yy yy y y y y y y y

40. Show that

(The improper triple integral is defined as the limit of a triple integral over a solid sphere as the radius of the sphere increases indefinitely.)

41. (a) Use cylindrical coordinates to show that the volume of the solid bounded above by the sphere and below by the cone (or ), where

, is

(b) Deduce that the volume of the spherical wedge given by

, , is

(c) Use the Mean Value Theorem to show that the volume in part (b) can be written as

where lies between and , lies between and , , ,   ²¹   ²¹ and .  ²¹

^{2  1 2} ^ ¹

V ^{ 2} sin ^  

V  ²³¹³

3 cos¹ cos²²¹

¹  ²

¹  ²

^{1  2}

V 2a³

3 1  cos⁰ 0 ⁰ 2 z  r cot⁰   ⁰

r² z² a² e^x2y2z2dx dy dz  2

y

y

y

tive -axis passes through the point where the prime merid- ian (the meridian through Greenwich, England) intersects the equator. Then the latitude of is and the longitude is . Find the great-circle distance from Los Angeles (lat. N, long. W) to Mont- réal (lat. N, long. W). Take the radius of the Earth to be 3960 mi. (A great circle is the circle of inter- section of a sphere and a plane through the center of the sphere.)

39. The surfaces have been used as models for tumors. The “bumpy sphere” with and

is shown. Use a computer algebra system to find the volume it encloses.

n 5

m 6

  1 ¹5 sin m sin n

CAS

73.60 45.50

118.25 34.06

  360   P   90   x

CHANGE OF VARIABLES IN MULTIPLE INTEGRALS

In one-dimensional calculus we often use a change of variable (a substitution) to sim- plify an integral. By reversing the roles of and , we can write the Substitution Rule (5.5.6) as

where and , . Another way of writing Formula 1 is as follows:

A change of variables can also be useful in double integrals. We have already seen one example of this: conversion to polar coordinates. The new variables and are related to the old variables and by the equations

and the change of variables formula (12.3.2) can be written as

where is the region in the S r-plane that corresponds to the region in theR xy-plane.

yy

R

fx, y dA 

yy

S

fr cos, r sin  r dr d

y r sin x r cos

y x

 r

y

y

2

b td

a tc

x tu

y

y

1

u x

12.8

More generally, we consider a change of variables that is given by a transforma- tion from the -plane to the -plane:

where and are related to and by the equations

or, as we sometimes write,

We usually assume that is a C transformation, which means that and have con- tinuous first-order partial derivatives.

A transformation is really just a function whose domain and range are both sub- sets of . If , then the point is called the image of the point . If no two points have the same image, is called one-to-one. Figure 1 shows the effect of a transformation on a region in the -plane. transforms into a region in the -plane called the image of S, consisting of the images of all points in .

If is a one-to-one transformation, then it has an inverse transformation from the -plane to the -plane and it may be possible to solve Equations 3 for and in terms of and :

EXAMPLE 1 A transformation is defined by the equations

Find the image of the square , .

SOLUTION The transformation maps the boundary of into the boundary of the image. So we begin by finding the images of the sides of . The first side, , is given by . (See Figure 2.) From the given equations we have , , and so . Thus is mapped into the line segment from to in the -plane. The second side, is and, put- ting in the given equations, we get

Eliminating , we obtain

0 x 1 x 1  y²

4 4

v

y 2^v x 1 ^v2

u 11, 0 xy S2, u 10 ^v 1

0, 0 y 0 0 x 1 S1

x u^{2 v} 00 u 1 S S1

S

0 ^v 1 S u,^v



y 2u^v x u²^v2

V

v Hx, y

u Gx, y

y x v

u uv

xy

T^1 T

FIGURE 1

0

√

0 y

u x

(u¡, √¡)

(x¡, y¡)

S R

T –!

T

S

xy R

S T

uv S

T

u¹,v1⺢² Tu¹,v1  xT ¹, y1 Tx¹, y1

t h T 1

y yu,^v x xu,^v

y hu,^v x tu,^v

3

u v y

x

Tu,^v  x, y

xy uv

T

FIGURE 2

T 0

√

u

(0, 1) (1, 1)

(1, 0) S S£

S¡

S™

S¢

0 y

(_1, 0) x

(0, 2)

(1, 0) R

x=1-^¥₄ x=      -1^¥₄

which is part of a parabola. Similarly, is given by , whose image is the parabolic arc

Finally, is given by whose image is , , that is,

. (Notice that as we move around the square in the counterclockwise direction, we also move around the parabolic region in the counterclockwise direc- tion.) The image of is the region (shown in Figure 2) bounded by the -axis and

the parabolas given by Equations 4 and 5. _■

Now let’s see how a change of variables affects a double integral. We start with a small rectangle in the -plane whose lower left corner is the point and whose dimensions are and . (See Figure 3.)

