)

a cos(t2)

= 2ta2.

As r(t) = a, we nally get

={ez r(t)} r(t)

r(t)3 = 2ta2

a3= 2 t

a= 2s

a2.


r(t) =(

12t2 ln t, 2 sin t, 2 cos t

), t [1, 2].

Prove that r(t) = t + 1t, and nd the length of K.

A Arc length.

D Compute r(t) og = 10r(t) dt.



Calculus 2c-7

52

0.8

0.6

0.4

0.2

0

0.2

0.4

0.6

0.8

1

1.81.9

20.6

0.81

1.2


I It follows from

r(t) =(t 1

t, 2 cos t,2 sin t

), t [1, 2],

that

r(t)2 =(t 1

t

)2+ 4 cos2 t + 4 sin2 t = t2 2 + 1

t2=(t +

1t

)2,

thus

r(t) =t + 1t

= t + 1t ,and accordingly,

= 21

r(t) dt = 21

(1 +

1t

)dt =

[t2

2+ ln t

]21

=32

+ ln 2.

Example 3.7 A space curve K is given by the parametric descriptionr(t) =

(e3t, e3t,

18 t

), t [1, 1].

Prove that r(t) = 3 (e3t + 33t), and nd nd the arc length of K.A Arc length.

D Find r(t).

I We get by dierentiation

r(t) =(3 e3t,3 e3t, 3

2)

= 3(e3t,e3t,

2),

thus

r(t) = 3

(e3t)2 + (e3t)2 + 2 = 3

(e3t + e3t)2 = 3(e3t + e3t

),



Calculus 2c-7

53

42024

5

10

15

20

5

10

15

20


and we get the arc length

(K) = 11r(t) dt =

11

3(e3t + e3t

)dt

= 2 10

3 2 cosh 3t dt = 4[sinh 3t]10 = 4 sinh 3.


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Calculus 2c-7

54


r(t) =(

13t3 t, 1

3t3 + t, t2

), t [1, 1].

Find the arc length of K.

A Arc length.

D Find r(t).

0

0.5

1

1

0.5

0.5

1

0.60.4

0.2

0.20.4

0.6


I It follows from

r(t) = (t2 1, t2 + 1, 2t),

that

r(t)2 = (t21)2+(t2+1)2+4t2 = 2t4+2+ 4t2 = 2(t2 + 1)2,

hence

(K) = 11r(t) dt = 2

10

2(t2 + 1) dt = 2

2(

13

+ 1)

=8

23

.


r(t) =(6t2, 4

2 t3, 3t4

), t [1, 1].

Explain why the curve is symmetric with respect to the (X,Z)-plane. Then nd the arc length of K.

A Arc length.

D Replace t by t. Then nd r(t).



Calculus 2c-7

55

0

1

2

34

2

2

4

12

34

56


I It follows from r(t) = (6t2,42 t3, 3t4), that the curve is symmetric with respect to the (X,Z)-plane.

From

r(t) =(12t, 12

2 t2, 12t3

)= 12t

(1,

2 t, t2)

follows that

r(t) = 12|t|

1 + 2t2 + t4 = 12|t| (1 + t2).Finally, when we exploit the symmetry above and put u = t2, we nd the arc length

(K) = 2 10

r(t) dt = 2 10

12t(1 + t2) dt = 12 10

(1 + u) du = 12(1 +

12

)= 18.

Example 3.10 A space curve K is given by the parametric descriptionr(t) = (t + sin t,

2 cos t, t sin t), t [1, 1].

1) Find a parametric description of the tangent of K at the point corresponding tot = 0.

2) Compute the arc length of K.A Space curve.

D Follow the standard method.

I 1) As r(0) = (0,

2, 0), and

r(t) = (1 + cos t,

2 sin t, 1 cos t), r(0) = (2, 0, 0),it follows that a parametric description of the tangent of K at (0,2, 0) is given by

x(u) = (0,

2, 0) + (2u, 0, 0) = (2u,

2, 0), u R.



Calculus 2c-7

56

0.15

0.1

0.05

0.05

0.1

0.15

0.91

1.11.2

1.31.4

2

1

0

1

2

Figure 49: The curve K and its tangent at (0,2, 0).

