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

section 16.9 The Divergence Theorem 1145
Another application of the Divergence Theorem occurs in fluid flow. Let vsx, y, zd be
the velocity field of a fluid with constant density . Then F − v is the rate of flow per
unit area. If P 0 sx 0 , y 0 , z 0 d is a point in the fluid and B a is a ball with center P 0 and
very small radius a, then div FsPd < div FsP 0 d for all points P in B a since div F is continuous.
We approximate the flux over the boundary sphere S a as follows:
yy F dS − y yy div F dV < y yy div FsP 0 d dV − div FsP 0 dVsB a d
S a B a B a
This approximation becomes better as a l 0 and suggests that
P
y
P¡
FIGURE 4
The vector field F − x 2 i 1 y 2 j
x
8 div FsP 0 d − lim
a l 0
1
VsB a d y S a
y F dS
Equation 8 says that div FsP 0 d is the net rate of outward flux per unit volume at P 0 . (This
is the reason for the name divergence.) If div FsPd . 0, the net flow is outward near P
and P is called a source. If div FsPd , 0, the net flow is inward near P and P is called
a sink.
For the vector field in Figure 4, it appears that the vectors that end near P 1 are shorter
than the vectors that start near P 1 . Thus the net flow is outward near P 1 , so div FsP 1 d . 0
and P 1 is a source. Near P 2 , on the other hand, the incoming arrows are longer than the
outgoing arrows. Here the net flow is inward, so div FsP 2 d , 0 and P 2 is a sink. We
can use the formula for F to confirm this impression. Since F − x 2 i 1 y 2 j, we have
div F − 2x 1 2y, which is positive when y . 2x. So the points above the line y − 2x
are sources and those below are sinks.
1–4 Verify that the Divergence Theorem is true for the vector field
F on the region E.
1. Fsx, y, zd − 3x i 1 xy j 1 2xz k,
E is the cube bounded by the planes x − 0, x − 1, y − 0,
y − 1, z − 0, and z − 1
2. Fsx, y, zd − y 2 z 3 i 1 2yz j 1 4z 2 k,
E is the solid enclosed by the paraboloid z − x 2 1 y 2 and the
plane z − 9
3. Fsx, y, zd − kz, y, xl,
E is the solid ball x 2 1 y 2 1 z 2 < 16
4. Fsx, y, zd − kx 2 , 2y, zl,
E is the solid cylinder y 2 1 z 2 < 9, 0 < x < 2
5–15 Use the Divergence Theorem to calculate the surface integral
yy S
F dS; that is, calculate the flux of F across S.
5. Fsx, y, zd − xye z i 1 xy 2 z 3 j 2 ye z k,
S is the surface of the box bounded by the coordinate planes
and the planes x − 3, y − 2, and z − 1
6. Fsx, y, zd − x 2 yz i 1 xy 2 z j 1 xyz 2 k,
S is the surface of the box enclosed by the planes x − 0,
x − a, y − 0, y − b, z − 0, and z − c, where a, b, and c are
positive numbers
7. Fsx, y, zd − 3xy 2 i 1 xe z j 1 z 3 k,
S is the surface of the solid bounded by the cylinder
y 2 1 z 2 − 1 and the planes x − 21 and x − 2
8. Fsx, y, zd − sx 3 1 y 3 d i 1 sy 3 1 z 3 d j 1 sz 3 1 x 3 d k,
S is the sphere with center the origin and radius 2
9. Fsx, y, zd − xe y i 1 sz 2 e y d j 2 xy k,
S is the ellipsoid x 2 1 2y 2 1 3z 2 − 4
10. Fsx, y, zd − z i 1 y j 1 zx k,
S is the surface of the tetrahedron enclosed by the coordinate
planes and the plane
x
a 1 y b 1 z c − 1
where a, b, and c are positive numbers
11. Fsx, y, zd − s2x 3 1 y 3 d i 1 sy 3 1 z 3 d j 1 3y 2 z k,
S is the surface of the solid bounded by the paraboloid
z − 1 2 x 2 2 y 2 and the xy-plane
12. Fsx, y, zd − sxy 1 2xzd i 1 sx 2 1 y 2 d j 1 sxy 2 z 2 d k,
S is the surface of the solid bounded by the cylinder
x 2 1 y 2 − 4 and the planes z − y 2 2 and z − 0
13. F − | r | r, where r − x i 1 y j 1 z k,
S consists of the hemisphere z − s1 2 x 2 2 y 2 and the disk
x 2 1 y 2 < 1 in the xy-plane
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.