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Prévia do material em texto

We now easily compute this last integral using the parametrization given by
, . Thus
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 is a vector field on
an open simply-connected region , that and have continuous first-order partial
derivatives, and that
If is any simple closed path in and is the region that encloses, then Green’s
Theorem gives
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 around these 
simple curves are all 0 and, adding these integrals, we see that for any
closed curve . Therefore is independent of path in by Theorem 16.3.3. It
follows that is a conservative vector field. MF
DxC F � drC
xC F � dr � 0
F
�y
C
 F � dr � �y
C
 P dx � Q dy � yy
R
 ��Q
�x
�
�P
�y � dA � yy
R
 0 dA � 0
CRDC
 throughout D
�P
�y
�
�Q
�x
QPD
F � P i � Q j
� y2�
0
 dt � 2�� y2�
0
 
��a sin t���a sin t� � �a cos t��a cos t�
a 2 cos2t � a 2 sin2t
 dt
 y
C
 F � dr � y
C�
 F � dr � y2�
0
 F�r�t�� � r��t� dt
0 � t � 2�r�t� � a cos t i � a sin t j
1060 | | | | CHAPTER 16 VECTOR CALCULUS
6. ,
is the rectangle with vertices , , , and 
,
is the boundary of the region enclosed by the parabolas
and 
8. ,
is the boundary of the region between the circles
and 
, is the circle 
10. , is the ellipse 
11–14 Use Green’s Theorem to evaluate . (Check the 
orientation of the curve before applying the theorem.)
11. ,
consists of the arc of the curve from to 
and the line segment from to �0, 0���, 0�
��, 0��0, 0�y � sin xC
F�x, y� � �sx � y 3, x 2 � sy 
xC F � dr
x 2 � xy � y 2 � 1CxC sin y dx � x cos y dy
x 2 � y 2 � 4CxC y 3 dx � x 3 dy9.
x 2 � y 2 � 4x 2 � y 2 � 1
C
xC xe�2x dx � �x 4 � 2x 2y 2� dy
x � y 2y � x 2
C
xC (y � esx ) dx � �2x � cos y 2 � dy7.
�0, 2��5, 2��5, 0��0, 0�C
xC cos y dx � x 2 sin y dy1–4 Evaluate the line integral by two methods: (a) directly and (b) using Green’s Theorem.
1. ,
is the circle with center the origin and radius 2
2. ,
is the rectangle with vertices , , , and 
,
is the triangle with vertices , (1, 0), and (1, 2)
4. , consists of the line segments from 
to and from to and the parabola 
from to 
5–10 Use Green’s Theorem to evaluate the line integral along the
given positively oriented curve.
5. ,
is the triangle with vertices , , and �2, 4��2, 2��0, 0�C
xC xy 2 dx � 2x 2y dy
�0, 1��1, 0�
y � 1 � x 2�1, 0��0, 0��0, 0�
�0, 1�C�xC x dx � y dy
�0, 0�C
�xC xy dx � x 2 y 3 dy3.
�0, 1��3, 1��3, 0��0, 0�C
�xC xy dx � x 2 dy
C
�xC �x � y� dx � �x � y� dy
EXERCISES16.4
(c) Find the area of the pentagon with vertices , ,
, , and .
22. Let be a region bounded by a simple closed path in the 
-plane. Use Green’s Theorem to prove that the coordinates
of the centroid of are
where is the area of .
23. Use Exercise 22 to find the centroid of a quarter-circular
region of radius .
24. Use Exercise 22 to find the centroid of the triangle with 
vertices , , and , where and .
25. A plane lamina with constant density occupies a
region in the -plane bounded by a simple closed path .
Show that its moments of inertia about the axes are
26. Use Exercise 25 to find the moment of inertia of a circular
disk of radius with constant density about a diameter.
(Compare with Example 4 in Section 15.5.)
If is the vector field of Example 5, show that 
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 
formula for a double integral (Formula 15.9.9) for the case
where :
Here is the region in the -plane that corresponds to the
region in the -plane under the transformation given by
, .
[Hint: Note that the left side is and apply the first part
of Equation 5. Convert the line integral over to a line inte-
gral over and apply Green’s Theorem in the -plane.]uv�S
�R
A�R�
y � h�u, v�x � t�u, v�
uvS
xyR
yy
R
 dx dy � yy
S
 � ��x, y�
��u, v� � du dv
f �x, y� � 1
xC F � dr � 0F27.
a
Iy �
3
 
�y
C
 x 3 dyIx � �
3
 
�y
C
 y 3 dx
Cxy
�x, y� � 
b � 0a � 0�a, b��a, 0��0, 0�
a
DA
y � �
1
2A
 
�y
C
 y 2 dxx �
1
2A
 
�y
C
 x 2 dy
D�x, y �
xy
CD
��1, 1��0, 2��1, 3�
�2, 1��0, 0�12. ,
is the triangle from to to to 
13. ,
is the circle oriented clockwise
14. , is the circle
oriented counterclockwise
15–16 Verify Green’s Theorem by using a computer algebra sys-
tem to evaluate both the line integral and the double integral.
15. , ,
consists of the line segment from to followed
by the arc of the parabola from to 
16. , ,
is the ellipse 
Use Green’s Theorem to find the work done by the force
in moving a particle from the 
origin along the -axis to , then along the line segment 
to , and then back to the origin along the -axis.
18. A particle starts at the point , moves along the -axis
to , and then along the semicircle to the
starting point. Use Green’s Theorem to find the work done on
this particle by the force field .
19. Use one of the formulas in (5) to find the area under one arch
of the cycloid .
; 20. If a circle with radius 1 rolls along the outside of the 
circle , a fixed point on traces out a 
curve called an epicycloid, with parametric equations
, . Graph the epi-
cycloid and use (5) to find the area it encloses.
(a) If is the line segment connecting the point to the
point , show that 
(b) If the vertices of a polygon, in counterclockwise order,
are , , show that the area of
the polygon is
 A � � �xn�1 yn � xn yn�1 � � �xn y1 � x1 yn ��
 A � 12 ��x1 y2 � x2 y1 � � �x2 y3 � x3 y2 � � � � �
�xn , yn ��x2, y2 �, . . . , �x1, y1 �
y
C
 x dy � y dx � x1 y2 � x2 y1
�x2, y2�
�x1, y1�C21.
y � 5 sin t � sin 5tx � 5 cos t � cos 5t
CPx 2 � y 2 � 16
C
x � t � sin t, y � 1 � cos t
F�x, y� � �x, x 3 � 3xy 2 
y � s4 � x 2 �2, 0�
x��2, 0�
y�0, 1�
�1, 0�x
F�x, y� � x�x � y� i � xy 2 j
17.
