1060
||||
CHAPTER 16 VECTOR CALCULUS
We now easily compute this last integral using the parametrization given by
r共t兲 苷 a cos t i a sin t j , 0 t 2. Thus
y
C
F ⴢ dr 苷 y F ⴢ dr 苷 y
C
苷y
2
0
2
0
F共r共t兲兲 ⴢ r共t兲 dt
2
共a sin t兲共a sin t兲 共a cos t兲共a cos t兲
dt 苷 y dt 苷 2
2
2
2
2
0
a cos t a sin t
M
We end this section by using Green’s Theorem to discuss a result that was stated in the
preceding section.
SKETCH OF PROOF OF THEOREM 16.3.6 We’re assuming that F 苷 P i Q j is a vector field on
an open simply-connected region D, that P and Q have continuous first-order partial
derivatives, and that
Q
P
苷
y
x
throughout D
If C is any simple closed path in D and R is the region that C encloses, then Green’s
Theorem gives
y
䊊
C
䊊 P dx Q dy 苷
F ⴢ dr 苷 y
C
yy
R
冉
Q
P
x
y
冊
dA 苷 yy 0 dA 苷 0
R
A curve that is not simple crosses itself at one or more points and can be broken up into
a number of simple curves. We have shown that the line integrals of F around these
simple curves are all 0 and, adding these integrals, we see that xC F ⴢ dr 苷 0 for any
closed curve C. Therefore xC F ⴢ dr is independent of path in D by Theorem 16.3.3. It
M
follows that F is a conservative vector field.
16.4
EXERCISES
1– 4 Evaluate the line integral by two methods: (a) directly and
6.
xC cos y dx 7.
xC ( y e sx ) dx 共2x cos y 2 兲 dy,
(b) using Green’s Theorem.
1.
2.
共x y兲 dx 共x y兲 dy,
C is the circle with center the origin and radius 2
x
䊊
C
x xy dx x dy,
C is the rectangle with vertices 共0, 0兲, 共3, 0兲, 共3, 1兲, and 共0, 1兲
C is the boundary of the region enclosed by the parabolas
y 苷 x 2 and x 苷 y 2
2
䊊
C
3.
x y dx x 2 y 3 dy,
C is the triangle with vertices 共0, 0兲, (1, 0), and (1, 2)
4.
䊊
C
8.
x dx y dy , C consists of the line segments from 共0, 1兲
to 共0, 0兲 and from 共0, 0兲 to 共1, 0兲 and the parabola y 苷 1 x 2
from 共1, 0兲 to 共0, 1兲
xC xe2x dx 共x 4 2x 2 y 2 兲 dy,
C is the boundary of the region between the circles
x 2 y 2 苷 1 and x 2 y 2 苷 4
x
䊊
C
x
x 2 sin y dy,
C is the rectangle with vertices 共0, 0兲, 共5, 0兲, 共5, 2兲, and 共0, 2兲
9.
10.
xC y 3 dx x 3 dy, C is the circle x 2 y 2 苷 4
xC sin y dx x cos y dy, C is the ellipse x 2 xy y 2 苷 1
11–14 Use Green’s Theorem to evaluate xC F ⴢ d r. (Check the
5–10 Use Green’s Theorem to evaluate the line integral along the
orientation of the curve before applying the theorem.)
given positively oriented curve.
11. F共x, y兲 苷 具 sx y 3, x 2 sy 典,
5.
dx 2 x y dy,
C is the triangle with vertices 共0, 0兲, 共2, 2兲, and 共2, 4兲
xC xy
2
2
C consists of the arc of the curve y 苷 sin x from 共0, 0兲 to 共, 0兲
and the line segment from 共, 0兲 to 共0, 0兲
SECTION 16.5 CURL AND DIVERGENCE
C is the triangle from 共0, 0兲 to 共2, 6兲 to 共2, 0兲 to 共0, 0兲
22. Let D be a region bounded by a simple closed path C in the
C is the circle x 2 y 2 苷 25 oriented clockwise
xy-plane. Use Green’s Theorem to prove that the coordinates
of the centroid 共 x, y 兲 of D are
1
14. F共x, y兲 苷 具 y ln共x y 兲, 2 tan 共y兾x兲典, C is the circle
2
2
共x 2兲2 共 y 3兲2 苷 1 oriented counterclockwise
CAS
x苷
15–16 Verify Green’s Theorem by using a computer algebra system to evaluate both the line integral and the double integral.
15. P共x, y兲 苷 y 2e x,
Q共x, y兲 苷 x 2e y,
C consists of the line segment from 共1, 1兲 to 共1, 1兲 followed
by the arc of the parabola y 苷 2 x 2 from 共1, 1兲 to 共1, 1兲
Q共x, y兲 苷 x 3 y 8,
C is the ellipse 4x 2 y 2 苷 4
19. Use one of the formulas in (5) to find the area under one arch
circle x 2 y 2 苷 16, a fixed point P on C traces out a
curve called an epicycloid, with parametric equations
x 苷 5 cos t cos 5t, y 苷 5 sin t sin 5t. Graph the epicycloid and use (5) to find the area it encloses.
䊊
C
y 2 dx
region in the x y-plane bounded by a simple closed path C.
Show that its moments of inertia about the axes are
Ix 苷 3
y
䊊
C
Iy 苷
y 3 dx
3
y
䊊
C
x 3 dy
26. Use Exercise 25 to find the moment of inertia of a circular
disk of radius a with constant density about a diameter.
(Compare with Example 4 in Section 15.5.)
27. If F is the vector field of Example 5, show that xC F ⴢ dr 苷 0
for every simple closed path that does not pass through or
enclose the origin.
28. Complete the proof of the special case of Green’s Theorem
by proving Equation 3.
29. Use Green’s Theorem to prove the change of variables
x dy y dx 苷 x 1 y 2 x 2 y1
(b) If the vertices of a polygon, in counterclockwise order,
are 共x 1, y1 兲, 共x 2 , y 2 兲, . . . , 共x n , yn 兲, show that the area of
the polygon is
A 苷 12 关共x 1 y 2 x 2 y1 兲 共x 2 y 3 x 3 y 2 兲 共x n1 yn x n yn1 兲 共x n y1 x 1 yn 兲兴
16.5
y
25. A plane lamina with constant density 共x, y兲 苷 occupies a
point 共x 2 , y2兲, show that
A苷
1
2A
vertices 共0, 0兲, 共a, 0兲, and 共a, b兲, where a 0 and b 0.
21. (a) If C is the line segment connecting the point 共x 1, y1兲 to the
C
y苷
x 2 dy
24. Use Exercise 22 to find the centroid of the triangle with
of the cycloid x 苷 t sin t, y 苷 1 cos t.
; 20. If a circle C with radius 1 rolls along the outside of the
䊊
C
region of radius a.
F共x, y兲 苷 x共x y兲 i x y 2 j in moving a particle from the
origin along the x-axis to 共1, 0兲, then along the line segment
to 共0, 1兲, and then back to the origin along the y-axis.
to 共2, 0兲, and then along the semicircle y 苷 s4 x 2 to the
starting point. Use Green’s Theorem to find the work done on
this particle by the force field F共x, y兲 苷 具x, x 3 3x y 2 典 .
y
23. Use Exercise 22 to find the centroid of a quarter-circular
17. Use Green’s Theorem to find the work done by the force
18. A particle starts at the point 共2, 0兲, moves along the x-axis
1
2A
where A is the area of D.
