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

1132 Chapter 16 Vector Calculus
where K is an experimentally determined constant called the conductivity of the substance.
The rate of heat flow across the surface S in the body is then given by the surface
integral
y F dS − 2K yy =u dS
y
S
ExamplE 6 The temperature u in a metal ball is proportional to the square of the
distance from the center of the ball. Find the rate of heat flow across a sphere S of
radius a with center at the center of the ball.
SOLUTIon Taking the center of the ball to be at the origin, we have
usx, y, zd − Csx 2 1 y 2 1 z 2 d
where C is the proportionality constant. Then the heat flow is
Fsx, y, zd − 2K =u − 2KCs2x i 1 2y j 1 2z kd
where K is the conductivity of the metal. Instead of using the usual parametrization of
the sphere as in Example 4, we observe that the outward unit normal to the sphere
x 2 1 y 2 1 z 2 − a 2 at the point sx, y, zd is
n − 1 sx i 1 y j 1 z kd
a
S
and so
F n − 2 2KC
a
sx 2 1 y 2 1 z 2 d
But on S we have x 2 1 y 2 1 z 2 − a 2 , so F n − 22aKC. Therefore the rate of heat
flow across S is
y
S
y F dS − yy F n dS − 22aKC y
S
S
y dS
− 22aKCAsSd − 22aKCs4a 2 d − 28KCa 3
■
1. Let S be the surface of the box enclosed by the planes x − 61,
y − 61, z − 61. Approximate yy S
cossx 1 2y 1 3zd dS by
using a Riemann sum as in Definition 1, taking the patches S ij
to be the squares that are the faces of the box S and the points
P ij * to be the centers of the squares.
2. A surface S consists of the cylinder x 2 1 y 2 − 1, 21 < z < 1,
together with its top and bottom disks. Suppose you know that
f is a continuous function with
f s61, 0, 0d − 2 f s0, 61, 0d − 3 f s0, 0, 61d − 4
Estimate the value of yy S
f sx, y, zd dS by using a Riemann sum,
taking the patches S ij to be four quarter-cylinders and the top
and bottom disks.
3. Let H be the hemisphere x 2 1 y 2 1 z 2 − 50, z > 0, and
suppose f is a continuous function with f s3, 4, 5d − 7,
f s3, 24, 5d − 8, f s23, 4, 5d − 9, and f s23, 24, 5d − 12.
By dividing H into four patches, estimate the value of
yy H
f sx, y, zd dS.
4. Suppose that f sx, y, zd − t(sx 2 1 y 2 1 z 2 ), where t is a
function of one variable such that ts2d − 25. Evaluate
yy S
f sx, y, zd dS, where S is the sphere x 2 1 y 2 1 z 2 − 4.
5–20 Evaluate the surface integral.
5. yy S
sx 1 y 1 zd dS,
S is the parallelogram with parametric equations x − u 1 v,
y − u 2 v, z − 1 1 2u 1 v, 0 < u < 2, 0 < v < 1
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