Solucao_Calculo_5ed_CAP 16

Prévia do material em texto

F(x,y)=
1
2 (i+ j)
1
2
y=x.
F(x, y)=i+x j
i+x j 1+x2 y �
F(x, y)=y i+ 12 j
y i+ 12 j y
2
+
1
4 x �
F(x, y)=(x�y) i+x j
(x�y) i+x j (x�y)2+x2 y=x
1.
All vectors in this field are identical, with length and direction parallel to the line 
2.
The length of the vector is . Vectors are tangent to parabolas opening about the axis.
3. .
The length of the vector is . Vectors are tangent to parabolas opening about the 
axis.
4.
The length of the vector is . Vectors along the line are vertical.
 1
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.1 Vector Fields
F(x, y)= y i+x j
x
2
+y
2
y i+x j
x
2
+y
2
F(x, y)= y i�x j
x
2
+y
2
F(x, y) x
2
+y
2
F(x, y, z)= j
y � 1
F(x, y, z)=z j
(x, y, z) F(x, y, z) |z| z>0
y � z<0 y �
5.
The length of the vector is 1.
6.
All the vectors are unit vectors tangent to circles centered at the origin with radius .
7.
All vectors in this field are parallel to the axis and have length .
8.
At each point , is a vector of length . For , all point in the direction of the
positive axis while for , all are in the direction of the negative axis.
 2
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.1 Vector Fields
F(x, y, z)=y j
F(x, y, z) |y| xz � y=0
y=b
F(x, y, z)= j�i
2 xy �
F(x, y)= y, x x � y
� x � y �
x � y �
x � y �
F(x, y)= 1, sin y x �
x
F(x, y)= x�2, x+1 y
x � y �
F(x, y)= y, 1/x
x � y � x �
y � x � y �
9.
The length of is . No vectors emanate from the plane since there. In each plane
 , all the vectors are identical.
10.
All vectors in this field have length and point in the same direction, parallel to the plane.
11. corresponds to graph II. In the first quadrant all the vectors have positive and 
components, in the second quadrant all vectors have positive components and negative 
components, in the third quadrant all vectors have negative and components, and in the fourth
quadrant all vectors have negative components and positive components. In addition, the
vectors get shorter as we approach the origin.
12. corresponds to graph IV since the component of each vector is constant, the
vectors are independent of (vectors along horizontal lines are identical), and the vector field appears
to repeat the same pattern vertically.
13. corresponds to graph I since the vectors are independent of (vectors along
vertical lines are identical) and, as we move to the right, both the and the components get
larger.
14. corresponds to graph III. As in Exercise 11, all the vectors in the first quadrant
have positive and components, in the second quadrant all vectors have positive components
and negative components, in the third quadrant all vectors have negative and components,
 3
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.1 Vector Fields
x � y �
y �
F(x, y, z)=i+2 j+3 k
F(x, y, z)=i+2 j+zk
xy � xy �
F(x, y, z)=x i+y j+3 k xy �
x i+y j z �
3
F(x, y, z)=x i+y j+zk F(x, y, z)
(x, y, z)
y=2x F(x, y)= 0, 0
y
2
�2xy=0 3xy�6x
2
=0 y=0 y=2x x=0
y=2x F(x, y)=0 y=2x
|x|=2 F(x)=0�
r(r�2)=0�r=0 2 F(x)=0 |x|=2 |x|=0 r
2
�r<0
r
2
�r>0
and in the fourth quadrant all vectors have negative components and positive components.
Also, the vectors become longer as we approach the axis.
15. corresponds to graph IV, since all vectors have identical length and direction.
16. corresponds to graph I, since the horizontal vector components remain
constant, but the vectors above the plane point generally upward while the vectors below the 
plane point generally downward.
17. corresponds to graph III; the projection of each vector onto the plane is
 , which points away from the origin, and the vectors point generally upward because their 
components are all .
18. corresponds to graph II; each vector has the same length and
direction as the position vector of the point , and therefore the vectors all point directly away
from the origin.
19.
The vector field seems to have very short vectors near the line . For we must
have and . The first equation holds if or , and the second holds if 
or . So both equations hold [and thus ] along the line .
20.