16. P共x, y兲 苷 2x x 3 y 5,
y
1061
(c) Find the area of the pentagon with vertices 共0, 0兲, 共2, 1兲,
共1, 3兲, 共0, 2兲, and 共1, 1兲.
12. F共x, y兲 苷 具 y 2 cos x, x 2 2y sin x典,
13. F共x, y兲 苷 具e x x 2 y, e y xy 2 典,
||||
formula for a double integral (Formula 15.9.9) for the case
where f 共x, y兲 苷 1:
yy dx dy 苷 yy
R
S
冟
冟
共x, y兲
du dv
共u, v兲
Here R is the region in the xy-plane that corresponds to the
region S in the uv-plane under the transformation given by
x 苷 t共u, v兲, y 苷 h共u, v兲.
[Hint: Note that the left side is A共R兲 and apply the first part
of Equation 5. Convert the line integral over R to a line integral over S and apply Green’s Theorem in the uv-plane.]
CURL AND DIVERGENCE
In this section we define two operations that can be performed on vector fields and that
play a basic role in the applications of vector calculus to fluid flow and electricity and magnetism. Each operation resembles differentiation, but one produces a vector field whereas
the other produces a scalar field.
1078
||||
CHAPTER 16 VECTOR CALCULUS
Converting to polar coordinates, we obtain
A苷y
2
0
y
3
s1 4r 2 r dr d 苷 y
2
0
0
苷 2 ( 18 ) 23 共1 4r 2 兲3兾2
]
3
0
d
y
3
0
rs1 4r 2 dr
(37s37 1)
6
苷
M
The question remains whether our definition of surface area (6) is consistent with the
surface area formula from single-variable calculus (8.2.4).
We consider the surface S obtained by rotating the curve y 苷 f 共x兲, a x b, about
the x-axis, where f 共x兲 0 and f is continuous. From Equations 3 we know that parametric equations of S are
z 苷 f 共x兲 sin y 苷 f 共x兲 cos x苷x
axb
0 2
To compute the surface area of S we need the tangent vectors
rx 苷 i f 共x兲 cos j f 共x兲 sin k
r 苷 f 共x兲 sin j f 共x兲 cos k
ⱍ
Thus
i
j
rx r 苷 1 f 共x兲 cos 0 f 共x兲 sin k
f 共x兲 sin f 共x兲 cos ⱍ
苷 f 共x兲 f 共x兲 i f 共x兲 cos j f 共x兲 sin k
and so
ⱍr
x
ⱍ
r 苷 s关 f 共x兲兴 2 关 f 共x兲兴 2 关 f 共x兲兴 2 cos 2 关 f 共x兲兴 2 sin 2
苷 s关 f 共x兲兴 2 关1 关 f 共x兲兴 2 兴 苷 f 共x兲s1 关 f 共x兲兴 2
because f 共x兲 0. Therefore the area of S is
ⱍ
ⱍ
A 苷 yy rx r dA 苷 y
D
2
0
y
b
a
f 共x兲s1 关 f 共x兲兴 2 dx d
b
苷 2 y f 共x兲s1 关 f 共x兲兴 2 dx
a
This is precisely the formula that was used to define the area of a surface of revolution in
single-variable calculus (8.2.4).
16.6
EXERCISES
1–2 Determine whether the points P and Q lie on the given surface.
1. r共u, v兲 苷 具2u 3v, 1 5u v, 2 u v 典
5. r共s, t兲 苷 具s, t, t 2 s 2 典
6. r共s, t兲 苷 具s sin 2t, s 2, s cos 2t典
P共7, 10, 4兲, Q共5, 22, 5兲
2. r共u, v兲 苷 具u v, u 2 v, u v 2 典
; 7–12 Use a computer to graph the parametric surface. Get a printout
P共3, 1, 5兲, Q共1, 3, 4兲
and indicate on it which grid curves have u constant and which have
v constant.
3–6 Identify the surface with the given vector equation.
7. r共u, v兲 苷 具u 2 1, v 3 1, u v 典,
8. r共u, v兲 苷 具u v, u , v 典,
4. r共u, v兲 苷 2 sin u i 3 cos u j v k,
9. r共u, v兲 苷 具u cos v, u sin v, u 典 ,
0v2
2
1 u 1, 1 v 1
1 u 1, 1 v 1
3. r共u, v兲 苷 共u v兲 i 共3 v兲 j 共1 4u 5v兲 k
2
5
1 u 1, 0 v 2
SECTION 16.6 PARAMETRIC SURFACES AND THEIR AREAS
10. r共u, v兲 苷 具cos u sin v, sin u sin v, cos v ln tan共v兾2兲典 ,
0 u 2, 0.1 v 6.2
19. The plane that passes through the point 共1, 2, 3兲 and
contains the vectors i j k and i j k
y 苷 cos u sin 4 v, z 苷 sin 2u sin 4 v,
0 u 2, 兾2 v 兾2
y 苷 u cos u cos v,
1079
19–26 Find a parametric representation for the surface.
11. x 苷 sin v,
12. x 苷 u sin u cos v,
||||
20. The lower half of the ellipsoid 2 x 2 4y 2 z 2 苷 1
z 苷 u sin v
21. The part of the hyperboloid x 2 y 2 z 2 苷 1 that lies to the
right of the xz-plane
22. The part of the elliptic paraboloid x y 2 2z 2 苷 4 that lies
13–18 Match the equations with the graphs labeled I–VI and
in front of the plane x 苷 0
give reasons for your answers. Determine which families of grid
curves have u constant and which have v constant.
23. The part of the sphere x 2 y 2 z 2 苷 4 that lies above the
cone z 苷 sx 2 y 2
13. r共u, v兲 苷 u cos v i u sin v j v k
14. r共u, v兲 苷 u cos v i u sin v j sin u k,
24. The part of the sphere x 2 y 2 z 2 苷 16 that lies between
u the planes z 苷 2 and z 苷 2
15. r共u, v兲 苷 sin v i cos u sin 2v j sin u sin 2v k
25. The part of the cylinder y 2 z 2 苷 16 that lies between the
16. x 苷 共1 u兲共3 cos v兲 cos 4 u,
y 苷 共1 u兲共3 cos v兲 sin 4 u,
z 苷 3u 共1 u兲 sin v
17. x 苷 cos 3 u cos 3 v,
26. The part of the plane z 苷 x 3 that lies inside the cylinder
x2 y2 苷 1
y 苷 sin 3 u cos 3 v,
ⱍ ⱍ
18. x 苷 共1 u 兲 cos v,
z 苷 sin 3 v
ⱍ ⱍ
y 苷 共1 u 兲 sin v, z 苷 u
z
I
planes x 苷 0 and x 苷 5
II
CAS
27–28 Use a computer algebra system to produce a graph that
looks like the given one.
27.
z
28.