4x 2 � y 2 � 4C
Q�x, y� � x 3y 8P�x, y� � 2x � x 3y 5
��1, 1��1, 1�y � 2 � x 2
�1, 1���1, 1�C
Q�x, y� � x 2e yP�x, y� � y 2e x
CAS
�x � 2�2 � �y � 3�2 � 1
CF�x, y� � � y � ln�x 2 � y 2�, 2 tan�1�y�x�
x 2 � y 2 � 25C
F�x, y� � �e x � x 2 y, e y � xy 2 
�0, 0��2, 0��2, 6��0, 0�C
F�x, y� � � y 2 cos x, x 2 � 2y sin x
SECTION 16.5 CURL AND DIVERGENCE | | | | 1061
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 mag-
netism. Each operation resembles differentiation, but one produces a vector field whereas
the other produces a scalar field.
16.5
Converting to polar coordinates, we obtain
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 obtained by rotating the curve , , about
the -axis, where and is continuous. From Equations 3 we know that para-
metric equations of are
To compute the surface area of we need the tangent vectors
Thus
and so
because . Therefore the area of is
This is precisely the formula that was used to define the area of a surface of revolution in
single-variable calculus (8.2.4).
 � 2� yb
a
 f �x�s1 � � f ��x��2 dx
 A � yy
D
 � rx � r� � dA � y2�0 y
b
a
 f �x�s1 � � f ��x��2 dx d�
Sf �x� � 0
 � s� f �x��2�1 � � f ��x��2 � � f �x�s1 � � f ��x��2 
 � rx � r� � � s� f �x��2� f ��x��2 � � f �x��2 cos2� � � f �x��2 sin 2� 
 � f �x� f ��x� i � f �x� cos � j � f �x� sin � k
 rx � r� � � i10 j f ��x� cos � �f �x� sin � kf ��x� sin �f �x� cos � �
 r� � �f �x� sin � j � f �x� cos � k
 rx � i � f ��x� cos � j � f ��x� sin � k
S
0 	 � 	 2�a 	 x 	 bz � f �x� sin �y � f �x� cos �x � x
S
f �f �x� � 0x
a 	 x 	 by � f �x�S
 � 2� ( 18 ) 23 �1 � 4r 2 �3�2 ]03 � �6 (37s37 � 1)
 A � y2�
0
 y3
0
 s1 � 4r 2 r dr d� � y2�
0
 d� y3
0
 rs1 � 4r 2 dr1078 | | | | CHAPTER 16 VECTOR CALCULUS
5.
6.
; 7–12 Use a computer to graph the parametric surface. Get a printout
and indicate on it which grid curves have constant and which have 
constant.
7.
8.
9. , , 0 	 v 	 2��1 	 u 	 1r�u, v� � �u cos v, u sin v, u5 	
�1 	 v 	 1�1 	 u 	 1,r�u, v� � �u � v, u 2, v2 	 ,
�1 	 u 	 1, �1 	 v 	 1r�u, v� � �u 2 � 1, v3 � 1, u � v	 , 
v
u
r�s, t� � �s sin 2t, s 2, s cos 2t	
r�s, t� � �s, t, t 2 � s 2 	1–2 Determine whether the points and lie on the given surface.
1.
2.
3–6 Identify the surface with the given vector equation.
4. , 0 	 v 	 2r�u, v� � 2 sin u i � 3 cos u j � v k
r�u, v� � �u � v� i � �3 � v� j � �1 � 4u � 5v� k3.
P�3, �1, 5�, Q��1, 3, 4�
r�u, v� � �u � v, u 2 � v, u � v 2 	
P�7, 10, 4�, Q�5, 22, 5�
r�u, v� � �2u � 3v, 1 � 5u � v, 2 � u � v 	
QP
EXERCISES16.6
19–26 Find a parametric representation for the surface.
The plane that passes through the point and
contains the vectors and 
20. The lower half of the ellipsoid 
21. The part of the hyperboloid that lies to the
right of the -plane
22. The part of the elliptic paraboloid that lies
in front of the plane 
The part of the sphere that lies above the
cone 
24. The part of the sphere that lies between
the planes and 
25. The part of the cylinder that lies between the
planes and 
The part of the plane that lies inside the cylinder
27–28 Use a computer algebra system to produce a graph that
looks like the given one.
27. 28.
; 29. Find parametric equations for the surface obtained by rotating
the curve , , about the -axis and use them
to graph the surface.
; 30. Find parametric equations for the surface obtained by rotating
the curve , , about the -axis and
use them to graph the surface.
; 31. (a) What happens to the spiral tube in Example 2 (see Fig-
ure 5) if we replace by and by ?
(b) What happens if we replace by and 
by ?
; 32. The surface with parametric equations
where and , is called a Möbius
strip. Graph this surface with several viewpoints. What is
unusual about it?
0 	 � 	 2�� 12 	 r 	
1
2
 z � r sin���2�
 y � 2 sin � � r cos���2�
 x � 2 cos � � r cos���2�
sin 2u
sin ucos 2ucos u
cos usin usin ucos u
y�2 	 y 	 2x � 4y 2 � y 4
x0 	 x 	 3y � e �x
3
0
_3
_3
0
0 5
z
y
x
0
_1
_1
1
0
1
0
_1
z
y x
CAS
x 2 � y 2 � 1
z � x � 326.
x � 5x � 0
y 2 � z 2 � 16
z � 2z � �2
x 2 � y 2 � z 2 � 16
z � sx 2 � y 2 
x 2 � y 2 � z2 � 423.
x � 0
x � y 2 � 2z2 � 4
xz
x 2 � y2 � z2 � 1
2x 2 � 4y2 � z2 � 1
i � j � ki � j � k
�1, 2, �3�19.
10. ,
,
11. , , ,
,
12. , ,
13–18 Match the equations with the graphs labeled I–VI and 
give reasons for your answers. Determine which families of grid
curves have constant and which have constant.
14. ,
15.
