682
46.
■
yy e
CHAPTER 12
x 2y 2
MULTIPLE INTEGRALS
49. Evaluate xxD x 2 tan x y 3 4 dA, where
dA ,
D x, y x 2 y 2 2. [Hint: Exploit the fact that
D is symmetric with respect to both axes.]
D
D is the disk with center the origin and radius
■
■
■
■
■
■
■
■
■
1
2
■
■
50. Use symmetry to evaluate xxD 2 3x 4y dA, where D
■
is the region bounded by the square with vertices 5, 0
and 0, 5.
47. Prove Property 11.
51. Compute xxD s1 x 2 y 2 dA , where D is the disk
48. In evaluating a double integral over a region D, a sum of
x 2 y 2 1, by first identifying the integral as the volume
of a solid.
iterated integrals was obtained as follows:
yy f x, y dA y y
1
0
2y
0
f x, y dx dy y y
3
1
3y
0
f x, y dx dy
D
Sketch the region D and express the double integral as an
iterated integral with reversed order of integration.
12.3
CAS
52. Graph the solid bounded by the plane x y z 1 and
the paraboloid z 4 x 2 y 2 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.)
DOUBLE INTEGRALS IN POLAR COORDINATES
■ Polar coordinates were introduced in
Section 9.3.
Suppose that we want to evaluate a double integral xxR f x, y dA, where R is one of
the regions shown in Figure 1. In either case the description of R in terms of rectangular coordinates is rather complicated but R is easily described using polar coordinates.
y
y
≈+¥=4
≈+¥=1
R
R
0
x
0
FIGURE 1
y
P (r, ¨ ) =P (x, y)
r
(a) R=s(r, ¨) | 0¯r¯1, 0¯¨¯2πd
x
(b) R=s(r, ¨ ) | 1¯r¯2, 0¯¨¯πd
Recall from Figure 2 that the polar coordinates r, of a point are related to the rectangular coordinates x, y by the equations
x r cos r2 x2 y2
y
≈+¥=1
y r sin ¨
O
FIGURE 2
x
x
The regions in Figure 1 are special cases of a polar rectangle
R r, a r b, which is shown in Figure 3. In order to compute the double integral xxR f x, y dA,
where R is a polar rectangle, we divide the interval a, b into m subintervals ri1, ri with lengths ri ri ri1 and we divide the interval , into n subintervals
j1, j with lengths j j j1. Then the circles r ri and the rays j
divide the polar rectangle R into the small polar rectangles shown in Figure 4.
SECTION 12.3
DOUBLE INTEGRALS IN POLAR COORDINATES
■
683
¨=¨ j
¨=¨ j_ ¡
r=b
R ij
¨=∫
(r i*, ¨ j*)
R
Ψ j
r=a
¨=å
r=r i
r=r i _ ¡
∫
å
O
O
FIGURE 4 Dividing R into polar subrectangles
FIGURE 3 Polar rectangle
The “center” of the polar subrectangle
Rij r, r
i1
r ri , j1 j has polar coordinates
j* 12 j1 j ri* 12 ri1 ri We compute the area of Rij using the fact that the area of a sector of a circle with radius
1
r and central angle is 2 r 2. Subtracting the areas of two such sectors, each of which
has central angle j , we find that the area of Rij is
2
2
Aij 12 ri2 j 12 ri1
j 12 ri2 ri1
j
12 ri ri1 ri ri1 j ri* ri j
Although we have defined the double integral xxR f x, y dA in terms of ordinary
rectangles, it can be shown that, for continuous functions f , we always obtain the
same answer using polar rectangles. The rectangular coordinates of the center of Rij
are ri* cos j*, ri* sin j* , so a typical Riemann sum is
m
1
n
m
f r* cos *, r* sin * A
i
j
i
j
ij
i1 j1
n
f r* cos *, r* sin * r* r i
j
i
j
i
i
j
i1 j1
If we write tr, r f r cos , r sin , then the Riemann sum in Equation 1 can be
written as
m
n
tr*, * r i
j
i
i1 j1
which is a Riemann sum for the double integral
y y
b
a
tr, dr d
j
684
■
CHAPTER 12
MULTIPLE INTEGRALS
Therefore, we have
m
yy f x, y dA n
lim
f r* cos *, r* sin * A
lim
tr*, * r i
max ri , j l 0 i1 j1
R
m
y
b
i
j
n
max ri , j l 0 i1 j1
y
j
i
j
i
ij
y
j
y
b
a
tr, dr d
f r cos , r sin r dr d
a
2 CHANGE TO POLAR COORDINATES IN A DOUBLE INTEGRAL If f is continuous on a polar rectangle R given by 0 a r b, , where
0 2, then
yy f x, y dA y y
b
a
f r cos , r sin r dr d
R
dA
d¨
dr
r
The formula in (2) says that we convert from rectangular to polar coordinates in a
double integral by writing x r cos and y r sin , using the appropriate limits of
| integration for r and , and replacing dA by r dr d. Be careful not to forget the additional 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 r d and dr and therefore has “area” dA r dr d.
r d¨
EXAMPLE 1 Evaluate xxR 3x 4y 2 dA, where R is the region in the upper half-
plane bounded by the circles x 2 y 2 1 and x 2 y 2 4.