The image of is a region in the -plane, one of whose boundary points is . The vector

is the position vector of the image of the point . The equation of the lower side of is , whose image curve is given by the vector function . The tangent vector at to this image curve is

Similarly, the tangent vector at to the image curve of the left side of (namely, ) is

We can approximate the image region by a parallelogram determined by the secant vectors

shown in Figure 4. But

ru lim

u l 0

ru⁰ u,^v0  ru⁰,v0

u

b ru⁰,v0 ^v  ru⁰,v0 a ru⁰ u,^v0  ru⁰,v0

R TS

rv t^vu⁰,^v0 i  h^vu⁰,^v0 j  &x

&^v i &y

&^v j

u u⁰ x⁰, y0 S

ru t^uu⁰,^v0 i  h^uu⁰,^v0 j  &x

&u i &y

&u j

x⁰, y0 ru,^v0

v^v0 S

u,^v ru,^v  tu,^v i  hu,^v j

x⁰, y0  Tu⁰,v0

xy R

S

FIGURE 3

T

0 y

x (x¸, y¸) R

r (u, √ ¸) r (u ¸, √)

0

√

u Îu

Î√

√=√ ¸ u=u ¸

S (u¸, √ ¸)

^v

uuv u⁰,v0

S

x R

S

1 x 0S4 u 00 ^v 1 x ^v2 y 0

1 x 0 x y²

4  1

5

0 u 1

v 1 S3

FIGURE 4 r (u ¸, √ ¸)

r (u¸+Î u, √¸) a

r (u¸, √¸+Î√)

and so Similarly

This means that we can approximate R by a parallelogram determined by the vec- tors and . (See Figure 5.) Therefore, we can approximate the area of by the area of this parallelogram, which, from Section 10.4, is

Computing the cross product, we obtain i j k

The determinant that arises in this calculation is called the Jacobian of the transforma- tion and is given a special notation.

DEFINITION The Jacobian of the transformation given by and is

With this notation we can use Equation 6 to give an approximation to the area of :

where the Jacobian is evaluated at .

Next we divide a region in the -plane into rectangles and call their images in the -plane . (See Figure 6.)

FIGURE 6

T

0 y

x R

(x_i, y

R_ij

0

√

u

S Î √

Î u

(u_i, √_j) S_ij

j)

Rij

xy

Sij

u^v S

u⁰,^v0

A



&u,^v



8

R

A

&x, y

&u,^v 

&x

&u

&y

&u

&x

&^v

&y

&^v

 &x

&u

&y

&^v  &x

&^v

&y

&u

y hu,^v T x tu,^v

7

&x

&u

&y

&u

&x

&^v

&y

&^v

 k

&x

&u

&x

&^v

&y

&u

&y

&^v k

&x

&u

&x

&^v

&y

&u

&y

&^v 0 0 ru r^v









6

^vrv R

u r^u

ru⁰,^v0 ^v  ru⁰,^v0 ^vrv

ru⁰ u,^v0  ru⁰,^v0 u r^u

r (u ¸, √ ¸) Î √ r√

FIGURE 5

■ The Jacobian is named after the German mathematician Carl Gustav Jacob Jacobi (1804–1851). Although the French mathemati- cian Cauchy first used these special determi- nants involving partial derivatives, Jacobi developed them into a method for evaluating multiple integrals.

Applying the approximation (8) to each we approximate the double integral of over as follows:

where the Jacobian is evaluated at . Notice that this double sum is a Riemann sum for the integral

The foregoing argument suggests that the following theorem is true. (A full proof is given in books on advanced calculus.)

CHANGE OF VARIABLES IN A DOUBLE INTEGRAL Suppose that is a transformation whose Jacobian is nonzero and that maps a region in the - plane onto a region in the -plane. Suppose that is continuous on and that and are type I or type II plane regions. Suppose also that is one-to- one, except perhaps on the boundary of . Then

Theorem 9 says that we change from an integral in and to an integral in and by expressing and in terms of and and writing

Notice the similarity between Theorem 9 and the one-dimensional formula in Equa- tion 2. Instead of the derivative , we have the absolute value of the Jacobian, that is, .