3
x
y
y
III
z
z 0
x
_3
_3
IV
z
z
0
y
0 5
0
_1
_1
x
_1
y
0
0
1 1
x
; 29. Find parametric equations for the surface obtained by rotating
the curve y 苷 e x, 0 x 3, about the x-axis and use them
to graph the surface.
; 30. Find parametric equations for the surface obtained by rotating
x
y
x
y
the curve x 苷 4y 2 y 4, 2 y 2, about the y-axis and
use them to graph the surface.
; 31. (a) What happens to the spiral tube in Example 2 (see Figz
V
VI
ure 5) if we replace cos u by sin u and sin u by cos u ?
(b) What happens if we replace cos u by cos 2u and sin u
by sin 2u?
z
; 32. The surface with parametric equations
x 苷 2 cos r cos共兾2兲
x
y 苷 2 sin r cos共兾2兲
y
x
y
z 苷 r sin共兾2兲
where 12 r 12 and 0 2, is called a Möbius
strip. Graph this surface with several viewpoints. What is
unusual about it?
1080
||||
CHAPTER 16 VECTOR CALCULUS
33–36 Find an equation of the tangent plane to the given para-
metric surface at the specified point. If you have software that
graphs parametric surfaces, use a computer to graph the surface
and the tangent plane.
33. x 苷 u v,
34. x 苷 u ,
2
y 苷 3u ,
2
y苷v ,
2
z 苷 u v;
51. (a) Use the Midpoint Rule for double integrals (see Sec-
CAS
共2, 3, 0兲
z 苷 u v ; u 苷 1, v 苷 1
35. r共u, v兲 苷 u i 2u sin v j u cos v k;
u 苷 1, v 苷 0
2
CAS
u 苷 0, v 苷 36. r共u, v兲 苷 u v i u sin v j v cos u k;
first octant
54. (a) Set up, but do not evaluate, a double integral for the area
of the surface with parametric equations x 苷 au cos v,
y 苷 bu sin v, z 苷 u 2, 0 u 2, 0 v 2.
38. The part of the plane 2x 5y z 苷 10 that lies inside the
cylinder x 2 y 2 苷 9
39. The surface z 苷 3 共x 3兾2 y 3兾2 兲, 0 x 1, 0 y 1
2
;
CAS
41. The part of the surface z 苷 xy that lies within the cylinder
x2 y2 苷 1
triangle with vertices 共0, 0兲, 共0, 1兲, and 共2, 1兲
between the cylinders x 2 y 2 苷 1 and x 2 y 2 苷 4
;
44. The part of the paraboloid x 苷 y 2 z 2 that lies inside the
cylinder y 2 z 2 苷 9
planes x 苷 0, x 苷 1, z 苷 0, and z 苷 1
47. The surface with parametric equations x 苷 u 2 , y 苷 u v,
z 苷 12 v 2, 0 u 1, 0 v 2
represent an ellipsoid.
(b) Use the parametric equations in part (a) to graph the ellipsoid for the case a 苷 1, b 苷 2, c 苷 3.
(c) Set up, but do not evaluate, a double integral for the surface area of the ellipsoid in part (b).
56. (a) Show that the parametric equations x 苷 a cosh u cos v,
y 苷 b cosh u sin v, z 苷 c sinh u, represent a hyperboloid
45. The part of the surface y 苷 4x z 2 that lies between the
46. The helicoid (or spiral ramp) with vector equation
r共u, v兲 苷 u cos v i u sin v j v k, 0 u 1, 0 v (b) Eliminate the parameters to show that the surface is an
elliptic paraboloid and set up another double integral for
the surface area.
(c) Use the parametric equations in part (a) with a 苷 2 and
b 苷 3 to graph the surface.
(d) For the case a 苷 2, b 苷 3, use a computer algebra
system to find the surface area correct to four decimal
places.
55. (a) Show that the parametric equations x 苷 a sin u cos v,
y 苷 b sin u sin v, z 苷 c cos u, 0 u , 0 v 2,
42. The part of the surface z 苷 1 3x 2y 2 that lies above the
43. The part of the hyperbolic paraboloid z 苷 y 2 x 2 that lies
53. Find the exact area of the surface z 苷 1 2x 3y 4y 2,
1 x 4, 0 y 1.
37. The part of the plane 3x 2y z 苷 6 that lies in the
40. The part of the plane with vector equation
r共u, v兲 苷 具1 v, u 2v, 3 5u v 典 that is given by
0 u 1, 0 v 1
52. Find the area of the surface with vector equation
r共u, v兲 苷 具cos 3u cos 3v, sin 3u cos 3v, sin 3v 典 , 0 u ,
0 v 2. State your answer correct to four decimal
places.
CAS
37– 47 Find the area of the surface.
tion 15.1) with six squares to estimate the area of the
surface z 苷 1兾共1 x 2 y 2 兲, 0 x 6, 0 y 4.
(b) Use a computer algebra system to approximate the
surface area in part (a) to four decimal places. Compare
with the answer to part (a).
;
of one sheet.
(b) Use the parametric equations in part (a) to graph the
hyperboloid for the case a 苷 1, b 苷 2, c 苷 3.
(c) Set up, but do not evaluate, a double integral for the surface area of the part of the hyperboloid in part (b) that lies
between the planes z 苷 3 and z 苷 3.
57. Find the area of the part of the sphere x 2 y 2 z 2 苷 4z that
48 – 49 Find the area of the surface correct to four decimal places
by expressing the area in terms of a single integral and using your
calculator to estimate the integral.
48. The part of the surface z 苷 cos共x y 兲 that lies inside the
2
2
lies inside the paraboloid z 苷 x 2 y 2.
58. The figure shows the surface created when the cylinder
y 2 z 2 苷 1 intersects the cylinder x 2 z 2 苷 1. Find the
area of this surface.
cylinder x 2 y 2 苷 1
z
x 2y 2
49. The part of the surface z 苷 e
that lies above the disk
x2 y2 4
CAS
50. Find, to four decimal places, the area of the part of the
surface z 苷 共1 x 2 兲兾共1 y 2 兲 that lies above the square
x y 1. Illustrate by graphing this part of the surface.
ⱍ ⱍ ⱍ ⱍ
x
y
SECTION 16.7 SURFACE INTEGRALS
16.7
||||
1091
EXERCISES
1. Let S be the boundary surface of the box enclosed by the
planes x 苷 0, x 苷 2, y 苷 0, y 苷 4, z 苷 0, and z 苷 6. Approximate xxS e0.1共xyz兲 dS by using a Riemann sum as in Definition 1, taking the patches Sij to be the rectangles that are the
faces of the box S and the points Pij* to be the centers of the
rectangles.
13.