2) The arc length of K is 11r(t) dt =

11

(1+cos t)2+2 sin2 t+(1cos t)2 dt

= 11

2 + 2 cos2 t + 2 sin2 t dt =

11

4 dt = 2 2 = 4.


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Calculus 2c-7

57

4 Tangential line integrals

Example 4.1 Calculate in each of the following cases the tangential line integralKV(x) dx

of the vector eld V along the den plane curve K. This curve will either be given by a parametricdescription or by an equation. First sketch the curve.

1) The vector eld V(x, y) = (x2+y2, x2y2) along the curve K given by y = 1|1x| for x [0, 2].2) The vector eld V(x, y) = (x2 2xy, y2 2xy) along the curve K given by y = x2 for x [1, 1].3) The vector eld V(x, y) = (2a y, x) along the curve K given by r(t) = a(t sin t, 1 cos t) for

t [0, 2].

4) The vector eld V(x, y) =(

x + yx2 + y2

,y x

x2 + y2

)along the curve K given by x2 + y2 = a2 and run

through in the positive orientation of the plane.

5) The vector eld V(x, y) = (x2 y2,(x + y)) along the curve K given by r(t) = (a cos t, b sin t)for t

[0,

2

].

6) The vector eld V(x, y) = (x2 y2,(x + y)) along the curve K given by r(t) = (a(1 t), b t) fort [0, 1].

7) The vector eld V(x, y) = (y3, x3) along the curve K given by r(t) = (1 + cos t, sin t) for t [2, ].

8) The vector eld V(x, y) =(y2, a2 sinh x

a

)along the curve K given by y = a cosh x

afor x [a, 2a].

A Tangential line integrals.

D First sketch the curve. Then compute the tangential line integral.

0

0.2

0.4

0.6

0.8

1

y

0.5 1 1.5 2

x


Tangential line integrals


Calculus 2c-7

58

I 1) Here the parametric description of the curve can also be written

y ={

x for x [0, 1],2 x for x [1, 2].


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Calculus 2c-7

59

0

0.2

0.4

0.6

0.8

1

1 0.8 0.6 0.4 0.2 0.2 0.4 0.6 0.8 1

x


This gives the following computation of the tangential line integralKV(x) dx =

K

{(x2 + y2

)dx +

(x2 y2) dy}

= 10

{(x2 + x2

)dx +

(x2 x2) dx}

+ 21

{(x2 + (2 x)2

)dx +

(x2 (2 x)2

)(dx)

}

= 10

2x2 dx + 21

2 (2 x)2 dx = 23[x3]10+

23[(x 2)3]2

1

=23

+23

=43.

2) HereKV(x) dx =

K

{(x2 2xy) dx + (y2 2xy) dy}

= 11

{(x2 2x3) dx + (x4 2x3) 2x dx}

= 11

(x2 2x3 + 2x5 4x4) dx

= 11

(x2 4x4) dx + 0 = 2 [1

3x3 4

5x5]10

= 2(

13 4

5

)=

215

(5 12) = 1415

.



Calculus 2c-7

60

0

0.5

1

1.5

2

1 2 3 4 5 6


1

0.5

0.5

1

1 0.5 0.5 1


3) Similarly we getKV(x) dx =

K{(2a y) dx + x dy}

= 20

{(2aa(1cos t))a(1cos t)+a(tsin t)a sin t}dt

= a2 20

{(1+cos t)(1cos t)+(tsin t) sin t}dt

= a2 20

{1cos2 t+t sin tsin2 t}dt = a2 20

t sin t dt

= a2[t cos t + sin t]20 = 2a2.

4) We split the curve K into two pieces, K = K1 +K2, where K1 lies in the upper half plane, andK2 lies in the lower half plane, i.e. y > 0 inside K1, and y < 0 inside K2. Then we get the



Calculus 2c-7

61

tangential line integralKV(x) dx =

K

(x + y

x2 + y2dx +

y xx2 + y2

dy

)

=K

1x2 + y2

12d(x2 + y2

)+K

1x2 + y2

(y dx x dy)

=K

12d ln

(x2 + y2

)+K1

1

1 +(

x

y

)2(

1y

dx + x( 1

y2

)dy

)

+K2

1

1 +(

x

y

)2(

1y

dx + x( 1

y2

)dy

)

= 0 +K1

1

1 +(

x

y

)2 d(

x

y

)+K2

1

1 +(

x

y

)2 d(

x

y

)

=K1

dArctan(

x

y

)+K2

dArctan(

x

y

)

= [Arctan t]+ + [Arctan t]+ = = 2.