As a first illustration of Theorem 9, we show that the formula for integration in polar coordinates is just a special case. Here the transformation from the -plane to the -plane is given by

and the geometry of the transformation is shown in Figure 7. maps an ordinary rect- angle in the -plane to a polar rectangle in the -plane. The Jacobian of is

&x, y

&r, 

&x

&r

&y

&r

&x

&

&y

&







T xy

r T

y hr,  r sin  x tr,  r cos 

xy

r T





dA



&u,^v



u v y

x v

u y

x

yy

R

fx, y dA 

yy

S

fxu,^v, yu,^v



&u,^v



S

T S

R

R f

xy R

u^v S

C¹ T

9

yy

S

ftu,^v, hu,^v



&u,^v



uⁱ,^vj

_i







&u,^v



yy

R

fx, y dA _i





R f

Rij,

FIGURE 7

The polar coordinate transformation 0

y

x

¨=∫ r=b

r=a ¨=å

∫ å

R 0

¨

∫

å

a b r

¨=∫

r=a

¨=å

r=b S

T

Thus Theorem 9 gives

which is the same as Formula 12.3.2.

EXAMPLE 2 Use the change of variables , to evaluate the integral , where is the region bounded by the -axis and the parabolas

and , .

SOLUTION The region is pictured in Figure 2 (on page 714). In Example 1 we discovered that , where is the square . Indeed, the reason for making the change of variables to evaluate the integral is that is a much sim- pler region than . First we need to compute the Jacobian:

Therefore, by Theorem 9,

■ NOTE Example 2 was not a very difficult problem to solve because we were given a suitable change of variables. If we are not supplied with a transformation, then the first step is to think of an appropriate change of variables. If is difficult to inte- grate, then the form of may suggest a transformation. If the region of integra- tion is awkward, then the transformation should be chosen so that the corresponding region in the -plane has a convenient description.

EXAMPLE 3 Evaluate the integral , where is the trapezoidal

region with vertices , , , and .

SOLUTION Since it isn’t easy to integrate , we make a change of variables suggested by the form of this function:

These equations define a transformation from the -plane to the -plane.

Theorem 9 talks about a transformation from the T u^v-plane to the xy-plane. It is u^v

xy T^1

v x  y u x  y

10

exyxy

0, 1

0, 2

2, 0

1, 0

xx

S R

fx, y fx, y



y

[

]

 8

y

y

y

[

]

yy

R

y dA

yy

S

2u^v



&u,^v



^y

^y

&x, y

&u,^v 

&x

&u

&y

&u

&x

&^v

&y

&^v







R

S 0, 1 0, 1 S

TS  RR

y 0 y² 4  4x

y² 4  4x

xx

y 2u^v x u²^v2

V



y

y

yy

R

fx, y dx dy 

yy

S

fr cos, r sin 



&r,



obtained by solving Equations 10 for and :

The Jacobian of is

To find the region in the -plane corresponding to , we note that the sides of lie on the lines

and, from either Equations 10 or Equations 11, the image lines in the -plane are

Thus the region is the trapezoidal region with vertices , , , and shown in Figure 8. Since

Theorem 9 gives

■

TRIPLE INTEGRALS

There is a similar change of variables formula for triple integrals. Let be a transfor- mation that maps a region in -space onto a region in -space by means of the equations

The Jacobian of is the following determinant:

&x, y, z

&u,^v,^w 

&x

&u

&y

&u

&z

&u

&x

&^v

&y

&^v

&z

&^v

&x

&^w

&y

&^w

&z

&^w

12

3 3 T

z  ku,^v,w y hu,^v,w

x tu,^v,w

x yz R u^vw

S

T

¹2

y



y

y

(

)

y

[

]

d^v

yy

R

exyxydA

yy

S

e^uv



&u,^v



S



1, 1 S 1, 1 2, 2 2, 2

v 1 u ^v

v 2 u^v

uv x y  1 x 0

x y  2 y 0

R R

u^v S

&x, y

&u,^v 

&x

&u

&y

&u

&x

&^v

&y

&^v







T

y¹2u ^v x¹2u ^v

11

y x

FIGURE 8

T T –!

0

√

u

(_2, 2) (2, 2)

(_1, 1) (1, 1)

√=2

√=1 u=√

u=_√ S

0 y

_1

_2

x

1 2

x-y=2 x-y=1

R

Under hypotheses similar to those in Theorem 9, we have the following formula for triple integrals:

EXAMPLE 4 Use Formula 13 to derive the formula for triple integration in spherical coordinates.