S is the part of the paraboloid y 苷 x 2 z 2 that lies inside the
cylinder x 2 z 2 苷 4
14.
together with its top and bottom disks. Suppose you know that
f is a continuous function with
15.
f 共1, 0, 0兲 苷 2
16.
f 共0, 0, 1兲 苷 4
Estimate the value of xxS f 共x, y, z兲 dS by using a Riemann sum,
taking the patches Sij to be four quarter-cylinders and the top
and bottom disks.
xxS 共x 2 z y 2 z兲 dS,
S is the hemisphere x 2 y 2 z 2 苷 4, z 0
xxS xz dS,
S is the boundary of the region enclosed by the cylinder
y 2 z 2 苷 9 and the planes x 苷 0 and x y 苷 5
17.
xxS 共z x 2 y兲 dS,
S is the part of the cylinder y 2 z 2 苷 1 that lies between the
planes x 苷 0 and x 苷 3 in the first octant
3. Let H be the hemisphere x 2 y 2 z 2 苷 50, z 0, and
suppose f is a continuous function with f 共3, 4, 5兲 苷 7,
f 共3, 4, 5兲 苷 8, f 共3, 4, 5兲 苷 9, and f 共3, 4, 5兲 苷 12.
By dividing H into four patches, estimate the value of
xxH f 共x, y, z兲 dS.
xxS y 2 dS,
S is the part of the sphere x 2 y 2 z 2 苷 4 that lies
inside the cylinder x 2 y 2 苷 1 and above the xy-plane
2. A surface S consists of the cylinder x 2 y 2 苷 1, 1 z 1,
f 共0, 1, 0兲 苷 3
xxS y dS,
18.
xxS 共x 2 y 2 z 2 兲 dS,
S is the part of the cylinder x 2 y 2 苷 9 between the planes
z 苷 0 and z 苷 2, together with its top and bottom disks
4. Suppose that f 共x, y, z兲 苷 t(sx 2 y 2 z 2 ), where t is a
function of one variable such that t共2兲 苷 5. Evaluate
xxS f 共x, y, z兲 dS, where S is the sphere x 2 y 2 z 2 苷 4.
5–18 Evaluate the surface integral.
5.
xxS x 2 yz dS,
19. F共x, y, z兲 苷 x y i yz j zx k,
xxS x y dS,
20. F共x, y, z兲 苷 y i x j z 2 k,
S is the part of the plane z 苷 1 2x 3y that lies above the
rectangle 关0, 3兴 关0, 2兴
6.
S is the triangular region with vertices (1, 0, 0), (0, 2, 0),
and (0, 0, 2)
7.
xxS yz dS,
S is the part of the plane x y z 苷 1 that lies in the
first octant
8.
xxS y dS,
S is the surface z 苷 23 共x 3兾2 y 3兾2 兲, 0 x 1, 0 y 1
9.
19–30 Evaluate the surface integral xxS F ⴢ dS for the given vector
field F and the oriented surface S. In other words, find the flux of F
across S. For closed surfaces, use the positive (outward) orientation.
S is the part of the
paraboloid z 苷 4 x 2 y 2 that lies above the square
0 x 1, 0 y 1, and has upward orientation
S is the helicoid of Exercise 10 with upward orientation
21. F共x, y, z兲 苷 xze y i xze y j z k,
S is the part of the plane x y z 苷 1 in the first octant and
has downward orientation
22. F共x, y, z兲 苷 x i y j z 4 k,
S is the part of the cone z 苷 sx 2 y 2 beneath the plane z 苷 1
with downward orientation
23. F共x, y, z兲 苷 x i z j y k,
xxS yz dS,
S is the surface with parametric equations x 苷 u , y 苷 u sin v,
z 苷 u cos v, 0 u 1, 0 v 兾2
2
S is the part of the sphere x 2 y 2 z 2 苷 4 in the first octant,
with orientation toward the origin
10.
xxS s1 x 2 y 2
24. F共x, y, z兲 苷 xz i x j y k,
11.
xxS x 2 z 2 dS,
25. F共x, y, z兲 苷 y j z k,
dS,
S is the helicoid with vector equation
r共u, v兲 苷 u cos v i u sin v j v k, 0 u 1, 0 v S is the part of the cone z 2 苷 x 2 y 2 that lies between the
planes z 苷 1 and z 苷 3
12.
xxS z dS,
S is the surface x 苷 y 2z , 0 y 1, 0 z 1
2
S is the hemisphere x 2 y 2 z 2 苷 25, y 0, oriented in the
direction of the positive y-axis
S consists of the paraboloid y 苷 x 2 z 2, 0 y 1,
and the disk x 2 z 2 1, y 苷 1
26. F共x, y, z兲 苷 x y i 4 x 2 j yz k,
S is the surface z 苷 xe y,
0 x 1, 0 y 1, with upward orientation
1092
||||
CHAPTER 16 VECTOR CALCULUS
27. F共x, y, z兲 苷 x i 2y j 3z k,
39. (a) Give an integral expression for the moment of inertia I z
S is the cube with vertices 共1, 1, 1兲
28. F共x, y, z兲 苷 x i y j 5 k,
S is the boundary of the region
enclosed by the cylinder x 2 z 2 苷 1 and the planes y 苷 0
and x y 苷 2
29. F共x, y, z兲 苷 x 2 i y 2 j z 2 k,
S is the boundary of the
solid half-cylinder 0 z s1 y 2 , 0 x 2
30. F共x, y, z兲 苷 y i 共z y兲 j x k,
S is the surface of the tetrahedron with vertices 共0, 0, 0兲,
共1, 0, 0兲, 共0, 1, 0兲, and 共0, 0, 1兲
CAS
31. Evaluate xxS xyz dS correct to four decimal places, where S is
the surface z 苷 xy, 0 x 1, 0 y 1.
CAS
32. Find the exact value of xxS x 2 yz dS, where S is the surface in
Exercise 31.
CAS
33. Find the value of xxS x 2 y 2z 2 dS correct to four decimal places,
where S is the part of the paraboloid z 苷 3 2x 2 y 2 that
lies above the x y-plane.
CAS
about the z -axis of a thin sheet in the shape of a surface S
if the density function is .
(b) Find the moment of inertia about the z -axis of the funnel
in Exercise 38.
40. Let S be the part of the sphere x 2 y 2 z 2 苷 25 that lies
above the plane z 苷 4. If S has constant density k, find
(a) the center of mass and (b) the moment of inertia about
the z-axis.
41. A fluid has density 870 kg兾m3 and flows with velocity
v 苷 z i y 2 j x 2 k , where x, y, and z are measured in
meters and the components of v in meters per second. Find
the rate of flow outward through the cylinder x 2 y 2 苷 4 ,
0 z 1.
42. Seawater has density 1025 kg兾m3 and flows in a velocity field
v 苷 y i x j, where x, y, and z are measured in meters and
the components of v in meters per second. Find the rate of
flow outward through the hemisphere x 2 y 2 z 2 苷 9 ,
z 0.
43. Use Gauss’s Law to find the charge contained in the solid
34. Find the flux of
F共x, y, z兲 苷 sin共x yz兲 i x 2 y j z 2e x兾5 k
across the part of the cylinder 4y 2 z 2 苷 4 that lies above
the xy-plane and between the planes x 苷 2 and x 苷 2 with
upward orientation. Illustrate by using a computer algebra
system to draw the cylinder and the vector field on the same
screen.
35. Find a formula for xxS F ⴢ dS similar to Formula 10 for the
case where S is given by y 苷 h共x, z兲 and n is the unit normal
that points toward the left.