From the graph, it appears that all of the vectors in the field lie on lines through the origin, and that
the vectors have very small magnitudes near the circle and near the origin. Note that 
 or , so as we suspected, for and for . Note that where , the
vectors point towards the origin, and where , they point away from the origin.
 4
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.1 Vector Fields
� f (x, y)= f
x
(x, y)i+ f
y
(x, y) j= 1
x+2y i+
2
x+2y j
� f (x, y)= f
x
(x, y)i+ f
y
(x, y) j= x� (�� e�� x)+� x� �1e�� x i+0 j=(� �� x)x� �1e�� x i
� f (x, y, z) = f x(x, y, z) i+ f y(x, y, z) j+ f z(x, y, z) k
= x
x
2
+y
2
+z
2
i+ y
x
2
+y
2
+z
2
j+ z
x
2
+y
2
+z
2
k
� f (x, y, z) = f x(x, y, z) i+ f y(x, y, z) j+ f z(x, y, z) k
= cos
y
z
i�x sin y
z
1
z
j�x sin y
z
� y
z
2
k
= cos
y
z
i� x
z
sin
y
z
j+ xy
z
2
sin
y
z
k
f (x, y)=xy�2x�� f (x, y)=(y�2) i+x j
� f (x, y) (y�2)2+x2 � f (x, y) y=x+2
(x+y�2, x+y).
f (x, y)= 14 (x+y)
2
�
� f (x, y)= 12 (x+y) i+
1
2 (x+y) j
� f (x, y)
21.
22.
23.
24.
25. .
The length of is and terminates on the line at the point
26.
 .
The length of is
 5
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.1 Vector Fields
1
2 (x+y)
2
=
1
2
|x+y|. y=�x
y=�x
� f f
f
� f f
f
f (x, y)=xy�� f (x, y)=y i+x j
x � y �
� f
 The vectors are perpendicular to the line and point away from the line,
with length that increases as the distance from the line increases.
27. We graph along with a contour map of .
The graph shows that the gradient vectors are perpendicular to the level curves. Also, the gradient
vectors point in the direction in which is increasing and are longer where the level curves are closer
together.
28. We graph along with a contour map of .
The graph shows that the gradient vectors are perpendicular to the level curves. Also, the gradient
vectors point in the direction in which is increasing and are longer where the level curves are closer
together.
29. . In the first quadrant, both components of each vector are positive,
while in the third quadrant both components are negative. However, in the second quadrant each
vector’s component is positive while its component is negative (and vice versa in the fourth
quadrant). Thus, is graph IV.
 6
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.1 Vector Fields
f (x, y)=x2�y2�� f (x, y)=2x i�2y j x �
y � x �
x � y �
y � � f
f (x, y)=x2+y2�� f (x, y)=2x i+2y j � f (x, y)
(x, y)
� f
f (x, y)= x2+y2 �� f (x, y)= x
x
2
+y
2
i+ y
x
2
+y
2
j |� f (x, y)|= 1
x
2
+y
2
x
2
+y
2
=1
� f (x, y)
(x,y) � f
F(x, y)=x i�y j
y=�1/x
y=C /x
x=x(t) y=y(t)
(x, y) x
/
(t)i+y / t( ) j
x
/
(t)i+y /(t) j=x i�y j�dx/dt=x dy/dt=�y
dx/dt=x�dx/x=dt� ln |x|=t+C� x=�et +C=Aet A
dy/dt=�y�dy/y=�dt� ln |y|=�t+K� y=�e�t +K=Be�t B
xy=Ae
t
Be
�t
=AB= (1, 1) (1)(1)= =1� xy=1�
y=1/x x>0
F(x, y)=i+x j
30. . In the first quadrant, the component of each vector is
positive while the component is negative. The other three quadrants are similar, where the 
component of each vector has the same sign as the value of its initial point, and the component
has sign opposite that of the value of the initial point. Thus, is graph III.
31. . Thus, each vector has the same direction and twice
the length of the position vector of the point , so the vectorsall point directly away from the
origin and their lengths increase as we move away from the origin. Hence, is graph II.
32. . Then , so
all vectors are unit vectors. In addition, each vector has the same direction as the position
vector of the point , so the vectors all point directly away from the origin. Hence, is graph I.
