James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

1042 Chapter 15 Multiple Integrals
z
z=u(x, y)
Evaluating Triple Integrals with Cylindrical Coordinates
Suppose that E is a type 1 region whose projection D onto the xy-plane is conveniently
described in polar coordinates (see Figure 6). In particular, suppose that f is continuous
and
E − 5sx, y, zd | sx, yd [ D, u 1sx, yd < z < u 2 sx, yd6
where D is given in polar coordinates by
r=h¡(¨) 0
¨=a
x
FIGURE 6
z=u¡(x, y)
¨=b
D
r=h(¨)
y
D − 5sr, d | < < , h 1sd < r < h 2 sd6
We know from Equation 15.6.6 that
3 y yy f sx, y, zd dV − yy Fy
E
D
u2sx, yd
u1sx, yd
f sx, y, zd dzG dA
But we also know how to evaluate double integrals in polar coordinates. In fact, combining
Equation 3 with Equation 15.3.3, we obtain
4 y y f sx, y, zd dV − y yh2sd
E
u2sr cos , r sin d
y
h1sd u1sr cos , r sin d
f sr cos , r sin , zd r dz dr d
z
d¨
r
r d¨
dr
dz
FIGURE 7
Volume element in cylindrical
coordinates: dV=r dz dr d¨
z=4
z
(0 , 0, 4 )
Formula 4 is the formula for triple integration in cylindrical coordinates. It says that
we convert a triple integral from rectangular to cylindrical coordinates by writing
x − r cos , y − r sin , leaving z as it is, using the appropriate limits of integration for z,
r, and , and replacing dV by r dz dr d. (Figure 7 shows how to remember this.) It is
worthwhile to use this formula when E is a solid region easily described in cylindrical
coordinates, and especially when the function f sx, y, zd involves the expression x 2 1 y 2 .
ExamplE 3 A solid E lies within the cylinder x 2 1 y 2 − 1, below the plane z − 4, and
above the paraboloid z − 1 2 x 2 2 y 2 . (See Figure 8.) The density at any point is
proportional to its distance from the axis of the cylinder. Find the mass of E.
SOLUTION In cylindrical coordinates the cylinder is r − 1 and the paraboloid is
z − 1 2 r 2 , so we can write
E − 5sr, , zd | 0 < < 2, 0 < r < 1, 1 2 r 2 < z < 46
Since the density at sx, y, zd is proportional to the distance from the z-axis, the density
function is
f sx, y, zd − Ksx 2 1 y 2 − Kr
where K is the proportionality constant. Therefore, from Formula 15.6.13, the mass of
E is
x
FIGURE 8
0
(1, 0, 0 )
(0 , 0, 1 )
z=1-r @
y
m − y yy Ksx 2 1 y 2 dV − y 2
y 1
0 0 y4 sKrd r dz dr d
12r 2
E
− y 2
0
y 1
0
− 2KFr 3 1 r 5
Kr 2 f4 2 s1 2 r 2 dg dr d − K y 2
d y 1
s3r 2 1 r 4 d dr
1
5G0
− 12K
5
0
0
■
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