xx xx yy y y y y yy

46. ,

is the disk with center the origin and radius

■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■

47. Prove Property 11.

In evaluating a double integral over a region , a sum of iterated integrals was obtained as follows:

Sketch the region and express the double integral as an iterated integral with reversed order of integration.

D

yy

D

fx, y dA 

y

y

y

y

1

D 2

yy

D

e^x2y2dA ^49. Evaluate , where

[Hint: Exploit the fact that is symmetric with respect to both axes.]

50. Use symmetry to evaluate , where is the region bounded by the square with vertices and .

51. Compute , where is the disk

, by first identifying the integral as the volume of a solid.

52. Graph the solid bounded by the plane and the paraboloid and find its exact volume.

(Use your CAS to do the graphing, to find the equations of the boundary curves of the region of integration, and to evaluate the double integral.)

z  4  x² y²

x y  z  1

CAS

x² y² 1

xxDs1  x² y²dA D

0, 5 5, 0

xxD2  3x  4y dA D D

D x, y^xxD



2tan x y³ 4 dA

DOUBLE INTEGRALS IN POLAR COORDINATES

Suppose that we want to evaluate a double integral , where is one of the regions shown in Figure 1. In either case the description of in terms of rectan- gular coordinates is rather complicated but is easily described using polar coordinates.

Recall from Figure 2 that the polar coordinates of a point are related to the rect- angular coordinates by the equations

The regions in Figure 1 are special cases of a polar rectangle

which is shown in Figure 3. In order to compute the double integral , where is a polar rectangle, we divide the interval into subintervals with lengths and we divide the interval into subintervals

with lengths . Then the circles and the rays divide the polar rectangle R into the small polar rectangles shown in Figure 4.

  ^j r rⁱ

^j^j^j1

^j1,^j rⁱ rⁱ rⁱ1 ,  n

rⁱ1, ri a, b m

R

xx



y r sin x r cos

r² x² y²

x, y r,

FIGURE 1

0 x y

R

≈+¥=1

(a) R=s(r, ¨) | 0¯r¯1, 0¯¨¯2πd

x 0

y

R

≈+¥=4

≈+¥=1

(b) R=s(r, ¨) | 1¯r¯2, 0¯¨¯πd

R

R

xx

12.3

■ Polar coordinates were introduced in Section 9.3.

O y

x

¨ x r y

P(r, ¨ ) =P(x, y)

FIGURE 2

The “center” of the polar subrectangle

has polar coordinates

We compute the area of using the fact that the area of a sector of a circle with radius and central angle is . Subtracting the areas of two such sectors, each of which has central angle , we find that the area of is

Although we have defined the double integral in terms of ordinary rectangles, it can be shown that, for continuous functions , we always obtain the same answer using polar rectangles. The rectangular coordinates of the center of

are , so a typical Riemann sum is

If we write , then the Riemann sum in Equation 1 can be written as

which is a Riemann sum for the double integral

y

y



i1_j1



tr,  r f r cos , r sin 



i1_j







1

rⁱ* cos^j*, ri* sin ^j*

Rij

f

xx

¹2rⁱ rⁱ1rⁱ rⁱ1 ^j rⁱ*rⁱ^j

A^ij¹2ri2^j¹2ri21^j¹2rⁱ² rⁱ21 ^j Rij

^j

1 2r²

 r

Rij

^j*¹2^j1^j ri* ¹2rⁱ1 rⁱ

Rij



FIGURE 3 Polar rectangle O

∫ å

r=a ¨=å

¨=∫

r=b

R

r=ri

r=r_{i _}¡ Rij (r_i^*, ¨_j^*)

¨=¨_{j _}¡

¨=¨_j

Ψj

FIGURE 4 Dividing R into polar subrectangles O

Therefore, we have

CHANGE TO POLAR COORDINATES IN A DOUBLE INTEGRAL If is contin- uous on a polar rectangle given by , , where

, then

The formula in (2) says that we convert from rectangular to polar coordinates in a double integral by writing and , using the appropriate limits of

|integration for and , and replacing by . Be careful not to forget the addi- tional factor r on the right side of Formula 2.A classical method for remembering this is shown in Figure 5, where the “infinitesimal” polar rectangle can be thought of as an ordinary rectangle with dimensions and and therefore has “area”

EXAMPLE 1 Evaluate , where is the region in the upper half-

plane bounded by the circles and .

SOLUTION The region can be described as

It is the half-ring shown in Figure 1(b), and in polar coordinates it is given by , . Therefore, by Formula 2,

 7 sin   15 ■ 2  15

4 sin 2





y

[

]



y

[

]

r2d 

y



y

y

yy

R

3x  4y² dA 

y

y

0   1 r 2

R



R

x² y² 4 x² y² 1 R

xx

dA r dr d.

r d dr

r dr d

 dA r

y r sin x r cos

yy

R

fx, y dA 

y

y

0    2 R 0 a r b    f

2



y

y

 lim

maxri,jl 0_i





y

y

yy

R

fx, y dA  lim

maxri,jl 0_i





O

d¨

r d¨

dr dA

r

FIGURE 5

■ Here we use the trigonometric identity

as discussed in Section 6.2.

sin² ¹21  cos 2

EXAMPLE 2 Find the volume of the solid bounded by the plane and the paraboloid .

SOLUTION If we put in the equation of the paraboloid, we get . This means that the plane intersects the paraboloid in the circle , so the solid lies under the paraboloid and above the circular disk given by

[see Figures 6 and 1(a)]. In polar coordinates is given by , .