33. (a) We sketch the vector field along with several approximate flow lines.The flow
lines appear to be hyperbolas with shape similar to the graph of , so we might guess that the
flow lines have equations .
(b) If and are parametric equations of a flow line, then the velocity vector of the flow
line at the point is . Since the velocity vectors coincide with the vectors in the
vector field, we have , . To solve these differential
equations, we know for some constant , and
 for some constant . Therefore
 constant. If the flow line passes through then constant 
 , .
34. (a) We sketch the vector field along with several approximate flow lines.The flow
lines appear to be parabolas.
 7
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.1 Vector Fields
x=x(t) y=y(t)
(x, y) x
/
(t)i+y /(t) j
x
/
(t)i+y /(t) j=i+x j� dxdt =1
dy
dt =x
dy
dx =
dy/dt
dx/dt =
x
1 =x
dy/dx=x y= 12 x
2
+c
(0, 0) 0=0+c� c=0 y=
1
2 x
2
(b) If and are parametric equations of a flow line, then the velocity vector of the flow
line at the point is . Since the velocity vectors coincide with the vectors in the
vector field, we have , . Thus .
(c) From part (b), . Integrating, we have . Since the particle starts at the origin, we
know is on the curve, so and the path the particle follows is .
 8
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.1 Vector Fields
x=t
2
y=t 0� t� 2
�C y ds = �
2
0
t
dx
dt
2
+
dy
dt
2
dt=�2
0
t (2t)
2
+(1)
2
dt
= �2
0
t 4t
2
+1 dt=
1
12 (4t
2
+1)
3/2 2
0
=
1
12 (17 17�1)
�
C
y
x ds
=�1
1/2
t
3
t
4
(4t
3
)
2
+(3t
2
)
2
dt=�1
1/2
1
t 16t
6
+9t
4
dt=�1
1/2
t 16t
2
+9 dt
=
1
48 (16t
2
+9)
3/2 1
1/2
=
1
48 (25
3/2
�13
3/2
)=
1
48 (125�13 13)
C x=4cos t y=4sin t �
�
2 � t�
�
2
�Cxy
4
ds = �� /2
�� /2
(4cos t)(4sin t)
4
(�4sin t)
2
+(4cos t)
2
dt
= �� /2
�� /2
4
5
cos tsin
4
t 16(sin
2
t+cos
2
t) dt
= 4
5�� /2
�� /2
(sin
4
tcos t)(4)dt=(4)
6 1
5 sin
5
t
� /2
�� /2
=
2 � 4
6
5 =1638.4
C x=1+3t y=2+5t 0� t� 1
�Cye
x
ds=�1
0
(2+5t)e
1+3t
3
2
+5
2
dt= 34�1
0
(2+5t)e
1+3t
dt
u=2+5t � du=5dt dv=e
1+3t
� v=
1
3 e
1+3t
dt
�Cye
x
ds = 34
1
3 (2+5t)e
1+3t
�
5
9 e
1+3t 1
0
= 34
7
3 �
5
9 e
4
�
2
3 �
5
9 e =
34
9 (16e
4
�e)
x C x=x y=x
2
1� x� 3
1. and , , so by Formula 3
2.
3. Parametric equations for are , , . Then
4. Parametric equations for are , , . Then
Integrating by parts with , gives
5. If we choose as the parameter, parametric equations for are , for and
 1
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.2 Line Integrals
�
C
(xy+ln x)dy = �3
1
(x� x
2
+ln x)2xdx=�3
1
2(x
4
+xln x)dx
= 2
1
5 x
5
+
1
2 x
2
ln x�
1
4 x
2 3
1
= 2
243
5 +
9
2 ln 3�
9
4 �
1
5 +
1
4 =
464
5 +9ln 3
y x=e
y
y=y 0� y� 1
�cxe
y
dx=�1
0
e
y
(e
y
)e
y
dy=�1
0
e
3y
dy=
1
3 e
3y 1
0
=
1
3 (e
3
�1)
C=C
1
+C
2
C
1
x=x y=0�dy=0dx 0� x� 2
C
2
x=x y=2x�4�dy=2dx 2� x� 3
�Cxydx+(x�y)dy =�C1
xydx+(x�y)dy+�
C2
xydx+(x�y)dy
=�2
0
(0+0)dx+�3
2
dx
=�3
2
(2x
2
�6x+8)dx=
17
3
C=C
1
+C
2
C
1
x=cos t�dx=�sin t dt y=sin t�
 (by integrating by parts in the second term)
6. Choosing as the parameter, we have , , . Then
 .