36. Find a formula for xxS F ⴢ dS similar to Formula 10 for the
case where S is given by x 苷 k共 y, z兲 and n is the unit normal
that points forward (that is, toward the viewer when the axes
are drawn in the usual way).
37. Find the center of mass of the hemisphere x 2 y 2 z 2 苷 a 2,
z 0, if it has constant density.
E共x, y, z兲 苷 x i y j 2z k
44. Use Gauss’s Law to find the charge enclosed by the cube
with vertices 共1, 1, 1兲 if the electric field is
E共x, y, z兲 苷 x i y j z k
45. The temperature at the point 共x, y, z兲 in a substance with con-
ductivity K 苷 6.5 is u共x, y, z兲 苷 2y 2 2z 2. Find the rate of
heat flow inward across the cylindrical surface y 2 z 2 苷 6,
0 x 4.
46. The temperature at a point in a ball with conductivity K is
inversely proportional to 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.
ⱍ ⱍ
47. Let F be an inverse square field, that is, F共r兲 苷 cr兾 r
38. Find the mass of a thin funnel in the shape of a cone
z 苷 sx 2 y 2 , 1 z 4, if its density function is
共x, y, z兲 苷 10 z.
16.8
hemisphere x 2 y 2 z 2 a 2, z 0, if the electric field is
3
for
some constant c, where r 苷 x i y j z k. Show that the
flux of F across a sphere S with center the origin is independent of the radius of S.
STOKES’ THEOREM
Stokes’ Theorem can be regarded as a higher-dimensional version of Green’s Theorem.
Whereas Green’s Theorem relates a double integral over a plane region D to a line integral
around its plane boundary curve, Stokes’ Theorem relates a surface integral over a surface
S to a line integral around the boundary curve of S (which is a space curve). Figure 1 shows
SECTION 16.8 STOKES’ THEOREM
16.8
||||
1097
EXERCISES
10. F共x, y, z兲 苷 xy i 2z j 3y k,
1. A hemisphere H and a portion P of a paraboloid are shown.
C is the curve of intersection
of the plane x z 苷 5 and the cylinder x 2 y 2 苷 9
Suppose F is a vector field on ⺢3 whose components have continuous partial derivatives. Explain why
11. (a) Use Stokes’ Theorem to evaluate xC F ⴢ dr, where
yy curl F ⴢ dS 苷 yy curl F ⴢ dS
H
P
F共x, y, z兲 苷 x 2 z i x y 2 j z 2 k
z
z
4
4
;
P
H
;
and C is the curve of intersection of the plane
x y z 苷 1 and the cylinder x 2 y 2 苷 9 oriented
counterclockwise as viewed from above.
(b) Graph both the plane and the cylinder with domains
chosen so that you can see the curve C and the surface
that you used in part (a).
(c) Find parametric equations for C and use them to graph C.
12. (a) Use Stokes’ Theorem to evaluate xC F ⴢ dr, where
x
2
2
x
y
2
2
y
2–6 Use Stokes’ Theorem to evaluate xxS curl F ⴢ dS.
;
2. F共x, y, z兲 苷 2y cos z i e sin z j xe k,
x
y
S is the hemisphere x 2 y 2 z 2 苷 9, z 0, oriented
upward
3. F共x, y, z兲 苷 x 2 z 2 i y 2z 2 j xyz k,
S is the part of the paraboloid z 苷 x 2 y 2 that lies inside the
cylinder x 2 y 2 苷 4, oriented upward
4. F共x, y, z兲 苷 x y z i sin共x yz兲 j x yz k,
2
3
S is the part of the cone y 苷 x z that lies between the
planes y 苷 0 and y 苷 3, oriented in the direction of the
positive y-axis
2
2
2
5. F共x, y, z兲 苷 x yz i x y j x 2 yz k,
S consists of the top and the four sides (but not the bottom)
of the cube with vertices 共1, 1, 1兲, oriented outward
[Hint: Use Equation 3.]
;
F共x, y, z兲 苷 x 2 y i 13 x 3 j x y k and C is the curve of
intersection of the hyperbolic paraboloid z 苷 y 2 x 2 and
the cylinder x 2 y 2 苷 1 oriented counterclockwise as
viewed from above.
(b) Graph both the hyperbolic paraboloid and the cylinder with
domains chosen so that you can see the curve C and the
surface that you used in part (a).
(c) Find parametric equations for C and use them to graph C.
13–15 Verify that Stokes’ Theorem is true for the given vector
field F and surface S.
13. F共x, y, z兲 苷 y 2 i x j z 2 k,
S is the part of the paraboloid z 苷 x 2 y 2 that lies below the
plane z 苷 1, oriented upward
14. F共x, y, z兲 苷 x i y j x yz k,
S is the part of the plane 2x y z 苷 2 that lies in the first
octant, oriented upward
15. F共x, y, z兲 苷 y i z j x k,
S is the hemisphere x 2 y 2 z 2 苷 1, y 0, oriented in the
direction of the positive y-axis
6. F共x, y, z兲 苷 e xy cos z i x 2 z j x y k,
S is the hemisphere x 苷 s1 y 2 z 2 , oriented in the direction of the positive x-axis [Hint: Use Equation 3.]
7–10 Use Stokes’ Theorem to evaluate xC F ⴢ dr. In each case C is
oriented counterclockwise as viewed from above.
7. F共x, y, z兲 苷 共x y 2 兲 i 共 y z 2 兲 j 共z x 2 兲 k,
16. Let C be a simple closed smooth curve that lies in the plane
x y z 苷 1. Show that the line integral
xC z dx 2x dy 3y dz
depends only on the area of the region enclosed by C and not
on the shape of C or its location in the plane.
C is the triangle with vertices (1, 0, 0), (0, 1, 0), and (0, 0, 1)
8. F共x, y, z兲 苷 ex i e x j e z k,
C is the boundary of the part of the plane 2x y 2z 苷 2
in the first octant
9. F共x, y, z兲 苷 yz i 2 xz j e xy k,
C is the circle x 2 y 2 苷 16, z 苷 5
17. A particle moves along line segments from the origin to
the points 共1, 0, 0兲, 共1, 2, 1兲, 共0, 2, 1兲, and back to the
origin under the influence of the force field
F共x, y, z兲 苷 z 2 i 2xy j 4y 2 k
Find the work done.
1098
||||
CHAPTER 16 VECTOR CALCULUS
20. Suppose S and C satisfy the hypotheses of Stokes’ Theorem
18. Evaluate
xC 共 y ⫹ sin x兲 dx ⫹ 共z
2
⫹ cos y兲 dy ⫹ x dz
3
where C is the curve r共t兲 苷 具sin t, cos t, sin 2t典 , 0 艋 t 艋 2␲.
[Hint: Observe that C lies on the surface z 苷 2 x y.]
19. If S is a sphere and F satisfies the hypotheses of Stokes’
Theorem, show that xxS curl F ⴢ dS 苷 0.
WRITING
PROJECT
The photograph shows a stained-glass
window at Cambridge University in honor of
George Green.
N
Courtesy of the Masters and Fellows of Gonville and
Caius College, University of Cambridge, England
and f , t have continuous second-order partial derivatives. Use
Exercises 24 and 26 in Section 16.5 to show the following.