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Calculus 2c-7

62

0

0.5

1

1.5

2

0.2 0.4 0.6 0.8 1

Figure 54: The curve K of Example 4.1.5 for a = 1 and b = 2.

Alternatively we get by using the parametric description

(x, y) = a (cos t, sin t), t [0, 2],

thatKV(x) dx =

K

(x + y

x2 + y2dx +

y xx2 + y2

dy

)

= 20

a2

a2{(cos t+sin t)(sin t)+(sin tcos t) cos t}dt

= 20

{cos t sin tsin2 t+cos t sin tcos2 t}dt

= 20

dt = 2.

5) HereKV(x) dx =

K{(x2 y2)dx (x + y)dy}

=

2

0

{(a2 cos2 tb2 sin2 t)(a sin t)(a cos t+b sin t)b cos t}dt

=

2

0

{a[(a2+b2) cos2 tb2] sin tab cos2 tb sin t cos t}dt

=[+a(a2+b2)

13cos3 tab2 cos t ab

2(t+

12sin 2t) 1

2b2 sin2 t

]2

0

= ab2 2 b

2

2 a(a

2 + b2)3

+ ab2 =a

3(2b2 a2) b

4(2b + a).



Calculus 2c-7

63

0

0.5

1

1.5

2

0.2 0.4 0.6 0.8 1

Figure 55: The curve K of Example 4.1.6 for a = 1 and b = 2.

6) HereVV(x) dx =

K{(x2 y2)dx (x + y)dy}

= 10

{[a2(1 t)2 b2t2](a) [a at + bt] b}dt

= 10

{a3(t 1)2 + ab2t2 + b(a b)t ab}dt

=[a

3

3(t 1)3 + ab

2

3t3 +

12b(a b)t2 abt

]10

=ab2

3+

12(a b)b ab a

3

3

=a

3(b2 a2) b

2(a + b).

Remark. The vector eld V(x) is identical to that in Example 4.1.5 and in Example 4.1.6.Furthermore, the curves of these two examples have the same initial point and end point.Nevertheless the two tangential line integrals give dierent results. We shall later be interestedin those vector elds V(x), for which the tangential line integral only depends on the initial andend points of the curve K. (In Physics such vector elds correspond to the so-called conservativeelds.) We have here an example in which this ideal property is not satised.



Calculus 2c-7

64

0

0.2

0.4

0.6

0.8

1

0.2 0.4 0.6 0.8 1


7) We getKV(x) dx =

K{y3dx + x3dy}

=

2

{ sin3 t ( sin t) + (1 + cos t)3 cos t}dt

=

2

{sin4 t + cos t + 3 cos2 t + 3 cos3 t + cos4 t}dt

=

2

{sin4 t + cos4 t +

(2 cos2 t sin2 t 1

2sin2 2t

)+ cos t +

32

+32

cos 2t + 3 cos3 y}

dt

=

2

{(sin2 t + cos2 t)2 1

4+

14

cos 4t + cos t +32

cos 2t + 3 cos t 3 sin2 t cos t}

dt

=[t t

4+

116

sin 4t + sin t +32t +

34

sin 2t + 3 sin t sin3 t]

2

=(1 1

4+

32

)

2 4 + 1 = 9

8 3.

8) We getKV(x) dx =

K

{y2 dx + a2 sinh x

ady}

= 2a

a

{a2 cosh2 x

adx + a2 sinh

x

a sinh x

adx}

= a2 2a

a

{cosh2

(xa

) sinh2

(xa

)}dx = a3.



Calculus 2c-7

65

0

1

2

3

4

y

0.5 1 1.5 2

x


0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

0.2 0.4 0.6 0.8 1

x

Figure 58: The curves y =

3x, y =

3x and y =

3x2.