7.
On : , , .
On : , , .
Then
8.
On : , 
 2
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.2 Line Integrals
dy=cos t dt 0� t��
C
2
x=�1�t�dx=�dt y=3t�
dy=3dt 0� t� 1
�
C
sin xdx+cos ydy =�C1
sin xdx+cos ydy+�
C2
sin xdx+cos ydy
=��
0
sin (cos t)(�sin t dt)+cos (sin t)cos t dt
+�1
0
sin (�1�t)(�dt)+cos (3t)(3dt)
= �cos (cos t)+sin (sin t)
�
0
+ �cos (�1�t)+sin (3t)
1
0
=�cos (cos� )+sin (sin� )+cos (cos 0)�sin (sin 0)
�cos (�2)+sin (3)+cos (�1)�sin (0)
=�cos (�1)+sin 0+cos (1)�sin 0�cos (�2)+sin 3+cos (�1)
=�cos 1+cos 1�cos 2+sin 3+cos 1=cos 1�cos 2+sin 3
cos (�� )=cos�
x=4sin t y=4cos t z=3t 0� t�
�
2
�
C
xy
3
ds = �� /2
0
(4sin t)(4cos t)
3 dx
dt
2
+
dy
dt
2
+
dz
dt
2
dt
= �� /2
0
4
4
cos
3
tsin t (4cos t)
2
+(�4sin t)
2
+(3)
2
dt
= �� /2
0
256cos
3
tsin t 16(cos
2
t+sin
2
t +9)dt
= 1280�� /2
0
cos
3
tsin t dt= �320cos
4
t
� /2
0
=320
C x=4t y=6�5t z=�1+6t 0� t� 1
�
C
x
2
zds = �1
0
(4t)
2
(6t�1) 4
2
+(�5)
2
+6
2
dt= 77�1
0
(96t
3
�16t
2
)dt
= 77 96�
t
4
4 �16�
t
3
3
1
0
=
56
3 77
 , .
On : , 
 , .
Then
where we have used the identity .
9. , , , . Then by Formula 9,
10. Parametric equations for are , , , . Then
 3
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.2 Line Integrals
C x=t y=2t z=3t 0� t� 1
�Cxe
yz
ds = �1
0
te
(2t)(3t)
1
2
+2
2
+3
2
dt= 14�1
0
te
6t
2
dt
= 14
1
12 e
6t
2 1
0
=
14
12 (e
6
�1)
(dx/dt)
2
+(dy/dt)
2
+(dz/dt)
2
= 1
2
+(2t)
2
+(3t
2
)
2
= 1+4t
2
+9t
4
�
C
(2x+9z)ds =�1
0
(2t+9t
3
) 1+4t
2
+9t
4
dt u=1+4t
2
+9t
4
�
1
4 du=(2t+9t
3
)dt
=�14
1
1
4 u du=
1
6 u
3/2 14
1
=
1
6 (14
3/2
�1)
�Cx
2
y z dz=�1
0
(t
3
)
2
(t) t
2
� 2t dt=�1
0
2t
9
dt=
1
5 t
10 1
0
=
1
5
�Czdx+xdy+ydz =�
1
0
t
2
� 2t dt+t
2
� 3t
2
dt+t
3
� 2t dt=�1
0
(2t
3
+5t
4
)dt
=
1
2 t
4
+t
5 1
0
=
1
2 +1=
3
2
C
1
x=1+t�dx=dt y=3t�dy=3dt z=1
�dz=0dt 0� t� 1
C
2
x=2�dx=0dt y=3+2t�
dy=2dt z=1+t�dz=dt 0� t� 1
�C (x+yz)dx+2xdy+xyzdz
=�
C1
(x+yz)dx+2xdy+xyzdz+�
C2
(x+yz)dx+2xdy+xyzdz
11. Parametric equations for are , , , . Then
12. . Then
 [ let ]
13.
14.
15.