16. ,
,
17. , ,
18. , ,
y
x
x
y
y
z
z
x
z
z
x y
III
V
x
y
z
IV
I II
VI
y
z
x
z � uy � �1 � � u ��sin vx � �1 � � u ��cos v
z � sin3vy � sin3u cos3vx � cos3u cos3v
z � 3u � �1 � u� sin v
y � �1 � u��3 � cos v� sin 4�u
x � �1 � u��3 � cos v� cos 4�u
r�u, v� � sin v i � cos u sin 2v j � sin u sin 2v k
�� 	 u 	 �r�u, v� � u cos v i � u sin v j � sin u k
r�u, v� � u cos v i � u sin v j � v k13.
vu
z � u sin vy � u cos u cos vx � u sin u cos v
���2 	 v 	 ��20 	 u 	 2�
z � sin 2u sin 4vy � cos u sin 4vx � sin v
0.1 	 v 	 6.20 	 u 	 2�
r�u, v� � �cos u sin v, sin u sin v, cos v � ln tan�v�2�	
SECTION 16.6 PARAMETRIC SURFACES AND THEIR AREAS | | | | 1079
51. (a) Use the Midpoint Rule for double integrals (see Sec-
tion 15.1) with six squares to estimate the area of the 
surface , , .
(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).
52. Find the area of the surface with vector equation
, ,
. State your answer correct to four decimal
places.
53. Find the exact area of the surface ,
, .
54. (a) Set up, but do not evaluate, a double integral for the area
of the surface with parametric equations ,
, , , .
(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 and
to graph the surface.
(d) For the case , , use a computer algebra 
system to find the surface area correct to four decimal
places.
(a) Show that the parametric equations ,
, , , ,
represent an ellipsoid.
; (b) Use the parametric equations in part (a) to graph the ellip-
soid for the case , , .
(c) Set up, but do not evaluate, a double integral for the sur-
face area of the ellipsoid in part (b).
56. (a) Show that the parametric equations ,
, , represent a hyperboloid
of one sheet.
; (b) Use the parametric equations in part (a) to graph the
hyperboloid for the case , , .
(c) Set up, but do not evaluate, a double integral for the sur-
face area of the part of the hyperboloid in part (b) that lies
between the planes and .
Find the area of the part of the sphere that
lies inside the paraboloid .
58. The figure shows the surface created when the cylinder
intersects the cylinder . Find the 
area of this surface.
z
y
x
x 2 � z 2 � 1y 2 � z 2 � 1
z � x 2 � y 2
x 2 � y 2 � z2 � 4z57.
z � 3z � �3
c � 3b � 2a � 1
z � c sinh uy � b cosh u sin v
x � a cosh u cos v
c � 3b � 2a � 1
0 	 v 	 2�0 	 u 	 �z � c cos uy � b sin u sin v
x � a sin u cos v55.
b � 3a � 2CAS
b � 3
a � 2
0 	 v 	 2�0 	 u 	 2z � u 2y � bu sin v
x � au cos v
0 	 y 	 11 	 x 	 4
z � 1 � 2x � 3y � 4y 2CAS
0 	 v 	 2�
0 	 u 	 �r�u, v� � �cos3u cos3v, sin3u cos3v, sin3v 	
CAS
CAS
0 	 y 	 40 	 x 	 6z � 1��1 � x 2 � y 2�
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.
, , ;
34. , , ;
35. ;
36. ;
37–47 Find the area of the surface.
The part of the plane that lies in the 
first octant
38. The part of the plane that lies inside the
cylinder 
39. The surface , ,
40. The part of the plane with vector equation
that is given by
, 
The part of the surface that lies within the cylinder
42. The part of the surface that lies above the
triangle with vertices , , and 
43. The part of the hyperbolic paraboloid that lies
between the cylinders and 
44. The part of the paraboloid that lies inside the
cylinder 
45. The part of the surface that lies between the
planes , , , and 
46. The helicoid (or spiral ramp) with vector equation
, , 
The surface with parametric equations , ,
, , 
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 that lies inside the
cylinder 
49. The part of the surface that lies above the disk
50. Find, to four decimal places, the area of the part of the
surface that lies above the square
. Illustrate by graphing this part of the surface.� x � � � y � 	 1
z � �1 � x 2 ���1 � y 2 �
CAS
x 2 � y 2 	 4
z � e�x
2�y2
x 2 � y 2 � 1
z � cos�x 2 � y 2�
0 	 v 	 20 	 u 	 1z � 12v 2
y � uvx � u247.
0 	 v 	 �0 	 u 	 1r�u, v� � u cos v i � u sin v j � v k
z � 1z � 0x � 1x � 0
y � 4x � z2
y 2 � z2 � 9
x � y 2 � z2
x 2 � y 2 � 4x 2 � y 2 � 1
z � y 2 � x 2
�2, 1��0, 1��0, 0�
z � 1 � 3x � 2y 2
x 2 � y 2 � 1
z � xy41.
0 	 v 	 10 	 u 	 1
r�u, v� � �1 � v, u � 2v, 3 � 5u � v	
0 	 y 	 10 	 x 	 1z � 23 �x 3�2 � y 3�2 �
x 2 � y 2 � 9
2x � 5y � z � 10
3x � 2y � z � 637.
u � 0, v � �r�u, v� � uv i � u sin v j � v cos u k
u � 1, v � 0r�u, v� � u 2 i � 2u sin v j � u cos v k
u � 1, v � 1z � uvy � v 2x � u2
�2, 3, 0�z � u � vy � 3u2x � u � v33.
1080 | | | | CHAPTER 16 VECTOR CALCULUS
SECTION 16.7 SURFACE INTEGRALS | | | | 1091
13. ,
is the part of the paraboloid that lies inside the
cylinder 
14. ,
is the part of the sphere that lies 
inside the cylinder and above the -plane
,
is the hemisphere , 
16. ,
is the boundary of the region enclosed by the cylinder
and the planes and 
17. ,
is thepart of the cylinder that lies between the
planes and in the first octant
18. ,
is the part of the cylinder between the planes
and , together with its top and bottom disks
19–30 Evaluate the surface integral for the given vector
field and the oriented surface . In other words, find the flux of 
across . For closed surfaces, use the positive (outward) orientation.