(a) xC 共 f ⵜt兲 ⴢ dr 苷 xxS 共ⵜ f ⫻ ⵜt兲 ⴢ dS
(b)
xC 共 f ⵜ f 兲 ⴢ dr 苷 0
(c)
xC 共 f ⵜt ⫹ t ⵜ f 兲 ⴢ dr 苷 0
THREE MEN AND TWO THEOREMS
Although two of the most important theorems in vector calculus are named after George Green
and George Stokes, a third man, William Thomson (also known as Lord Kelvin), played a large
role in the formulation, dissemination, and application of both of these results. All three men
were interested in how the two theorems could help to explain and predict physical phenomena
in electricity and magnetism and fluid flow. The basic facts of the story are given in the margin
notes on pages 1056 and 1093.
Write a report on the historical origins of Green’s Theorem and Stokes’ Theorem. Explain the
similarities and relationship between the theorems. Discuss the roles that Green, Thomson, and
Stokes played in discovering these theorems and making them widely known. Show how both
theorems arose from the investigation of electricity and magnetism and were later used to study a
variety of physical problems.
The dictionary edited by Gillispie [2] is a good source for both biographical and scientific
information. The book by Hutchinson [5] gives an account of Stokes’ life and the book by
Thompson [8] is a biography of Lord Kelvin. The articles by Grattan-Guinness [3] and Gray [4]
and the book by Cannell [1] give background on the extraordinary life and works of Green.
Additional historical and mathematical information is found in the books by Katz [6] and
Kline [7].
1. D. M. Cannell, George Green, Mathematician and Physicist 1793–1841: The Background to
His Life and Work (Philadelphia: Society for Industrial and Applied Mathematics, 2001).
2. C. C. Gillispie, ed., Dictionary of Scientific Biography (New York: Scribner’s, 1974). See the
article on Green by P. J. Wallis in Volume XV and the articles on Thomson by Jed Buchwald
and on Stokes by E. M. Parkinson in Volume XIII.
www.stewartcalculus.com
The Internet is another source of information for this project. Click on History
of Mathematics. Follow the links to the
St. Andrew’s site and that of the British
Society for the History of Mathematics.
3. I. Grattan-Guinness, “Why did George Green write his essay of 1828 on electricity and
magnetism?” Amer. Math. Monthly, Vol. 102 (1995), pp. 387–396.
4. J. Gray, “There was a jolly miller.” The New Scientist, Vol. 139 (1993), pp. 24–27.
5. G. E. Hutchinson, The Enchanted Voyage and Other Studies (Westport, CT : Greenwood
Press, 1978).
6. Victor Katz, A History of Mathematics: An Introduction (New York: HarperCollins, 1993),
pp. 678–680.
7. Morris Kline, Mathematical Thought from Ancient to Modern Times (New York: Oxford
University Press, 1972), pp. 683–685.
8. Sylvanus P. Thompson, The Life of Lord Kelvin (New York: Chelsea, 1976).
SECTION 16.9 THE DIVERGENCE THEOREM
||||
1103
the origin. [This is a special case of Gauss’s Law (Equation 16.7.11) for a single charge.
The relationship between ␧ and ␧0 is ␧ 苷 1兾共4␲ ␧0 兲.]
Another application of the Divergence Theorem occurs in fluid flow. Let v共x, y, z兲 be
the velocity field of a fluid with constant density ␳. Then F 苷 ␳ v is the rate of flow per
unit area. If P0共x 0 , y0 , z0 兲 is a point in the fluid and Ba is a ball with center P0 and very small
radius a, then div F共P兲 ⬇ div F共P0 兲 for all points in Ba since div F is continuous. We
approximate the flux over the boundary sphere Sa as follows:
yy F ⴢ dS 苷 yyy div F dV 苷 yyy div F共P 兲 dV 苷 div F共P 兲V共B 兲
0
Sa
Ba
0
a
Ba
This approximation becomes better as a l 0 and suggests that
y
8
P¡
x
P™
FIGURE 4
The vector field F=≈ i+¥ j
16.9
div F共P0 兲 苷 lim
al0
1
V共Ba 兲
yy F ⴢ dS
Sa
Equation 8 says that div F共P0 兲 is the net rate of outward flux per unit volume at P0. (This
is the reason for the name divergence.) If div F共P兲 ⬎ 0, the net flow is outward near P and
P is called a source. If div F共P兲 ⬍ 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 P1 are shorter
than the vectors that start near P1. Thus the net flow is outward near P1, so div F共P1兲 ⬎ 0
and P1 is a source. Near P2 , on the other hand, the incoming arrows are longer than the
outgoing arrows. Here the net flow is inward, so div F共P2 兲 ⬍ 0 and P2 is a sink. We
can use the formula for F to confirm this impression. Since F 苷 x 2 i ⫹ y 2 j, we have
div F 苷 2x ⫹ 2y, which is positive when y ⬎ ⫺x. So the points above the line y 苷 ⫺x
are sources and those below are sinks.
EXERCISES
1– 4 Verify that the Divergence Theorem is true for the vector field
F on the region E.
1. F共x, y, z兲 苷 3x i ⫹ x y j ⫹ 2 xz k,
E is the cube bounded by the planes x 苷 0, x 苷 1, y 苷 0,
y 苷 1, z 苷 0, and z 苷 1
2. F共x, y, z兲 苷 x 2 i ⫹ x y j ⫹ z k,
E is the solid bounded by the paraboloid z 苷 4 ⫺ x 2 ⫺ y 2
and the xy-plane
3. F共x, y, z兲 苷 x y i ⫹ yz j ⫹ zx k,
E is the solid cylinder x 2 ⫹ y 2 艋 1, 0 艋 z 艋 1
4. F共x, y, z兲 苷 x i ⫹ y j ⫹ z k,
E is the unit ball x 2 ⫹ y 2 ⫹ z 2 艋 1
5–15 Use the Divergence Theorem to calculate the surface integral
xxS F ⴢ dS; that is, calculate the flux of F across S.