On : , , 
 , .
On : , 
 , , .
Then
 4
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.2 Line Integrals
=�1
0
(1+t+(3t)(1))dt+2(1+t)� 3dt+(1+t)(3t)(1)� 0dt
+�1
0
(2+(3+2t)(1+t))� 0dt+2(2)� 2dt+(2)(3+2t)(1+t)dt
=�1
0
(10t+7)dt+�1
0
(4t
2
+10t+14)dt
= 5t
2
+7t
1
0
+
4
3 t
3
+5t
2
+14t
1
0
=12+
61
3 =
97
3
C
1
x=t�dx=dt y=2t�dy=2dt z=�t
�dz=�dt 0� t� 1
C
2
x=1+2t�dx=2dt y=2�
dy=0dt z=�1+t�dz=dt 0� t� 1
�
C
x
2
dx+y
2
dy+z
2
dz
=�
C1
x
2
dx+y
2
dy+z
2
dz+�
C2
x
2
dx+y
2
dy+z
2
dz
=�1
0
t
2
dt+(2t)
2
� 2dt+(�t)
2
(�dt)+�1
0
(1+2t)
2
� 2dt+2
2
� 0dt+(�1+t)
2
dt
=�1
0
8t
2
dt+�1
0
(9t
2
+6t+3)dt=
8
3 t
3 1
0
+ 3t
3
+3t
2
+3t
1
0
=
35
3
x=�3 F y �
F�T �
C1
F� dr=�
C1
F�T ds
3
F�T
�
C2
F� dr=�
C2
F�T ds
16.
On : , , 
 , .
On : , 
 , , .
Then
17. (a) Along the line , the vectors of have positive components, so since the path goes
upward, the integrand is always positive. Therefore is positive.
(b) All of the (nonzero) field vectors along the circle with radius are pointed in the clockwise
direction, that is, opposite the direction tothe path. So is negative, and therefore
 is negative.
18. Vectors starting on
 5
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.2 Line Integrals
C
1
C
1
F�T
�
C1
F� dr=�
C1
F�T ds C
2
C
2
�
C2
F� dr=�
C2
F�T ds
r(t)=t
2
i�t
3
j F(r(t))=(t
2
)
2
(�t
3
)
3
i�(�t
3
) t
2
j=�t
13
i+t
4
j r
/
(t)=2t i�3t
2
j
�CF� dr=�
1
0
F(r(t))� r
/
(t)dt=�1
0
(�2t
14
�3t
6
)dt= �
2
15 t
15
�
3
7 t
7 1
0
=�
59
105
F(r(t))=(t
2
)(t
3
)i+(t)(t
3
) j+(t)(t
2
)k=t
5
i+t
4
j+t
3
k r
/
(t)=i+2t j+3t
2
k
�CF� dr=�
2
0
F(r(t))� r
/
(t)dt=�2
0
(t
5
+2t
5
+3t
5
)dt= t
6 2
0
=64
�CF� dr =�
1
0
sin t
3
, cos (�t
2
), t
4
� 3t
2
, �2t, 1 dt
=�1
0
(3t
2
sin t
3
�2tcos t
2
+t
4
)dt= �cos t
3
�sin t
2
+
1
5 t
5 1
0
=
6
5 �cos 1�sin 1
�CF� d r =�
�
0
cos t, sin t, �t � 1, cos t, �sin t dt=��
0
(cos t+sin tcos t+tsin t)dt
= sin t+
1
2 sin
2
t+(sin t�tcos t)
�
0
=�
F(x, y)=(x�y) i+xy j C C
C C F�T
C
C
�CF� dr=�CF�T ds
�CF� dr C r(t)=2cos t i+2sin t j 0� t�
3�
2
F(r(t))=(2cos t�2sin t) i+4cos tsin t j r
/
(t)=�2sin t i+2cos t j
 point in roughly the same direction as , so the tangential component is positive. Then
 is positive. On the other hand, no vectors starting on point in the same
direction as , while some vectors point in roughly the opposite direction, so we would expect
 to be negative.
19. , so and .
Thus .
20. , .
Thus .
21.
22.