Example 4.2 Compute the tangential line integral of the vector eld

V(x, y) = (2xy, x6y2)

along the curve K give n by y = a xb, x [0, 1]. Then nd a such that the line integral becomesindependent of b.

A Tangential line integral.

D Just use the standard method.

I We compute the line integral

KV(x, y) dx =

K(2xy dx + x6y2 dy) =

10

{2x axb + x6a2x2b ab xb1} dx

= 10

{2a xb+1 + a3b x3b+5

}dx =

2ab + 2

+a3b

3(b + 2)=

a(a2b + 6)3(b + 2)

.

Assume that this result is independent of b. Then b+2 must be proportional to a2b+6, so a2 = 3.



Calculus 2c-7

66

According to the convention a > 0, thus a =

3. By choosing this a we getKV(x, y) dx =

3(3b + 6)3(b + 2)

=

3,

which is independent of b.


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Calculus 2c-7

67

Example 4.3 Compute in each of the following cases the tangential line integralKV(x) dx

of the vector eld V along the space curve K, which is given by the parametric description

K = {x R3 | x = r(t), t I} .1) The vector eld is V(x, y, z) = (y2 z2, 2yz,x2), and the curve K is given by r(t) = (t, t2, t3) for

t I.

2) The vector eld is V(x, y, z) =(

1x + z

, y + z ,2

x + y + z

), and the curve K is given by r(t) =

(t, t2, t3) for t [1, 2].3) The vector eld is V(x, y, z) = (3x2 6yz, 2y + 3xz, 1 4xyz2), and the curve K is given by

r(t) = (t, t2, t3) for t [0, 1].4) The vector eld is V(x, y, z) = (3x2 6yz, 2y + 3xz, 1 4xyz2), and the curve K is given by

r(t) = (t, t, t) for t [0, 1].5) The vector eld is V(x, y, z) = (3x2 6yz, 2y + 3xz, 1 4xyz2), and the curve K is given by

r(t) =

(0, 0, t), for t [0, 1],(0, t 1, 1), for t [1, 2],(t 2, 1, 1), for t [2, 3].

6) The vector eld is V(x, y, z) = (x, y, xz y), and the curve K is given by r(t) = (t, 2t, 4t) fort [0, 1].

7) The vector eld is V(x, y, z) = (2x + yz, 2y + xz, 2z + xy), and the curve K is given by

r(t) = (a(cosh t) cos t, a(cosh t) sin t, at) for t [0, 2].

8) The vector eld is V(x, y, z) = (y2 z2, 2yz,x2), and the curve K is given by r(t) = (t, t, t) fort [0, 1].

A Tangential line integrals in space.

D Insert the parametric descriptions and compute the tangential line integral. Notice that Exam-ple 4.3.7 is a gradient eld, so it is in this case possible to nd the integral directly.

I 1) We getKV(x) dx =

K

{(y2 z2)dx + 2yz dy x2dz}

= 10

{(t4 t6) + 2t2 t3 2t t3 3t2} dt

= 10

{3t6 2t4}dt = 37 2

5=

135

.



Calculus 2c-7

68

2) HereKV(x) dx =

K

{1

x + zdx + (y + z) dy +

2x + y + z

dz

}

= 21

{1

t + t3+ (t + t3) +

6t2

2t + t3

}dt

= 21

{1t t

1 + t2+

6t2 + t2

+ t + t3}

dt

=[ln t 1

2ln(1 + t2) + 3 ln(2 + t2) +

t2

2+

t4

4

]21

= ln 2 12ln 5 + 3 ln 6 1

2ln 2 3 ln 3 + 4

2+

164 1

2 1

4

=92

ln 2 +12

ln 5 +214

=214

+12

ln5125

.

3) First notice that for any curve,KV(x) dx =

K{(3x26yz)dx+(2y+3xz)dy+(14xyz2)dz}

=K

d(x3+y2+z)K

z{6ydx3xdy+4xyzdz}.(2)

Such a rearrangement can also be used advantageously to Example 4.3.3, Example 4.3.4and Example 4.3.5.When we apply (2), we get

KV(x) d(x) = [x3+y2+z](1,1,1)

(0,0,0) 10

t3{6t2 3t 2t + t6 3t2}dt

= 3 10

12 t11dt = 3 1 = 2.