O
SOLUTION The region R can be described as
FIGURE 5
R x, y y 0, 1 x 2 y 2 4 It is the half-ring shown in Figure 1(b), and in polar coordinates it is given by
1 r 2, 0 . Therefore, by Formula 2,
yy 3x 4y
2
dA y
y
y
0
y
2
y
2
1
3r cos 4r 2 sin 2 r dr d
R
0
0
■
1
[r
3r 2 cos 4r 3 sin 2 dr d
3
cos r 4 sin 2
]
r2
r1
d y 7 cos 15 sin 2 d
0
Here we use the trigonometric identity
sin2 12 1 cos 2 y [7 cos 0
15
2
]
1 cos 2 d
as discussed in Section 6.2.
15
15
7 sin sin 2
2
4
0
15
2
■
SECTION 12.3
DOUBLE INTEGRALS IN POLAR COORDINATES
■
685
V EXAMPLE 2 Find the volume of the solid bounded by the plane z 0 and the
paraboloid z 1 x 2 y 2.
SOLUTION If we put z 0 in the equation of the paraboloid, we get x 2 y 2 1.
z
This means that the plane intersects the paraboloid in the circle x 2 y 2 1, so the
solid lies under the paraboloid and above the circular disk D given by x 2 y 2 1
[see Figures 6 and 1(a)]. In polar coordinates D is given by 0 r 1, 0 2.
Since 1 x 2 y 2 1 r 2, the volume is
(0, 0, 1)
V yy 1 x 2 y 2 dA y
y
x
2
0
D
FIGURE 6
y
2
0
d y r r 3 dr 2
1
0
y
1
0
1 r 2 r dr d
r2
r4
2
4
1
0
2
If we had used rectangular coordinates instead of polar coordinates, then we would
have obtained
V yy 1 x 2 y 2 dA y
1
1
D
y
s1x 2
s1x 2
1 x 2 y 2 dy dx
which is not easy to evaluate because it involves finding x 1 x 2 32 dx .
r=h™(¨)
¨=∫
D
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 Section 12.2. In fact, by combining Formula 2 in this section with Formula 12.2.5, we
obtain the following formula.
3
å
O
r=h¡(¨)
If f is continuous on a polar region of the form
D r, ¨=å
∫
■
then
,
yy f x, y dA y y
h 2 h1 D
h1 r h2 f r cos , r sin r dr d
FIGURE 7
D=s(r, ¨) | 寨¯∫, h¡(¨)¯r¯h™(¨)d
In particular, taking f x, y 1, h1 0, and h2 h in this formula, we
see that the area of the region D bounded by , , and r h is
AD yy 1 dA y y
D
h 0
y
and this agrees with Formula 9.4.3.
r2
2
h 0
r dr d
d y
1
2
h 2 d
686
■
CHAPTER 12
MULTIPLE INTEGRALS
y
V EXAMPLE 3 Find the volume of the solid that lies under the paraboloid
z x 2 y 2, above the xy-plane, and inside the cylinder x 2 y 2 2x.
(x-1)@+¥=1
(or r=2 cos ¨)
SOLUTION The solid lies above the disk D whose boundary circle has equation
x 2 y 2 2x or, after completing the square,
D
0
1
x 12 y 2 1
x
2
(See Figures 8 and 9.) In polar coordinates we have x 2 y 2 r 2 and x r cos ,
so the boundary circle becomes r 2 2r cos , or r 2 cos . Thus the disk D is
given by
FIGURE 8
D r, 2 2, 0 r 2 cos z
and, by Formula 3, we have
V yy x y dA y
2
2
4y
2
2
2y
2
0
y
cos 4 d 8 y
2
0
r r dr d cos 4 d 8 y
0
5
5.
1 cos 2
2
2
d
2
3
2
3
2
■
■
R
7.