SOLUTION Here the change of variables is given by

We compute the Jacobian as follows:

Since , we have . Therefore

and Formula 13 gives

which is equivalent to Formula 12.7.3. _■

yyy

R

fx, y, z dV 

yyy

S

f sin  cos ,  sin  sin ,  cos  ²sin d d d



&, , 



^

^

sin  0 0  

 ²sin cos²  ²sin sin²  ²sin

  sin   sin² cos²   sin² sin²

 cos ²sin cos  sin²  ²sin cos  cos²

 cos



 sin  cos 

 cos  cos 

 cos  sin 





 sin  cos 



&x, y, z

&, ,  



 sin  cos  0

 cos  cos 

 cos  sin 

 sin 



z  cos  y sin  sin 

x sin  cos 

V



yyy

S

fxu,^v,w, yu,^v,w, zu,^v,w



&u,^v,^w



yyy

R

fx, y, z dV

13

;^16. , where is the region bounded by the curves

, , , ; , .

Illustrate by using a graphing calculator or computer to draw .

■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■

17. (a) Evaluate , where is the solid enclosed by the

ellipsoid . Use the transfor-

mation , , .

(b) The Earth is not a perfect sphere; rotation has resulted in flattening at the poles. So the shape can be approxi- mated by an ellipsoid with km and

km. Use part (a) to estimate the volume of the Earth.

18. Evaluate , where is the solid of Exercise 17(a).

19–23 ^■ Evaluate the integral by making an appropriate change of variables.

19. , where is the parallelogram enclosed by

the lines , , , and

20. , where is the rectangle enclosed by

the lines , , , and

, where is the trapezoidal region with vertices , , , and

22. , where is the region in the first quadrant bounded by the ellipse

23. , where is given by the inequality

■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■

24. Let be continuous on and let be the triangular region with vertices , , and . Show that

yy

R

fx  y dA 

y

0, 1

1, 0

0, 0

0, 1 R f









xxRe^xydA

9x² 4y² 1

xxR sin9x² 4y² dA R

0, 1

0, 2

2, 0

1, 0

yy

R

cos





21.

x y  3 x y  0

x y  2 x y  0

xxRx  ye^x2y2dA R 3x y  8

3x y  1 x 2y  4

x 2y  0

yy

R

x 2y 3x y dA

xxxEx²y dV E c 6356

a b  6378 z  c^w y b^v x au

x²a² y²b² z²c² 1

xxxEdV E R

v xy² u xy

xy² 2 xy² 1

xy 2 xy 1

xxRy²dA R 1–6 ^■ Find the Jacobian of the transformation.

1. ,

,

3. ,

4. ,

, ,

6. , ,

■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■

7–10 ^■ Find the image of the set under the given transformation.

;

8. is the square bounded by the lines , , ,

; ,

9. is the triangular region with vertices , , ; ,

10. is the disk given by ; ,

■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■

11–16 ^■ Use the given transformation to evaluate the integral.

11. , where is the triangular region with

vertices , , and ; ,

12. , where is the parallelogram with

vertices , , , and ;

,

, where is the region bounded by the ellipse

; ,

14. , where is the region bounded

by the ellipse ;

,

15. , where is the region in the first quadrant bounded by the lines and and the hyperbolas

, ;xy 3 x u^v, y^v xy 1

y 3x y x

xxRxy dA R

y s2 u  s23^v x s2 u  s23^v

x² xy  y² 2

xxRx² xy  y² dA R

y 3^v x 2u 9x² 4y² 36

xxRx²dA R 13.

y¹4^v 3u

x¹4u 1, 3^v 1, 3 3, 1 1, 5

xxR4x  8y dA R

y u  2^v x 2u ^v

1, 2

2, 1

0, 0

xxRx  3y dA R

y b^v x au u²^v2 1

S

y^v

x u² 0, 0 1, 1 0, 1

S

y u1 ^v2 x^v

v 1

v 0 u 1 u 0 S

x 2u  3^v, y u ^v

S u,^v



S z  e^uvw y e^uv

x e^uv

z  u^w y^vw

x u^v 5.

y cos  x sin 

y ^v u^v x u

u^v

y u²^v2 x u²^v2

2.

y 3u  2^v x u  4^v

EXERCISES

12.8