, is the part of the 
paraboloid that lies above the square
, and has upward orientation
20. ,
is the helicoid of Exercise 10 with upward orientation
21. ,
is the part of the plane in the first octant and
has downward orientation
22. ,
is the part of the cone beneath the plane 
with downward orientation
23. ,
is the part of the sphere in the first octant,
with orientation toward the origin
24. ,
is the hemisphere , , oriented in the
direction of the positive -axis
,
consists of the paraboloid , , 
and the disk , 
26. , S is the surface ,
, , with upward orientation0 	 y 	 10 	 x 	 1
z � xe yF�x, y, z� � xy i � 4x 2 j � yz k
y � 1x 2 � z2 	 1
0 	 y 	 1y � x 2 � z2S
F�x, y, z� � y j � z k25.
y
y � 0x 2 � y 2 � z 2 � 25S
F�x, y, z� � xz i � x j � y k
x 2 � y 2 � z 2 � 4S
F�x, y, z� � x i � z j � y k
z � 1z � sx 2 � y 2 S
F�x, y, z� � x i � y j � z4 k
x � y � z � 1S
F�x, y, z� � xze y i � xze y j � z k
S
F�x, y, z� � y i � x j � z2 k
0 	 y 	 10 	 x 	 1,
z � 4 � x 2 � y 2
SF�x, y, z� � xy i � yz j � zx k19.
S
FSF
xxS F � dS
z � 2z � 0
x 2 � y2 � 9S
xxS �x 2 � y 2 � z2 � dS
x � 3x � 0
y2 � z2 � 1S
xxS �z � x 2 y� dS
x � y � 5x � 0y2 � z2 � 9
S
xxS xz dS
z � 0x 2 � y 2 � z2 � 4S
xxS �x 2z � y 2z� dS15.
xyx 2 � y2 � 1
x 2 � y2 � z2 � 4S
xxS y2 dS
x 2 � z2 � 4
y � x 2 � z2S
xxS y dS1. Let be the boundary surface of the box enclosed by the
planes , , , , , and . Approx-
imate by using a Riemann sum as in Defini-
tion 1, taking the patches to be the rectangles that are the
faces of the box and the points to be the centers of the
rectangles.
2. A surface consists of the cylinder , ,
together with its top and bottom disks. Suppose you know that
is a continuous function with 
Estimate the value of by using a Riemann sum,
taking the patches to be four quarter-cylinders and the top
and bottom disks.
3. Let be the hemisphere , and 
suppose is a continuous function with 
, and . 
By dividing into four patches, estimate the value of
.
Suppose that , where is a 
function of one variable such that . Evaluate
, where is the sphere .
5–18 Evaluate the surface integral.
,
is the part of the plane that lies above the
rectangle 
6. ,
is the triangular region with vertices (1, 0, 0), (0, 2, 0), 
and (0, 0, 2)
7. ,
is the part of the plane that lies in the 
first octant
8. ,
is the surface , , 
9. ,
is the surface with parametric equations , ,
, 
10. ,
is the helicoid with vector equation
, , 
11. ,
is the part of the cone that lies between the
planes and 
12. ,
is the surface , , 0 	 z 	 10 	 y 	 1x � y � 2z 2S
xxS z dS
z � 3z � 1
z2 � x 2 � y 2S
xxS x 2z2 dS
0 	 v 	 �0 	 u 	 1r�u, v� � u cos v i � u sin v j � v k
S
xxS s1 � x 2 � y 2 dS
0 	 u 	 1, 0 	 v 	 ��2z � u cos v
y � u sin vx � u 2S
xxS yz dS
0 	 y 	 10 	 x 	 1z � 23 �x 3�2 � y 3�2 �S
xxS y dS
x � y � z � 1S
xxS yz dS
S
xxS xy dS
�0, 3� � �0, 2�
z � 1 � 2x � 3yS
xxS x 2yz dS5.
x 2 � y 2 � z2 � 4SxxS f �x, y, z� dS
t�2� � �5
tf �x, y, z� � t(sx 2 � y 2 � z 2 )4.
xxH f �x, y, z� dS
H
f ��3, �4, 5� � 12f �3, �4, 5� � 8, f ��3, 4, 5� � 9
f �3, 4, 5� � 7,f
x 2 � y 2 � z2 � 50, z � 0H
Sij
xxS f �x, y, z� dS
f �0, 0, 
1� � 4f �0, 
1, 0� � 3f �
1, 0, 0� � 2
f
�1 	 z 	 1x 2 � y 2 � 1S
Pij*S
Sij
xxS e�0.1�x�y�z� dS
z � 6z � 0y � 4y � 0x � 2x � 0
S
EXERCISES16.7
39. (a) Give an integral expression for the moment of inertia 
about the -axis of a thin sheet in the shape of a surface 
if the density function is .
(b) Find the moment of inertia about the -axis of the funnel
in Exercise 38.
40. Let be the part of the sphere that lies
above the plane . If has constant density , find 
(a) the center of mass and (b) the moment of inertia about 
the -axis.
41. A fluid has density and flows with velocity
, where and are measured in
meters and the components of in meters per second. Find
the rate of flow outward through the cylinder ,
.
42. Seawater has density and flows in a velocity field
, where and are measured in meters and
the components of in meters per second. Find the rate of
flow outward through the hemisphere ,
.
43. Use Gauss’s Law to find the charge contained in the solid
hemisphere , , if the electric field is 
44. Use Gauss’s Law to find the charge enclosed by the cube 
with vertices if the electric field is 
The temperature at the point in a substance with con-
ductivity is . Find the rate of
heat flow inward across the cylindrical surface ,
.
46. The temperature at a point in a ball with conductivity is
inversely proportional to the distance from the center of the
ball. Find the rate of heat flow across a sphere of radius 
with center at the center of the ball.
47. Let be an inverse square field, that is, for
some constant , where . Show that the
flux of across a sphere with center the origin is inde-
pendent of the radius of .S
SF
r � x i � y j � z kc
F�r� � cr�
 r 
3F
aS
K
0 	 x 	 4
y 2 � z2 � 6
u�x, y, z� � 2y 2 � 2z2K � 6.5
�x, y, z�45.