2
z
3
7. F共x, y, z兲 苷 3x y i ⫹ xe j ⫹ z k,
S is the surface of the solid bounded by the cylinder
y 2 ⫹ z 2 苷 1 and the planes x 苷 ⫺1 and x 苷 2
8. F共x, y, z兲 苷 x 3 y i ⫺ x 2 y 2 j ⫺ x 2 yz k,
S is the surface of the solid bounded by the hyperboloid
x 2 ⫹ y 2 ⫺ z 2 苷 1 and the planes z 苷 ⫺2 and z 苷 2
9. F共x, y, z兲 苷 x y sin z i ⫹ cos共x z兲 j ⫹ y cos z k,
S is the ellipsoid x 2兾a 2 ⫹ y 2兾b 2 ⫹ z 2兾c 2 苷 1
10. F共x, y, z兲 苷 x 2 y i ⫹ x y 2 j ⫹ 2 x yz k,
S is the surface of the tetrahedron bounded by the planes
x 苷 0, y 苷 0, z 苷 0, and x ⫹ 2y ⫹ z 苷 2
11. F共x, y, z兲 苷 共cos z ⫹ x y 2 兲 i ⫹ xe⫺z j ⫹ 共sin y ⫹ x 2 z兲 k,
S is the surface of the solid bounded by the paraboloid
z 苷 x 2 ⫹ y 2 and the plane z 苷 4
5. F共x, y, z兲 苷 e x sin y i ⫹ e x cos y j ⫹ yz 2 k,
12. F共x, y, z兲 苷 x 4 i ⫺ x 3z 2 j ⫹ 4 x y 2z k,
6. F共x, y, z兲 苷 x 2z 3 i ⫹ 2 x yz 3 j ⫹ xz 4 k,
13. F共x, y, z兲 苷 4 x 3z i ⫹ 4 y 3z j ⫹ 3z 4 k,
S is the surface of the box bounded by the planes x 苷 0,
x 苷 1, y 苷 0, y 苷 1, z 苷 0, and z 苷 2
S is the surface of the box with vertices 共⫾1, ⫾2, ⫾3兲
S is the surface of the solid bounded by the cylinder
x 2 ⫹ y 2 苷 1 and the planes z 苷 x ⫹ 2 and z 苷 0
S is the sphere with radius R and center the origin
1104
||||
CHAPTER 16 VECTOR CALCULUS
ⱍ ⱍ
14. F 苷 r兾 r , where r 苷 x i ⫹ y j ⫹ z k,
S consists of the hemisphere z 苷 s1 ⫺ x ⫺ y and the
disk x 2 ⫹ y 2 艋 1 in the xy-plane
CAS
2
15. F共x, y, z兲 苷 e y tan z i ⫹ y s3 ⫺ x 2 j ⫹ x sin y k,
S is the surface of the solid that lies above the xy-plane
and below the surface z 苷 2 ⫺ x 4 ⫺ y 4, ⫺1 艋 x 艋 1,
⫺1 艋 y 艋 1
CAS
22. F共x, y兲 苷 具x 2, y 2 典
2
16. Use a computer algebra system to plot the vector field
F共x, y, z兲 苷 sin x cos 2 y i ⫹ sin 3 y cos 4z j ⫹ sin 5z cos 6x k
in the cube cut from the first octant by the planes x 苷 ␲兾2,
y 苷 ␲兾2, and z 苷 ␲兾2. Then compute the flux across the
surface of the cube.
17. Use the Divergence Theorem to evaluate xxS F ⴢ dS, where
1
F共x, y, z兲 苷 z 2 x i ⫹ ( 3 y 3 ⫹ tan z) j ⫹ 共x 2z ⫹ y 2 兲 k
and S is the top half of the sphere x 2 ⫹ y 2 ⫹ z 2 苷 1.
[Hint: Note that S is not a closed surface. First compute
integrals over S1 and S2, where S1 is the disk x 2 ⫹ y 2 艋 1,
oriented downward, and S2 苷 S 傼 S1.]
18. Let F共x, y, z兲 苷 z tan⫺1共 y 2 兲 i ⫹ z 3 ln共x 2 ⫹ 1兲 j ⫹ z k.
Find the flux of F across the part of the paraboloid
x 2 ⫹ y 2 ⫹ z 苷 2 that lies above the plane z 苷 1 and is
oriented upward.
19. A vector field F is shown. Use the interpretation of diver-
gence derived in this section to determine whether div F
is positive or negative at P1 and at P2.
2
23. Verify that div E 苷 0 for the electric field E共x兲 苷
␧Q
x.
x 3
ⱍ ⱍ
24. Use the Divergence Theorem to evaluate xxS 共2 x ⫹ 2y ⫹ z 2 兲 dS
where S is the sphere x 2 ⫹ y 2 ⫹ z 2 苷 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 ⴢ n dS 苷 0,
where a is a constant vector
S
26. V共E 兲 苷
1
3
yy F ⴢ dS,
where F共x, y, z兲 苷 x i ⫹ y j ⫹ z k
S
27.
yy curl F ⴢ dS 苷 0
S
28.
yy D
S
29.
n
f dS 苷 yyy ⵜ 2 f dV
E
yy 共 f ⵜt兲 ⴢ n dS 苷 yyy 共 f ⵜ t ⫹ ⵜ f ⴢ ⵜt兲 dV
2
S
30.
E
yy 共 f ⵜt ⫺ t ⵜ f 兲 ⴢ n dS 苷 yyy 共 f ⵜ t ⫺ t ⵜ
2
S
2
f 兲 dV
E
31. Suppose S and E satisfy the conditions of the Divergence The-
orem and f is a scalar function with continuous partial derivatives. Prove that
P¡
_2
2
P™
S
_2
20. (a) Are the points P1 and P2 sources or sinks for the vector
field F shown in the figure? Give an explanation based
solely on the picture.
(b) Given that F共x, y兲 苷 具x, y 2 典 , use the definition of divergence to verify your answer to part (a).
2
P¡
_2
yy f n dS 苷 yyy ⵜ f dV
2
E
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 苷 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 苷 ␳ tz, where t is
the acceleration due to gravity (see Section 6.5). The total
buoyant force on the solid due to the pressure distribution is
given by the surface integral
F 苷 ⫺yy pn dS
P™
S
_2
CAS
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. F共x, y兲 苷 具xy, x ⫹ y 2 典
where n is the outer unit normal. Use the result of Exercise 31
to show that F 苷 ⫺W k, 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.
A126
||||
APPENDIX I ANSWERS TO ODD-NUMBERED EXERCISES
EXERCISES 16.3
N
PAGE 1053
11.
1. 40
3. f 共x, y兲 苷 x 2 3xy 2y 2 8y K
5. f 共x, y兲 苷 e x sin y K
7. f 共x, y兲 苷 ye x x sin y K
9. f 共x, y兲 苷 x ln y x 2 y 3 K
1
(b) 2
13. (a) f 共x, y兲 苷 2 x 2 y 2
11. (b) 16
2
15. (a) f 共x, y, z兲 苷 xyz z
(b) 77
17. (a) f 共x, y, z兲 苷 xy 2 cos z
(b) 0
19. 2
21. 30
23. No
25. Conservative
29. (a) Yes
(b) Yes
(c) Yes
31. (a) Yes
(b) Yes
(c) No
EXERCISES 16.4
N
PAGE 1060
3.
5. 12
7. 3
9. 24
11. 3 2
1
9
15. 8e 48e1
17. 12
19. 3
21. (c) 2
1. 8
625
13. 2 2
3
1
4
23. 共4a兾3 , 4a兾3兲 if the region is the portion of the disk
x 2 y 2 苷 a 2 in the first quadrant
EXERCISES 16.5
N
N
z 0
√ constant
_1
_1
y
_1
0
0
1 1
PAGE 1078
1. P: no; Q: yes
3. Plane through 共0, 3, 1兲 containing vectors 具1, 0, 4 典 , 具1, 1, 5典
5. Hyperbolic paraboloid
7.
u constant
15. II
17. III
IV
x 苷 1 u v, y 苷 2 u v, z 苷 3 u v
x 苷 x, z 苷 z, y 苷 s1 x 2 z 2
x 苷 2 sin cos , y 苷 2 sin sin ,
z 苷 2 cos , 0 兾4, 0 2
or x 苷 x, y 苷 y, z 苷 s4 x 2 y 2, x 2 y 2 2
25. x 苷 x, y 苷 4 cos , z 苷 4 sin , 0 x 5, 0 2
[
]
1
z 苷 ex sin , 0 x 3,
0 2
z 0
1
1
0
1 0
y
x
2
(b) Number of coils doubles
31. (a) Direction reverses
33. 3x y 3z 苷 3
35. x 2z 苷 1
37. 3 s14
4
39. 15 共3 5兾2 2 7兾2 1兲
41. 共2兾3兲(2 s2 1)
43. 共兾6兲(17 s17 5 s5 )
[
45. 2 s21 4 ln(2 s21 ) ln s17
49. 13.9783
51. (a) 24.2055
(b) 24.2476
1
53.