Alternatively, it follows by a direct insertion thatKV(x) dx =

K{(3x2 6yz)dx + (2y + 3xz)dy + (1 4xyz2)dz}

= 10

{(3t26t2 t3)+(2t2+3t t3)2t+(14t t2 t6)3t2}dt

= 10

{3t26t5+4t3+6t5+3t212t11}dt

= 10

(6t2 + 4t3 12t11)dt = [2t3 + t4 t12]10= 2.

4) The vector eld is the same as in Example 4.3.3. Then we get by (2),KV(x) dx = [x3+y2+z](1,1,1)

(0,0,0) 10

(6t23t2+4t4)dt

= 3 10

(3t2 + 4t4)dt = 3 1 45

=65.



Calculus 2c-7

69

Alternatively, it follows by a direct insertion thatKV(x) dx) =

10

{(3t26t2)+(2t+3t2)+14t4}dt

= 10

(1 + 2t 4t4)dt = 1 + 1 45

=65.

5) The vector eld is the same as in Example 4.3.3. When we apply (2) and just check that r(t)is a continuous curve, we get

KV(x) dx = [x3+y2+z](1,1,1)

(0,0,0)K

z{6ydx3xdy+4xyzdx}

= 3 10

0 dt1

+20 dt 32

1 6 dt = 3 6 = 3.

Alternatively, it follows by direct insertion thatKV(x) dx =

K{(3x26yz)dx+(2y+3xz)dy+(14xyz2)dz}

= 10

(14 0)dt+ 21

{2(t1)+0}dt+ 32

{3(t2)26}dt

= [t]10 + 2[12(t 1)2

]21

+ 3[13(t 2)3 2t

]32

= 1 + 1 + 1 3 2 3 + 3 2 2 = 3(1 6 + 4) = 3.6) Here we get by insertion,

KV(x) dx =

K{xdx + ydy + (xz y)dz}

= 10

{t + 2t 2 + (t 4t 2t) 4}dt

= 10

(t + 4t + 16t2 8t)dt = 10

(16t2 3t)dt

=163 3

2=

32 96

=236

.

7) It follows immediately thatKV(x) dx =

K{(2x+yz)dx+(2y+xz)dy+(2zxy)dz}

=K{d(x2+y2+ z2)+(yzdx+xzdy+xydz)}

=K

d(x2+y2+z2+xyz) =[x2+y2+z2+xyz

]a(cosh 2,0,2)(x,y,z)=(a,0,0)

= a2 cosh2 2 + 4a22 a2 = a2(42 + sinh2 2).Alternatively, we get by the parametric description

r(t) = a(cosh t cos t, cosh t sin t, t), t [0, 2],that

r(t) = a(sinh t cos tcosh t sin t, sinh t sin t+cosh t cos t, 1),



Calculus 2c-7

70

thusKV(x) dx =

K{(2x + yz)dx + (2y + xz)dy + (2z + xy)dz}

= 20

(2a cosh t cos t+a2t cosh t sin t)a(sinh t cos tcosh t sin t)dt

+ 20

(2a cosh t sin t+a2 cosh t cos t)a(sinh t sin t+cosh t cos t)dt

+ 20

(2at+a2 cosh2 t cos t sin t)a dt

= a2 ( ) + a3 ( ).Then the easiest method is to reduce and use that

cos t =12(eit + eit

), sin t =

12i(eit eit) ,

and similarly for cosh t and sinh t. We nally obtain the result by a partial integration.


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Calculus 2c-7

71

The variants above are somewhat sophisticated, so we proceed here by rst calculating thecoecient of a2: 2

0

2 cosh t t(sinh t cos t cosh t sin t)dt

+ 20

2 cosh t sin t(sinh t sin t + cosh t cos t)dt + 20

2t dt

= 2 20

cosh t sinh t dt + 20

2t dt =[sinh2 t + t2

]20

= 42 + sinh2 2.