E�x, y, z� � x i � y j � z k
�
1, 
1, 
1�
E�x, y, z� � x i � y j � 2z k
z � 0x 2 � y 2 � z2 	 a 2
z � 0
x 2 � y 2 � z 2 � 9
v
zy,x,v � y i � x j
1025 kg�m3
0 	 z 	 1
x 2 � y 2 � 4
v
zy,x,v � z i � y 2 j � x 2 k
870 kg�m3
z
kSz � 4
x 2 � y2 � z2 � 25S
z
�
Sz
Iz27. ,
is the cube with vertices 
28. , is the boundary of the region
enclosed by the cylinder and the planes 
and 
29. , is the boundary of the
solid half-cylinder , 
30. ,
is the surface of the tetrahedron with vertices ,
, , and 
31. Evaluate correct to four decimal places, where is
the surface , , .
32. Find the exact value of , where is the surface in
Exercise 31.
33. Find the value of correct to four decimal places,
where is the part of the paraboloid that
lies above the -plane.
34. Find the flux of 
across the part of the cylinder that lies above 
the -plane and between the planes and 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 similar to Formula 10 for the
case where is given by and is the unit normal
that points toward the left.
36. Find a formula for similar to Formula 10 for the
case where is given by and is the unit normal
that points forward (that is, toward the viewer when the axes
are drawn in the usual way).
Find the center of mass of the hemisphere 
, if it has constant density.
38. Find the mass of a thin funnel in the shape of a cone
, , if its density function is
.��x, y, z� � 10 � z
1 	 z 	 4z � sx 2 � y 2 
z � 0
x 2 � y 2 � z2 � a 2,37.
nx � k�y, z�S
xxS F � dS
ny � h�x, z�S
xxS F � dS
x � 2x � �2xy
4y 2 � z2 � 4
F�x, y, z� � sin�xyz� i � x 2 y j � z2e x�5 k
CAS
xy
z � 3 � 2x 2 � y 2S
xxS x 2 y 2z2 dSCAS
SxxS x 2 yz dSCAS
0 	 y 	 10 	 x 	 1z � xy
SxxS xyz dSCAS
�0, 0, 1��0, 1, 0��1, 0, 0�
�0, 0, 0�S
F�x, y, z� � y i � �z � y� j � x k
0 	 x 	 20 	 z 	 s1 � y 2 
SF�x, y, z� � x 2 i � y 2 j � z2 k
x � y � 2
y � 0x 2 � z2 � 1
SF�x, y, z� � x i � y j � 5 k
�
1, 
1, 
1�S
F�x, y, z� � x i � 2y j � 3z k
1092 | | | | CHAPTER 16 VECTOR CALCULUS
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 to a line integral
around its plane boundary curve, Stokes’ Theorem relates a surface integral over a surface
to a line integral around the boundary curve of (which is a space curve). Figure 1 shows SS
D
16.8
SECTION 16.8 STOKES’ THEOREM | | | | 1097
10. , is the curve of intersection
of the plane and the cylinder 
11. (a) Use Stokes’ Theorem to evaluate , where
and is the curve of intersection of the plane
and the cylinder orientedcounterclockwise as viewed from above.
; (b) Graph both the plane and the cylinder with domains 
chosen so that you can see the curve and the surface 
that you used in part (a).
; (c) Find parametric equations for and use them to graph .
12. (a) Use Stokes’ Theorem to evaluate , where
and is the curve of
intersection of the hyperbolic paraboloid and
the cylinder 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 and the
surface that you used in part (a).
; (c) Find parametric equations for and use them to graph .
13–15 Verify that Stokes’ Theorem is true for the given vector 
field and surface .
13. ,
is the part of the paraboloid that lies below the
plane oriented upward
14. ,
is the part of the plane that lies in the first
octant, oriented upward
,
is the hemisphere , , oriented in the
direction of the positive -axis
16. Let be a simple closed smooth curve that lies in the plane
. Show that the line integral
depends only on the area of the region enclosed by and not
on the shape of or its location in the plane.
17. A particle moves along line segments from the origin to 
the points , , , and back to the 
origin under the influence of the force field 
Find the work done.
F�x, y, z� � z 2 i � 2xy j � 4y 2 k
�0, 2, 1��1, 2, 1��1, 0, 0�
C
C
xC z dx � 2x dy � 3y dz
x � y � z � 1
C
y
y � 0x 2 � y 2 � z 2 � 1S
F�x, y, z� � y i � z j � x k15.
2x � y � z � 2S
F�x, y, z� � x i � y j � xyz k
z � 1,
z � x 2 � y 2S
F�x, y, z� � y 2 i � x j � z2 k
SF
CC
C
x 2 � y 2 � 1
z � y 2 � x 2
CF�x, y, z� � x 2 y i � 13 x 3 j � xy k
xC F � dr
CC
C
x 2 � y 2 � 9x � y � z � 1
C
F�x, y, z� � x 2z i � xy 2 j � z2 k
xC F � dr
x 2 � y 2 � 9x � z � 5
CF�x, y, z� � xy i � 2z j � 3y kA hemisphere and a portion of a paraboloid are shown.
Suppose is a vector field on whose components have con-
tinuous partial derivatives. Explain why
2–6 Use Stokes’ Theorem to evaluate .
2. ,
is the hemisphere , , oriented 
upward
3. ,
is the part of the paraboloid that lies inside the
cylinder , oriented upward
4. ,
is the part of the cone that lies between the
planes and , oriented in the direction of the
positive -axis
,
consists of the top and the four sides (but not the bottom) 
of the cube with vertices , oriented outward
[Hint: Use Equation 3.]
6. ,
is the hemisphere , oriented in the direc-
tion of the positive -axis [Hint: Use Equation 3.]
7–10 Use Stokes’ Theorem to evaluate . In each case is
oriented counterclockwise as viewed from above.
,
is the triangle with vertices (1, 0, 0), (0, 1, 0), and (0, 0, 1)
8. ,
is the boundary of the part of the plane 
in the first octant
9. ,
is the circle x 2 � y 2 � 16, z � 5C
F�x, y, z� � yz i � 2xz j � e xy k
2x � y � 2z � 2C
F�x, y, z� � e�x i � e x j � e z k
C
F�x, y, z� � �x � y 2 � i � �y � z2 � j � �z � x 2 � k7.