√ constant
x
13.
19.
21.
23.
29. x 苷 x, y 苷 ex cos ,
PAGE 1068
1. (a) x 2 i 3xy j xz k
(b) yz
(b) z 1兾 (2sz )
3. (a) 共x y兲 i y j k
(b) 2兾sx 2 y 2 z 2
5. (a) 0
(b) 1兾x 1兾y 1兾z
7. (a) 具1兾y, 1兾x, 1兾x典
9. (a) Negative
(b) curl F 苷 0
11. (a) Zero
(b) curl F points in the negative z-direction
13. f 共x, y, z兲 苷 xy 2z 3 K
15. f 共x, y, z兲 苷 x 2y y 2z K
17. Not conservative
19. No
EXERCISES 16.6
1
45
8
17
[
]
47. 4
]
s14 16 ln (11s5 3s70 )兾(3s5 s70 )
15
55. (b)
2
2
z 0
1
_2
0
u constant
1.5
y1
2
z 0
x
2
2
9.
u constant
2
y
1
√ constant
0
2
1
1 0x
(c) x02 x0 s36 sin 4u cos 2v 9 sin 4u sin 2v 4 cos 2u sin 2u du dv
57. 4
59. 2a 2共 2兲
z 0
EXERCISES 16.7
_1
_1
_1
y
0
0
1 1
x
N
PAGE 1091
1. 49.09
3. 900
5. 171 s14
7. s3兾24
9. 5s5兾48 1兾240
11. 364 s2 兾3
13. 共兾60兲(391s17 1)
15. 16
17. 12
APPENDIX I ANSWERS TO ODD-NUMBERED EXERCISES
23. 3 21. 6
713
180
19.
1
29. 2 8
3
4
31. 0.1642
25. 0
33. 3.4895
EXERCISES 17.1
[
1
All solutions approach either
0 or as x l .
10
g
9. 80
f
_3
3
(b)
5
_10
z 0
5
2
0
(c) x 苷 3 cos t, y 苷 3 sin t,
z 苷 1 3共cos t sin t兲,
0 t 2
0
2
2
y
17. y 苷 2e3x兾2 ex
19. y 苷 e x/2 2xe x兾2
21. y 苷 3 cos 4x sin 4x
23. y 苷 ex共2 cos x 3 sin x兲
2
25. y 苷 3 cos( 2 x) 4 sin( 2 x)
1
x
0
EXERCISES 17.2
y
0
2
2
0
_2
x
PAGE 1124
3. y 苷 c1 c2 e 2x 3
7
4
cos 4x 20 sin 4x
1
5. y 苷 e 共c1 cos x c2 sin x兲 10 ex
3
11
1 x
7. y 苷 2 cos x 2 sin x 2 e x 3 6x
1
9. y 苷 e x ( 2 x 2 x 2)
3
11.
The solutions are all asymptotic
to yp 苷 101 cos x 103 sin x as
x l . Except for yp , all
_3
8 solutions approach either yp
or as x l .
1
40
1
2x
17. 3
PAGE 1103
7. 9兾2
11. 32兾3
13. 0
arcsin(s3兾3)
17. 13兾20
19. Negative at P1 , positive at P2
21. div F 0 in quadrants I, II; div F 0 in quadrants III, IV
15. 341 s2兾60 N
1. y 苷 c1 e2x c2 ex 2 x 2 2 x 1
_2
N
e 2x
e x3
e 1
1 e3
3
2
_2
EXERCISES 16.9
27. y 苷
1
29. No solution
31. y 苷 e2x 共2 cos 3x e sin 3x兲
33. (b) 苷 n 2 2兾L2, n a positive integer; y 苷 C sin共n x兾L兲
4
z
5. 2
9. 0
]
1
PAGE 1097
7. 1
3. 0
5. 0
11. (a) 81兾2
PAGE 1117
13. P 苷 et c1 cos (10 t) c 2 sin (10 t)
15.
EXERCISES 16.8
N
1. y 苷 c1 e 3x c 2 e2x
3. y 苷 c1 cos 4x c 2 sin 4x
5. y 苷 c1 e 2x兾3 c 2 xe 2x兾3
7. y 苷 c1 c 2 e x兾2
2x
9. y 苷 e 共c1 cos 3x c 2 sin 3x兲
11. y 苷 c1 e (s31) t兾2 c 2 e (s31) t兾2
where D 苷 projection of S on xz-plane
37. 共0, 0, a兾2兲
39. (a) Iz 苷 xxS 共x 2 y 2 兲 共x, y, z兲 dS
(b) 4329 s2 兾5
8
41. 0 kg兾s
43. 3 a 30
45. 1248
N
A127
CHAPTER 17
27. 48
xxS F ⴢ dS 苷 xxD 关P共h兾x兲 Q R共h兾z兲兴 dA,
35.
||||
81
20
_3
13. yp 苷 Ae 共Bx 2 Cx D兲 cos x 共Ex 2 Fx G兲 sin x
15. yp 苷 Ax 共Bx C 兲e 9x
17. yp 苷 xex 关共Ax 2 Bx C 兲 cos 3x 共Dx 2 Ex F兲 sin 3x兴
2x
CHAPTER 16 REVIEW
N
PAGE 1106
True-False Quiz
1. False
19. y 苷 c1 cos ( 2 x) c 2 sin ( 2 x) 3 cos x
1
3. True
5. False
7. True
7.
110
3
9.
4兾e
33. 2
1
25.
3. 6 s10
5.
11. f 共x, y兲 苷 e y xe xy
(27 5 s5 )
29.
共兾60兲(391 s17 1)
17. 8
27.
11
12
(b) Positive
1
21. y 苷 c1e x c2 xe x e 2x
23. y 苷 c1 sin x c 2 cos x sin x ln共sec x tan x兲 1
25. y 苷 关c1 ln共1 ex 兲兴e x 关c2 ex ln共1 ex 兲兴e 2x
Exercises
1. (a) Negative
1
4
15
13. 0
[
]
27. y 苷 e x c1 c 2 x 2 ln共1 x 2 兲 x tan1 x
1
1
6
37. 4
39. 21
64兾3
EXERCISES 17.3
N
PAGE 1132
1. x 苷 0.35 cos (2 s5 t)
3. x 苷 5 e6t 5 et
1
6
5.
49
12
kg