Then we nd the coecient of a3: 20

t{cosh t sinh t sin t cos tcosh2 t sin2}dt

+ 20

t{cosh t sinh t sin t cos t+cosh2 t cos2 t}dt + 20

cosh2 t cos t sin t dt

= 20

t(cosh t sinh t sin 2t+cosh2 t cos 2t)dt +12

20

cosh2 t sin 2t dt.

Notice that

d

dt

{12

cosh2 t sin 2t}

= cosh t sinh t sin 2t+cosh2 t cos 2t,

so the whole expression can then be written 20

{t

d

dt

(12

cosh2 t sin 2t)+

dt

dt(

12

cosh2 t sin 2t)}

dt

= 20

d

dy

(t

2cosh2 t sin 2t

)dt =

[t

2cosh2 t sin 2t

][20

= 0.

As a conclusion we getKV(x) dx = a2(42 + sinh2 2) + 0 a3 = a2(42 + sinh2 2).

8) Here we get [cf. also Example 4.3.1, where the vector eld is the same]KV(x) dx) =

K{(y2 z2)dx + 2yzdy x2dz}

= 10

{(t2 t2) + 2t2 t2}dt = 10

t2dt =13.



Calculus 2c-7

72

Example 4.4 Compute in each of the following cases the tangential line integral of the given vectoreld V along the given curve K.1) The vector eld is V(x, y) = (x + y, x y), and the curve K is the ellipse of centrum (0, 0) and

half axes a, b, run through in the positive orientation of the plane.

2) The vector eld is V(x, y) =(

1|x|+ |y| ,

1|x|+ |y|

), and the curve K is the square dened by its

vertices

(1, 0), (0, 1), (1, 0), (0,1),in the positive orientation of the plane.

3) The vector eld is V(x, y) = (x2 y, y2 + x), and the curve K is the line segment from (0, 1) to(1, 2).

4) The vector eld is V(x, y) = (x2 y2, y2 + x), and the curve K is the broken line from (0, 1) over(1, 1) to (1, 2).

5) The vector eld is V(x, y) = (x2 y, y2 + x), and the curve K is that part of the parabola ofequation y = 1 + x2, which has the initial point (0, 1) and the end point (1, 2).

6) The vector eld is V(x, y, z) = (yz , xz , x(y + 1)), and the curve K is the triangle given by itsvertices

(0, 0, 0), (1, 1, 1), (1, 1,1),and run through in given sequence.

7) The vector eld is V(x, y, z) = (sin y, sin z, sinx), and the curve K is the line segment from (0, 0, 0)to (, , ).

8) The vector eld is V(x, y, z) = (z , x , y), and the curve K is the quarter circle from (a, 0, 0) to(0, 0, a) followed by another quarter circle from (0, 0, a) to (0.a.0), both of centrum (0, 0, 0).

A Tangential line integrals in the 2-dimensional and the 3-dimensional space.

D Sketch in the 2-dimensional case the curve K. Then check if any part of V(x) dx can be sortedout as a total dierential. Finally, insert the parametric description and compute.

I 1) As K is a closed curve, we getKV(x) dx =

K{(x + y)dx + (x y)dy} =

K

d

(12x2 + xy 1

2y2)

= 0,

because V dx is a total dierential.

Alternatively, K has e.g. the parametric description(x, y) = r(t) = (a cos t, b sin t), t [0, 2],

thus

r(t) = (a sin t, b cos t).



Calculus 2c-7

73

1.5

1

0.5

0.5

1

1.5

y

1 0.5 0.5 1

x

Figure 59: A possible curve K in Example 4.4.1.

1.5

1

0.5

0

0.5

1

1.5

y

1.5 1 0.5 0.5 1 1.5

x

Figure 60: The curve of Example 4.4.2.

Then by insertion,KV dx =

K{(x + y)dx + (x y)dy}

= 20

{(a cos t+b sin t)(a sin t)+(a cos tb sin t)b cos t}dt

= 20

{a2 cos t sin tab sin2 t+ab cos2 tv2 sin t cos t}dt

= 20

{ab cos 2t 1

2(a2 + b2) sin 2t

}dt = 0.

2) Since |x|+ |y| = 1 on K, we haveKV dx =

K

1|x|+ |y| (dx + dy) =

K

1 d(x + y) = 0.