CxC F � dr
x
x � s1 � y 2 � z 2 S
F�x, y, z� � e xy cos z i � x 2z j � xy k
�
1, 
1, 
1�
S
F�x, y, z� � xyz i � xy j � x 2 yz k5.
y
y � 3y � 0
y 2 � x 2 � z 2S
F�x, y, z� � x 2 y 3z i � sin�xyz� j � xyz k
x 2 � y2 � 4
z � x 2 � y2S
F�x, y, z� � x 2z2 i � y2z2 j � xyz k
z � 0x 2 � y 2 � z2 � 9S
F�x, y, z� � 2y cos z i � e x sin z j � xe y k
xxS curl F � dS
H
4
z
x y22
P
4
z
x y22
yy
H
 curl F � dS � yy
P
 curl F � dS
�3F
PH1.
EXERCISES16.8
20. Suppose and satisfy the hypotheses of Stokes’ Theorem
and , have continuous second-order partial derivatives. Use
Exercises 24 and 26 in Section 16.5 to show the following.
(a)
(b)
(c) xC � f �t � t� f � � dr � 0
xC � f � f � � dr � 0
xC � f �t� � dr � xxS �� f � �t� � dS
tf
CS18. Evaluate 
where is the curve , .
[Hint: Observe that lies on the surface .]
If is a sphere and satisfies the hypotheses of Stokes’
Theorem, show that .xxS curl F � dS � 0
FS19.
z � 2xyC
0 � t � 2�r�t� � �sin t, cos t, sin 2t�C
xC �y � sin x� dx � �z2 � cos y� dy � x 3 dz
1098 | | | | CHAPTER 16 VECTOR CALCULUS
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.
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).
THREE MEN AND TWO THEOREMSW R I T I N G
P R O J E C T
N The photograph shows a stained-glass 
window at Cambridge University in honor of
George Green.
Courtesy of the Masters and Fellows of Gonville and 
Caius College, University of Cambridge, England
www.stewartcalculus.com
The Internet is another source of infor-
mation 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.
the origin. [This is a special case of Gauss’s Law (Equation 16.7.11) for a single charge.
The relationship between and is .]
Another application of the Divergence Theorem occurs in fluid flow. Let be
the velocity field of a fluid with constant density . Then is the rate of flow per
unit area. If is a point in the fluid and is a ball with center and very small
radius , then for all points in since is continuous. We
approximate the flux over the boundary sphere as follows:
This approximation becomes better as and suggests that
Equation 8 says that is the net rate of outward flux per unit volume at . (This
is the reason for the name divergence.) If , the net flow is outward near and
is called a source. If , the net flow is inward near and is called a sink.
For the vector field in Figure 4, it appears that the vectors that end near are shorter
than the vectors that start near Thus the net flow is outward near so 
and is a source. Near on the other hand, the incoming arrows are longer than the 
outgoing arrows. Here the net flow is inward, so and is a sink. We 
can use the formula for F toconfirm this impression. Since , we have
, which is positive when . So the points above the line 
are sources and those below are sinks.
y � �xy 
 �xdiv F � 2x � 2y
F � x 2 i � y 2 j
P2div F�P2 � � 0
P2,P1
div F�P1� 
 0P1,P1.
P1
PPdiv F�P� � 0P
Pdiv F�P� 
 0
P0div F�P0 �
div F�P0 � � lim 
al 0
 
1
V�Ba �
 yy
Sa
 F � dS8
a l 0
� div F�P0 �V�Ba �� yyy
Ba
 div F�P0 � dVyy
Sa
 F � dS � yyy
Ba
 div F dV
Sa
div FBadiv F�P� � div F�P0 �a
P0BaP0�x0, y0, z0 �
F � �v�
v�x, y, z�
	 � 1
�4�	0 �	0	
SECTION 16.9 THE DIVERGENCE THEOREM | | | | 1103
FIGURE 4
The vector field F=≈ i+¥ j
P¡
P™
y
x
,
is the surface of the solid bounded by the cylinder
and the planes and 
8. ,
is the surface of the solid bounded by the hyperboloid
and the planes and 
,
is the ellipsoid 
10. ,
is the surface of the tetrahedron bounded by the planes
, , and 
11. ,
is the surface of the solid bounded by the paraboloid
and the plane 
12. ,
is the surface of the solid bounded by the cylinder
and the planes and 
13. ,
is the sphere with radius and center the originRS
F�x, y, z� � 4x 3z i � 4y 3z j � 3z 4 k
z � 0z � x � 2x 2 � y 2 � 1
S
F�x, y, z� � x 4 i � x 3z 2 j � 4xy 2z k
z � 4z � x 2 � y 2
S
F�x, y, z� � �cos z � xy 2� i � xe�z j � �sin y � x 2z� k
x � 2y � z � 2z � 0y � 0x � 0,
S
F�x, y, z� � x 2y i � xy 2 j � 2xyz k
x 2
a2 � y 2
b2 � z 2
c2 � 1S
F�x, y, z� � xy sin z i � cos�xz� j � y cos z k9.
z � 2z � �2x 2 � y 2 � z2 � 1
S
F�x, y, z� � x 3y i � x 2y 2 j � x 2yz k
x � 2x � �1y 2 � z2 � 1
S
F�x, y, z� � 3xy 2 i � xe z j � z3 k7.1–4 Verify that the Divergence Theorem is true for the vector field
on the region .
,
is the cube bounded by the planes , , ,
, , and 
2. ,
is the solid bounded by the paraboloid 
and the -plane
3. ,
is the solid cylinder , 
4. ,
is the unit ball 
5–15 Use the Divergence Theorem to calculate the surface integral
; that is, calculate the flux of across .
5. ,
is the surface of the box bounded by the planes , 
, , , , and 
6. ,
is the surface of the box with vertices �
1, 
2, 
3�S
F�x, y, z� � x 2z3 i � 2xyz3 j � xz4 k
z � 2z � 0y � 1y � 0x � 1
x � 0S
F�x, y, z� � e x sin y i � e x cos y j � yz2 k
SFxxS F � dS
x 2 � y 2 � z2 � 1E
F�x, y, z� � x i � y j � z k
0 � z � 1x 2 � y 2 � 1E
F�x, y, z� � xy i � yz j � zx k
xy
z � 4 � x 2 � y 2E
F�x, y, z� � x 2 i � xy j � z k
z � 1z � 0y � 1
y � 0x � 1x � 0E
F�x, y, z� � 3x i � xy j � 2xz k1.
EF
EXERCISES16.9
22.
