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

1146 chapter 16 Vector Calculus
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14. F − | r | 2 r, where r − x i 1 y j 1 z k,
S is the sphere with radius R and center the origin
15. Fsx, y, zd − e y tan z i 1 ys3 2 x 2 j 1 x sin y k,
S is the surface of the solid that lies above the xy-plane
and below the surface z − 2 2 x 4 2 y 4 , 21 < x < 1,
21 < y < 1
16. Use a computer algebra system to plot the vector field
Fsx, y, zd − sin x cos 2 y i 1 sin 3 y cos 4 z j 1 sin 5 z cos 6 x k
in the cube cut from the first octant by the planes x − y2,
y − y2, and z − y2. Then compute the flux across the
surface of the cube.
17. Use the Divergence Theorem to evaluate yy S
F dS, where
Fsx, y, zd − z 2 x i 1 ( 1 3 y 3 1 tan z) j 1 sx 2 z 1 y 2 d k
and S is the top half of the sphere x 2 1 y 2 1 z 2 − 1.
[Hint: Note that S is not a closed surface. First compute
integrals over S 1 and S 2, where S 1 is the disk x 2 1 y 2 < 1,
oriented downward, and S 2 − S ø S 1.]
18. Let Fsx, y, zd − z tan 21 sy 2 d i 1 z 3 lnsx 2 1 1d j 1 z k.
Find the flux of F across the part of the paraboloid
x 2 1 y 2 1 z − 2 that lies above the plane z − 1 and is
oriented upward.
19. A vector field F is shown. Use the interpretation of divergence
derived in this section to determine whether div F
is positive or negative at P 1 and at P 2.
P¡
_2 2
P
2
_2
20. (a) Are the points P 1 and P 2 sources or sinks for the vector
field F shown in the figure? Give an explanation based
solely on the picture.
(b) Given that Fsx, yd − kx, y 2 l, use the definition of divergence
to verify your answer to part (a).
P¡
_2 2
P
21–22 Plot the vector field and guess where div F . 0 and
where div F , 0. Then calculate div F to check your guess.
21. Fsx, yd − kxy, x 1 y 2 l 22. Fsx, yd − kx 2 , y 2 l
2
_2
23. Verify that div E − 0 for the electric field Esxd − «Q
| x | 3 x.
24. Use the Divergence Theorem to evaluate
yy s2x 1 2y 1 z 2 d dS
S
where S is the sphere x 2 1 y 2 1 z 2 − 1.
25–30 Prove each identity, assuming that S and E satisfy the
conditions of the Divergence Theorem and the scalar functions
and components of the vector fields have continuous second-order
partial derivatives.
25. yy a n dS − 0, where a is a constant vector
S
26. VsEd − 1 3 yy F dS, where Fsx, y, zd − x i 1 y j 1 z k
S
27. yy curl F dS − 0
S
29. y
S
30. y
S
y s f =td n dS − y y
y s f =t 2 t=f d n dS − y y
E
28. y
S
y D n f dS − y y
y s f = 2 t 1 =f =td dV
E
y s f = 2 t 2 t= 2 f d dV
E
y = 2 f dV
31. Suppose S and E satisfy the conditions of the Divergence Theorem
and f is a scalar function with continuous partial derivatives.
Prove that
y
S
y f n dS − y y
E
y =f dV
These surface and triple integrals of vector functions are
vectors defined by integrating each component function.
[Hint: Start by applying the Divergence Theorem to F − f c,
where c is an arbitrary constant vector.]
32. A solid occupies a region E with surface S and is immersed
in a liquid with constant density . We set up a coordinate
system so that the xy-plane coincides with the surface of the
liquid, and positive values of z are measured downward into the
liquid. Then the pressure at depth z is p − tz, where t is the
acceleration due to gravity (see Section 8.3). The total buoyant
force on the solid due to the pressure distribution is given by
the surface integral
F − 2yy pn dS
S
where n is the outer unit normal. Use the result of Exer cise 31
to show that F − 2Wk, where W is the weight of the liquid
displaced by the solid. (Note that F is directed upward because
z is directed downward.) The result is Archimedes’ Principle:
The buoyant force on an object equals the weight of the displaced
liquid.
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