0
x
4.
y
R
x
■
■
■
r dr d
■
■
■
■
0
■
■
■
xxD xy dA,
xxR cosx 2 y 2 dA, where R is the region that lies above
the x-axis within the circle x 2 y 2 9
where R is the region that lies to the left of
the y-axis between the circles x 2 y 2 1 and x 2 y 2 4
xxR s4 x 2 y 2
where R x, y
1
■
■
9.
11.
3 x
■
■
■
dA,
x 2 y 2 4, x 0
xxR arctan yx dA,
where R x, y
12.
■
■
xxR x y dA,
R
1
■
■
4 cos 0
0
8.
10.
2
■
2
y y
6.
where D is the disk with center the origin and radius 3
x
2
y
3
2
0
r dr d
Evaluate the given integral by changing to polar
coordinates.
5
0
7
4
y
2
■
2
y y
7–12
2
■
d
0
■ Sketch the region whose area is given by the integral and
evaluate the integral.
2
3.
5–6
■
2.
y
R
2
2
2 cos r4
4
EXERCISES
A region R is shown. Decide whether to use polar coordinates or rectangular coordinates and write xxR f x, y dA
as an iterated integral, where f is an arbitrary continuous function on R.
1.
y
2
1 2 cos 2 12 1 cos 4 d
■
■
0
2
2
12.3
■
2 cos 2[ 32 sin 2 18 sin 4]0 2
FIGURE 9
1– 4
y
2
D
x
2
xxR ye x dA,
1x
2
y 2 4, 0 y x
where R is the region in the first quadrant
enclosed by the circle x 2 y 2 25
■
■
■
■
■
■
■
■
■
■
■
SECTION 12.3
13–19
Use polar coordinates to find the volume of the given
■
x2 y2 4
14. Below the paraboloid z 18 2x 2 2y 2 and above the
xy-plane
15. A sphere of radius a
16. Inside the sphere x 2 y 2 z 2 16 and outside the
29. Use polar coordinates to combine the sum
cylinder x 2 y 2 4
y y
17. Above the cone z sx 2 y 2 and below the sphere
1
x y z 1
2
18. Bounded by the paraboloid z 1 2x 2 2y 2 and the
■
■
■
■
■
■
■
■
2
al
■
■
■
■
■
■
■
23.
y y
24.
s9x 2
3
0
yy
0
0
sa 2 y 2
25.
yy
1
s2y 2
0
y
26.
2
yy
s2xx 2
0
0
■
xy dy dx
■
y
2
2
ex y dy dx
dA
2
2
ex y dA al
⺢2
■
yy e
x 2y 2 dA lim
dA
Sa
where Sa is the square with vertices a, a. Use this
to show that
y
ex dx y
2
2
2
ey dy (c) Deduce that
a
■
s4x 2
0
Da
x 2y 2 sinx y dy dx
2
x 2y 2 yy e
■
3
yy e
y y
Evaluate the iterated integral by converting to polar
coordinates.
23–26
2
s2
where Da is the disk with radius a and center the origin.
Show that
22. The region enclosed by the curve r 4 3 cos ■
y y
(b) An equivalent definition of the improper integral in
part (a) is
Use a double integral to find the area of the region.
■
xy dy dx lim
21. One loop of the rose r cos 3
■
2
⺢2
■
through the center of a sphere of radius r 2 . Find the volume of the ring-shaped solid that remains.
(b) Express the volume in part (a) in terms of the height h
of the ring. Notice that the volume depends only on h,
not on r 1 or r 2 .
■
x
0
I yy ex y dA y
20. (a) A cylindrical drill with radius r 1 is used to bore a hole
21–22
s2
1
2
4x 2 4y 2 z 2 64
■
y y
30. (a) We define the improper integral (over the entire plane ⺢ 2 19. Inside both the cylinder x y 4 and the ellipsoid
■
xy dy dx into one double integral. Then evaluate the double integral.
plane z 7 in the first octant
■
x
s1x 2
1s2
2
687
tern of radius 100 ft. It supplies water to a depth of er feet
per hour at a distance of r feet from the sprinkler.
(a) What is the total amount of water supplied per hour to
the region inside the circle of radius R 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 R.
13. Under the cone z sx 2 y 2 and above the disk
2
■
28. An agricultural sprinkler distributes water in a circular pat-
solid.
2
DOUBLE INTEGRALS IN POLAR COORDINATES
■
y
x 2 y dx dy
y
■
sx 2 y 2 dy dx
■
2
ex dx s
(d) By making the change of variable t s2 x, show that
x y dx dy
■
■
■
■
■
■
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.
2
ex 2 dx s2
(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
0
2
x 2ex dx
(b)
y
0
sx ex dx