Calculus 2c-7

74

1

1.2

1.4

1.6

1.8

2

0 0.2 0.4 0.6 0.8 1

Figure 61: The curve K of Example 4.4.3

Alternatively, and more dicult we can use the parametric description of K given by

r(t) =

(1 t, t), t [0, 1],(1 t, 2 t), t [1, 2],(t 3, 2 t), t [2, 3],(t 3, t 4), t [3, 4],

hence

r(t) =

(1, 1), t ]0, 1[,(1,1), t ]1, 2[,(1,1), t ]2, 3[,(1, 1), t ]3, 4[.

Since |x|+ |y| = 1 on K, we getKV dx =

K(dx + dy) =

10

(1 + 1)dt + 21

(1 1)dt + 32

(1 1)dt + 43

(1 + 1)dt

= 0 2 + 0 + 2 = 0.

3) First notice thatKV dx =

K{(x2 y)dx + (y2 + x)dy} = 1

3

K

d(x3 + y3) +K(y dx + x dy)

=13(8 + 1 1) +

K(y dx + x dy),

so KV dx = 8

3+K(y dx + x dy)(3)

=K{(x2 y)dx + (y2 + x)dy}.(4)

Now we compute Example 4.4.3, Example 4.4.4 and Example 4.4.5 in the two variantscorresponding to (3) and (4), respectively.



Calculus 2c-7

75

A parametric description of K is e.g.

r(t) = (t, 1 + t), t [0, 1],

and accordingly, r(t) = (1, 1).


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Calculus 2c-7

76

0

0.5

1

1.5

2

y

1 0.5 0.5 1

x


Then by (3),

KV dx = 8

3+ 10

{(1 + t) + t}dt = 83 1 = 5

3.

Alternatively, we get by (4) thatKV dx =

10

{t2 (1 + t) + (1 + t)2 + t}dt = 10

{(1 + t)2 + t2 1}dt

=[13(1 + t)3 +

13t3 t

]10

=8 + 1 1

3 1 = 5

3.

4) It follows from (3) thatKV dx = 8

3+ 10

(1)dx + 21

1 dy =83 1 + 1 = 8

3.

Alternatively, we get by (4),KV dx =

10

(x2 1)dx + 21

(y2 + 1)dy =[13x3 x

]10

+[13y3 + y

]21

=13 1 + 8

3+ 2 1

3 1 = 8

3.

5) By (3),KV dx = 8

3+ 10

{(1 x2) + x 2x}dx

=83

+ 10

(x2 1)dx = 83

+[13x3 x

]10

=83

+13 1 = 2.



Calculus 2c-7

77

1

1.2

1.4

1.6

1.8

2

0 0.2 0.4 0.6 0.8 1

x


Alternatively, by (4),KV dx =

K{x2 x2 1 + [(x2 + 1)2 + x] 2x}dx

=K{2x5 + 4x3 + 2x2 + 2x 1}dx

=[13x6 + x3 +

23x3 + x2 x

]10

=13

+ 1 +23

+ 1 1 = 2.

6) Here a parametric description is e.g. given by

r(t) =

(t, t, t), t [0, 1],(3 2t , 1 , 3 2t), t [1, 2],

(t 3, ,t + 3 , t 3), t [2, 3],hence

r(t) =

(1, 1, 1), t ]0, 1[,(2, 0,2), t ]1, 2[,(1,1, 1), t ]2, 3[.

First variant. We get by direct insertion,KV dx =

K{yz dx + xz dy + x(y + 1)dz}

= 10

(t2+t2+t2+t)dt+ 21

{1 (32t) (2)+(32t) 2 (2)}dt

+ 32

{(t+3)(t3) 1+(t3)2 (1)+(t3)(t+3) 1+t3}dt

= 10

(3t2 + t)dt 6 21

(3 2t)dt 32

{3(t 3)2 (t 3)}dt

=[t3 +

12t2]10

+ 6[t2 3t]2

1[(t 3)3 1

2(t 3)2

]32

= 1 +12

+ 6(4 6 1 + 3) +(1 1

2

)= 0.