23. Verify that for the electric field .
24. Use the Divergence Theorem to evaluate 
where is the sphere 
25–30 Prove each identity, assuming that and satisfy the con-
ditions of the Divergence Theorem and the scalar functions and
components of the vector fields have continuous second-order 
partial derivatives.
, where is a constant vector
26. , where 
27.
28.
29.
30.
31. Suppose and satisfy the conditions of the Divergence The-
orem and is a scalar function with continuous partial deriva-
tives. Prove that
These surface and triple integrals of vector functions are 
vectors defined by integrating each component function.
[Hint: Start by applying the Divergence Theorem to ,
where is an arbitrary constant vector.]
32. A solid occupies a region with surface and is immersed in
a liquid with constant density . We set up a coordinate
system so that the -plane coincides with the surface of the
liquid and positive values of are measured downward into
the liquid. Then the pressure at depth is , where 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
where is the outer unit normal. Use the result of Exercise 31
to show that , where is the weight of the liquid
displaced by the solid. (Note that is directed upward
because is directed downward.) The result is Archimedes’
principle: The buoyant force on an object equals the weight of
the displaced liquid.
z
F
WF � �Wk
n
F � �yy
S
 pn dS
tp � �tzz
z
xy
�
SE
c
F � f c
yy
S
 f n dS � yyy
E
 � f dV
f
ES
yy
S
 � f �t � t� f � � n dS � yyy
E
 � f � 2t � t� 2f � dV
yy
S
 � f �t� � n dS � yyy
E
 � f � 2t � � f � �t� dV
yy
S
 Dn f dS � yyy
E
 � 2f dV
yy
S
 curl F � dS � 0
F�x, y, z� � x i � y j � z kV�E � � 13 yy
S
 F � dS
ayy
S
 a � n dS � 025.
ES
x 2 � y 2 � z2 � 1.S
xxS �2x � 2y � z2� dS
E�x� �
	Q
	 x 	3 xdiv E � 0
F�x, y� � �x 2, y 2 �14. , where , 
consists of the hemisphere and the 
disk in the -plane
15. ,
is the surface of the solid that lies above the -plane 
and below the surface , 
16. Use a computer algebra system to plot the vector field
in the cube cut from the first octant by the planes ,
, and . Then compute the flux across the 
surface of the cube.
17. Use the Divergence Theorem to evaluate , where 
and is the top half of the sphere .
[Hint: Note that is not a closed surface. First compute 
integrals over and , where is the disk ,
oriented downward, and .]
18. Let . 
Find the flux of across the part of the paraboloid
that lies above the plane and is 
oriented upward.
A vector field is shown. Use the interpretation of diver-
gence derived in this section to determine whether 
is positive or negative at and at 
20. (a) Are the points and sources or sinks for the vector
field shown in the figure? Give an explanation based
solely on the picture.
(b) Given that , use the definition of diver-
gence to verify your answer to part (a).
21–22 Plot the vector field and guess where and
where . Then calculate to check your guess.
21. F�x, y� � �xy, x � y 2 �
div Fdiv F � 0
div F 
 0CAS
2
_2
_2 2
P¡
P™
F�x, y� � �x, y 2�
F
P2P1
2
_2
_2 2
P¡
P™
P2.P1
div F
F19.
z � 1x 2 � y 2 � z � 2
F
F�x, y, z� � z tan�1�y 2 � i � z3 ln�x 2 � 1� j � z k
S2 � S � S1
x 2 � y 2 � 1S1S2S1
S
x 2 � y 2 � z2 � 1S
F�x, y, z� � z2x i � ( 13 y 3 � tan z) j � �x 2z � y 2 � k
xxS F � dS
z � �
2y � �
2
x � �
2
F�x, y, z� � sin x cos2 y i � sin3y cos4z j � sin5z cos6x k
CAS
�1 � y � 1
�1 � x � 1,z � 2 � x 4 � y 4
xyS
F�x, y, z� � e y tan z i � ys3 � x 2 j � x sin y kCAS
xyx 2 � y2 � 1
z � s1 � x 2 � y 2 S
r � x i � y j � z kF � r
	 r 	
1104 | | | | CHAPTER 16 VECTOR CALCULUS
A126 || | | APPENDIX I ANSWERS TO ODD-NUMBERED EXERCISES
EXERCISES 16.3 N PAGE 1053
1. 40 3.
5. 7.
9.
11. (b) 16 13. (a) (b) 2
15. (a) (b) 77
17. (a) (b) 0
19. 2 21. 23. No 25. Conservative
29. (a) Yes (b) Yes (c) Yes
31. (a) Yes (b) Yes (c) No
EXERCISES 16.4 N PAGE 1060
1. 3. 5. 12 7. 9. 11.
13. 15. 17. 19. 21. (c)
23. if the region is the portion of the disk
in the first quadrant
EXERCISES 16.5 N PAGE 1068
1. (a) (b)
3. (a) (b)
5. (a) (b)
7. (a) (b)
9. (a) Negative (b)
11. (a) Zero (b) curl F points in the negative -direction
13. 15.
17. Not conservative 19. No 
EXERCISES 16.6 N PAGE 1078
1. : no; : yes
3. Plane through containing vectors ,
5. Hyperbolic paraboloid
7.
9.
_1
0
x
1
_1
0
y
z
u constant
√ constant
_1
0
1
1
2
0
_2
2
1
0
2
1.5
1
x
y
z
√ constant
u constant
�1, 	1, 5 ��1, 0, 4 ��0, 3, 1�
QP
f �x, y, z� � x 2y � y 2z � Kf �x, y, z� � xy 2z3 � K
z
curl F � 0
1�x � 1�y � 1�z�1�y, 	1�x, 1�x �
2�sx 2 � y 2 � z20
z 	 1�(2sz)�x 	 y� i 	 y j � k
yz	x 2 i � 3xy j 	 xz k
x 2 � y 2 � a 2
�4a�3�, 4a�3��
9
23�	
1
12	8e � 48e	1
625
2 �
4
3 	 2�	24�
1
3
2
38�
30
f �x, y, z� � xy 2 cos z
f �x, y, z� � xyz � z 2
f �x, y� � 12 x 2y 2
f �x, y� � x ln y � x 2y 3 � K
f �x, y� � ye x � x sin y � Kf �x, y� � e x sin y � K
f �x, y� � x 2 	 3xy � 2y 2 	 8y � K
11.
13. IV 15. II 17. III
19.
21.