Since , the volume is

If we had used rectangular coordinates instead of polar coordinates, then we would have obtained

which is not easy to evaluate because it involves finding . _■

What we have done so far can be extended to the more complicated type of region shown in Figure 7. It’s similar to the type II rectangular regions considered in Sec- tion 12.2. In fact, by combining Formula 2 in this section with Formula 12.2.5, we obtain the following formula.

If is continuous on a polar region of the form

then

In particular, taking , , and in this formula, we see that the area of the region bounded by , , and is

and this agrees with Formula 9.4.3.



y





^y

AD 

yy

D

1 dA

y

y

r h

  

   D

h2  h

h1  0 fx, y  1

yy

D

fx, y dA 

y

y

D



f

3

x

yy

D

1  x² y² dA 

y

y



y

y





V

yy

D

1  x² y² dA 

y

y

1 x² y² 1  r² D 0 r 1 0  2

x² y² 1 D

x² yx²² y 1² 1 z  0

z  1  x² y² z  0

V

FIGURE 6

y (0, 0, 1)

x

z

O

∫ å

r=h¡(¨)

¨=å

¨=∫ r=h™(¨)

D

FIGURE 7

D=s(r, ¨) | å¯¨¯∫, h¡(¨)¯r¯h™(¨)d

EXAMPLE 3 Find the volume of the solid that lies under the paraboloid , above the -plane, and inside the cylinder . SOLUTION The solid lies above the disk whose boundary circle has equation

or, after completing the square,

(See Figures 8 and 9.) In polar coordinates we have and , so the boundary circle becomes , or . Thus the disk is given by

and, by Formula 3, we have

 2

[

]

2 2







 2

y

 8

y

y





 4

y



y





V

yy

D

x² y² dA 

y

y

D

 

D r 2 cos 

r² 2r cos x² y² r² x r cos

x  1² y² 1

x² y² 2x D

x² y² 2x z  x² y² xy

V

FIGURE 8 0 y

1 2 x

D

(x-1)@+¥=1 (or  r=2 cos ¨)

FIGURE 9

y x

z

5–6 ^■ Sketch the region whose area is given by the integral and evaluate the integral.

5. 6.

■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■

7–12 ^■ Evaluate the given integral by changing to polar coordinates.

7. ,

where is the disk with center the origin and radius 3 8. , where is the region that lies to the left of

the -axis between the circles and

9. , where is the region that lies above the -axis within the circle

10. ,

where

, where

12. , where is the region in the first quadrant enclosed by the circle

■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■

x² y² 25

xxRye^xdA R

R x, y



xxR arctan yx dA 11.

R x, y



xxRs4  x² y²dA

x² y² 9 x

xxR cosx² y² dA R

x² y² 4 x² y² 1

y

xxRx  y dA R D

xxDxy dA

y

y

y

y

dinates or rectangular coordinates and write

as an iterated integral, where is an arbitrary continuous func- tion on .

2.

3. 4.

■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■

0 1 3

3

1 y

x R

0 2

2 y

x R

0 2

y

x R

0 2

5 2 5

2 y

x R

1.

R

f

xxRfx, y dA R

EXERCISES

12.3

28. An agricultural sprinkler distributes water in a circular pat- tern of radius 100 ft. It supplies water to a depth of feet per hour at a distance of feet from the sprinkler.

(a) What is the total amount of water supplied per hour to the region inside the circle of radius centered at the sprinkler?

(b) Determine an expression for the average amount of water per hour per square foot supplied to the region inside the circle of radius .

Use polar coordinates to combine the sum

into one double integral. Then evaluate the double integral.

30. (a) We define the improper integral (over the entire plane

where is the disk with radius and center the origin.

Show that

(b) An equivalent definition of the improper integral in part (a) is

where is the square with vertices . Use this to show that

(c) Deduce that

(d) By making the change of variable , show that

(This is a fundamental result for probability and statistics.)

31. Use the result of Exercise 30 part (c) to evaluate the follow- ing integrals.

(a)

y

y

y

t s2 x

y

y

y

a, a

Sa

yy

e^x2y2dA lim_al 

yy

Sa

e^x2y2dA

y

y

a Da

 lim_al 

yy

Da

e^x2y2dA I

yy

⺢²

e^x2y2dA

y

y

⺢²

y

y

y

y

y

y

R

R r

e^r 13–19 ^■ Use polar coordinates to find the volume of the given

solid.

13. Under the cone and above the disk

14. Below the paraboloid and above the -plane

15. A sphere of radius

16. Inside the sphere and outside the cylinder

Above the cone and below the sphere

18. Bounded by the paraboloid and the plane in the first octant

19. Inside both the cylinder and the ellipsoid

■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■

20. (a) A cylindrical drill with radius is used to bore a hole through the center of a sphere of radius . Find the vol- ume of the ring-shaped solid that remains.

(b) Express the volume in part (a) in terms of the height of the ring. Notice that the volume depends only on , not on or .

21–22 ^■ Use a double integral to find the area of the region.

One loop of the rose

22. The region enclosed by the curve

■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■

23–26 ^■ Evaluate the iterated integral by converting to polar coordinates.

23.

24.

25.

26.

■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■

27. A swimming pool is circular with a 40-ft diameter. The depth is constant along east-west lines and increases linearly from 2 ft at the south end to 7 ft at the north end.

Find the volume of water in the pool.

y

y

y

y

y

y

y

y

r 4  3 cos  r cos 3

21.

r2

r1

h h r2

r1

4x² 4y² z² 64

x² y² 4

z  7 z  1  2x² 2y² x² y² z² 1z  sx² y²

17.

x² y² 4

x² y² z² 16 a

xy

z  18  2x² 2y² x² y² 4 z  sx² y²