Calculus 2c-7

78

2. variant. Reduction by removing a total dierential.As

V dx = yz dx + xz dy + xy dz + x dz = d(xyz) + x dz,and as K is a closed curve, we have K d(xyz) = 0, so the calculations are simplied by removingd(xyz):

KV dx =

K

d(xyz) +K

x dz = 0 +K

x dz

= 10

t dt + 21

(3 2t) (2)dt + 32

(t 3)dt

=[12t2]10

+ 21

(4t 6)dt +[12(t 3)2

]32

=12

+[2t2 6t]2

1 1

2= 8 12 2 + 6 = 0.

Remark. The expressions will be even simpler, if we do not insist on that the parametricintervals [0, 1], [1, 2], [2, 3] should follow each other. Instead one can split K into three subcurves

K1 : r1(t) = (t, t, t), t [0, 1],K2 : r2(t) = (1 2t, 1, 1 2t), t [0, 1],K3 : r3(t) = (t 1, 1 t, t 1), t [0, 1],

where

K1 : r1(t) = (1, 1, 1), t ]0, 1[;K2 : r2(t) = (2, 0, 2), t ]0, 1[;K3 : r3(t) = (1,1, 1). t ]0, 1[.

We obtain that the three line integrals can be joined like in the second variant above:KV dx = =

K

x dz =K1

x dz +K2

x dz +K3

x dz

= 10

t dt + 10

(1 2t) (2)dt + 10

(t 1)dt

= 3 10

(2t 1)dt = 3 [t2 t]10= 0.

7) The most obvious parametric description is here

r(t) = t (1, 1, 1), med r(t) = (1, 1, 1), t [0, ].Thus we can put x = y = z = t everywhere. Then

KV dx =

K{sin y dx + sin z dy + sinx dz} =

0

3 sin t dt = [3 cos t]0 = 6.

8) If we call the two curve segments for K1 and K2, then the most obvious parametric descriptionis

K1 : a (cos t, 0, sin t), t [0,

2

],

K2 : a (0, sin t, cos t), t [0,

2

].



Calculus 2c-7

79

Then by insertion,KV dx =

K(z dx + x dy y dz)

= a2

2

0

(sin t ( sin t))dt + a2

2

0

( sin t) ( sin t)dt

= a2

2

0

sin2 t dt + a2

2

0

sin2 t dt = 0.


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Calculus 2c-7

80

Example 4.5 Find in each of the following cases a function

(x, y) =KV(x) dx,

to the given vector eld V : A R2, where K is the broken line which runs from (0, 0) over (x, 0) to(x, y). Check if is dened in all of A, and compute nally the gradient .1) The vector eld V(x, y) = (y2 2xy,x2 + 2xy) is dened in A = R2.

2) The vector eld V(x, y) =

(1

y2 x2 + 1 , x)

is dened in

A = {(x, y) |

1 + y2 < x 0,expx + ln(1 yz) for yz < 0,

so we must be very careful, whenever the curve K intersects one of the planes y = 0 or z = 0.In case of the rst curves this can occur, because the parametric description is

t(0, 1, 1) + (1 t)(,3, 2) = ((1 t), 4t 3, 2 t), t [0, 1],

and the same is true for the second curve, because it has the parametric description

t(,3, 2) + (1 t)(1,

3,

3) = (1 + t( 1),

3 t(3 +

3),

3 + t(2 +

3)),



Calculus 2c-7

86

for t [0, 1].The former curve intersects the plane y = 0 for t =

34, and the latter curve intersects both the

plane y = 0 and the plane z = 0. The point is, however, that in everyone of these intersectionpoints the dubious term ln(1+ |xy|) = 0, so they are of no importance. Hence we can concludethat

KV dx = [expx + ln(1 + |yz|](1,

3,3)

(0,1,1)

= e + ln 4 1 ln 2 = e 1 + ln 2.Remark. Always be very careful when either the absolute value or the square root occur. Oneshould at least give a note on them.


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Calculus 2c-7

87

Example 4.8 Given the vector eld

V(x, y) =(

2y2x + y

,y

2x + y+ ln(2x + y)

).

1. Sketch the domain of V, and explain why V is a gradient eld.

2. Find every integral of V.

Documents

Funciones reales de varias variablesº