23. , ,
, ,
25. , , ,
29. ,
, ,
31. (a) Direction reverses (b) Number of coils doubles
33. 35. 37.
39. 41.
43.
45. 47. 4
49.
51. (a) 24.2055 (b) 24.2476
53.
55. (b)
(c) 
57. 59.
EXERCISES 16.7 N PAGE 1091
1. 49.09 3. 5. 7.
9. 11.
13. 15. 17. 1216����60�(391s17 � 1)
364s2��35s5�48 � 1�240
s3�24171s14900�
2a 2�� 	 2�4�
x2�0 x�0 s36 sin4u cos2v � 9 sin4u sin2v � 4 cos2u sin2u du dv
2
0
	2
	2 	10 2 1 0
z
y x
45
8 s14 � 1516 ln[(11s5 � 3s70)�(3s5 � s70)]
13.9783
1
2s21 � 174 [ln(2� s21) 	 ln s17]
���6�(17s17 	 5s5)
�2��3�(2s2 	 1)415�35�2 	 27�2 � 1�
3s14	x � 2z � 13x 	 y � 3z � 3
0 � � � 2�
0 � x � 3z � e	x sin �
20	10
1
	1
0
1
y
z
x
x � x, y � e	x cos �
0 � x � 5, 0 � � � 2�z � 4 sin �y � 4 cos �x � x
[or x � x, y � y, z � s4 	 x 2 	 y 2, x 2 � y 2 � 2]0 � � � 2�0 � � � ��4z � 2 cos �
y � 2 sin � sin �x � 2 sin � cos �
x � x, z � z, y � s1 	 x 2 � z 2
x � 1 � u � v, y � 2 � u 	 v, z � 	3 	 u � v
_1
0
x
1
_1
0
y
1
_1
0z
1
√ constant
u constant
APPENDIX I ANSWERS TO ODD-NUMBERED EXERCISES | | | | A127
19. 21. 23. 25. 0 27. 48
29. 31. 33. 3.4895
35. ,
where projection of on -plane
37.
39. (a) (b)
41. 43. 45.
EXERCISES 16.8 N PAGE 1097
3. 0 5. 0 7. 	1 9.
11. (a) (b)
(c) ,
,
17. 3
EXERCISES 16.9 N PAGE 1103
5. 2 7.
9. 0 11. 13.
15. 17.
19. Negative at , positive at 
21. in quadrants I, II; in quadrants III, IV
CHAPTER 16 REVIEW N PAGE 1106
True-False Quiz
1. False 3. True 5. False 7. True
Exercises
1. (a) Negative (b) Positive 3. 5.
7. 9. 11. 13. 0
17. 25.
27. 29.
33. 37. 39. 21	4	 12
	64��3���60�(391s17 � 1)
1
6 (27 	 5s5)	8�
f �x, y� � e y � xe xy1112 	 4�e1103
4
156s10
div F � 0div F � 0
P2P1
13��20341s2�60 � 8120 arcsin(s3�3)
032��3
9��2
0 � t � 2�
z � 1 	 3�cos t � sin t�
_2
0
2
4
_2 0 2 2
0
_2
z
y
x
x � 3 cos t, y � 3 sin t
	2
5
0
	5
z
0
y 2
	2 2 0 x
81��2
80�
1248�83�a 3
00 kg�s
4329s2��5Iz � xxS �x 2 � y 2 ���x, y, z� dS
�0, 0, a�2�
xzSD �
xxS F � dS � xxD �P��h��x� 	 Q � R��h��z�	 dA
0.16422� � 83
	
4
3�	
1
6
713
180
CHAPTER 17
EXERCISES 17.1 N PAGE 1117
1. 3.
5. 7.
9.
11.
13.
15. All solutions approach either 
or as .
17. 19.
21. 23.
25. 27.
29. No solution
31.
33. (b) , n a positive integer; 
EXERCISES 17.2 N PAGE 1124
1.
3.
5.
7.
9.
11. The solutions are all asymptotic
to as
. Except for , all 
solutions approach either 
or as .
13.
15.
17.
19.
21.
23.
25.
27.
EXERCISES 17.3 N PAGE 1132
1. 3. 5. 4912 kgx � 	
1
5 e
	6 t �
6
5 e
	tx � 0.35 cos(2s5 t)
y � e x [c1 � c2 x 	 12 ln�1 � x 2� � x tan	1x]
y � �c1 � ln�1 � e	x �	e x � �c2 	 e	x � ln�1 � e	x �	e 2x
y � c1 sin x � c2 cos x � sin x ln�sec x � tan x� 	 1
y � c1e x � c2 xe x � e 2x
y � c1 cos(12 x) � c2 sin(12 x) 	 13 cos x
yp � xe	x ��Ax 2 � Bx � C � cos 3x � �Dx 2 � Ex � F� sin 3x	
yp � Ax � �Bx � C �e9x
yp � Ae 2x � �Bx 2 � Cx � D� cos x � �Ex 2 � Fx � G� sin x
xl	�	�
�
ypxl�
yp � 110 cos x �
3
10 sin x
3
_3
_3 8
yp
y � e x(12 x 2 	 x � 2)
y � 32 cos x �
11
2 sin x �
1
2 e
x � x 3 	 6x
y � e 2x�c1 cos x � c2 sin x� � 110 e	x
y � c1 � c2e 2x � 140 cos 4x 	
1
20 sin 4x
y � c1e	2x � c2e	x � 12 x 2 	
3
2 x �
7
4
y � C sin�n�x�L�� � n 2� 2�L2
y � e	2x�2 cos 3x 	 e� sin 3x�
y �
e x�3
e 3 	 1
�
e 2x
1 	 e 3
y � 3 cos(12 x) 	 4 sin(12 x)
y � e	x�2 cos x � 3 sin x� y � 3 cos 4x 	 sin 4x
y � e x /2 	 2xe x�2y � 2e	3x�2 � e	x
xl����0
10
_10
_3 3
g
f
P � e	t[c1 cos( 110 t) � c2 sin( 110 t)]
y � c1e (s3	1) t�2 � c2e	(s3�1) t�2
y � e 2x�c1 cos 3x � c2 sin 3x�
y � c1 � c2 e x�2y � c1e 2x�3 � c2 xe 2x�3
y � c1 cos 4x � c2 sin 4xy � c1e3x � c2e	2x

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