23. We graph and the curve . We see that most of the vectors starting on 
point in roughly the same direction as , so for these portions of the tangential component is
positive. Although some vectors in the third quadrant which start on point in roughly the opposite
direction, and hence give negative tangential components, it seems reasonable that the effect of these
portions of is outweighed by the positive tangential components. Thus, we would expect
 to be positive.
To verify, we evaluate . The curve can be represented by , 
, so and . Then
 6
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.2 Line Integrals
�
C
F� dr = �3� /2
0
F(r(t))� r
/
t( ) dt
= �3� /2
0
�2sin t(2cos t�2sin t)+2cos t(4cos tsin t) dt
= 4�3� /2
0
(sin
2
t�sin tcos t+2sin tcos
2
t)dt
= 3� +
2
3
F(x, y)= x
x
2
+y
2
i+ y
x
2
+y
2
j C
C C F�T
C C
F�T
�
C
F� dr=�
C
F�T ds
�CF� dr C r(t)=t i+(1+t
2
) j �1� t� 1
F(r(t))= t
t
2
+(1+t
2
)
2
i+ 1+t
2
t
2
+(1+t
2
)
2
j r
/
(t)=i+2t j
�CF� dr = �
1
�1
F(r(t))� r
/
(t)dt
= �
1
�1
t
t
2
+(1+t
2
)
2
+ 2t (1+t
2
)
t
2
+(1+t
2
)
2
dt
 [using a CAS]
24. We graph and the curve . In the first quadrant, each vector
starting on points in roughly the same direction as , so the tangential component is positive.
In the second quadrant, each vector starting on points in roughly the direction opposite to , so
 is negative. Here, it appears that the tangential components in the first and second quadrants
counteract each other, so it seems reasonable to guess that is zero. To verify, we
evaluate . The curve can be represented by , , so
 and . Then
 7
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.2 Line Integrals
= �1�1
t (3+2t
2
)
t
4
+3t
2
+1
dt=0
�
C
F� dr �1
0
e
t
2
�1
, t
5
� 2t, 3t
2
dt=�1
0
2te
t
2
�1
+3t
7( )dt= et2�1+ 38 t8
1
0
=
11
8 �1/e
r(0)=0 F(r(0))= e
�1
, 0
r
1
2
=
1
2 ,
1
2 2
F r
1
2
= e
�1/2
,
1
4 2
r(1)= 1, 1 F(r(1))= 1, 1
PLOT
plot
v1:=PLOT(CURVES( [ [0, 0], [evalf(1/exp(1)), 0] ] ) );
(0, 0) v1.
display ListPlot PlotJoined � > True
Show
�
C
F� dr=�1
�1
2t, t
2
, 3t � 2, 3, �2t dt=�1
�1
(4t+3t
2
�6t
2
)dt= 2t
2
�t
3 1
�1
=�2
F(r(t))= 2t, t
2
, 3t F(r(�1))= �2, 1, �3 F r �
1
2 = �1,
1
4 , �
3
2
F r
1
2 = 1,
1
4 ,
3
2 F(r(1))= 2, 1, 3
 [since the integrand is an odd function]
25. (a) = 
(b) , ;
 , ;
 , .
In order to generate the graph with Maple, we use the command (not to be confused with the
 command) to define each of the vectors. For example,
 generates the vector from the vector field at
the point (but without an arrowhead) and gives it the name To show everything on the same
screen, we use the command. In Mathematica, we use (with the 
option) to generate the vectors, and then to show everything on the same screen.
26. (a)
(b) Now , so , ,
 , and .
 8
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.2 Line Integrals
x=cos
3
t y=sin
3
t 0� t�
�
2
dx
dt =3cos
2
t (�sin t)
dy
dt =3sin
2
tcos t
dx
dt
2
+
dy
dt
2
= 9cos
4
tsin
2
t+9sin
4
tcos
2
t =3cos tsin t cos
2
t+sin
2
t =3cos tsin t
�
C
x
3
y
5
ds=�� /2
0
cos
9
tsin
15
t (3cos tsin t)dt=
945
16,777,216 �
(t)= 1, 2, 1 +t 6, 4, 5 � 1, 2, 1 =(1+5t)i+(2+2t) j+(1+4t)k
0� t� 1
�
C
F� dr =�1
0
(1+5t)
4
e
2+2t
, ln (1+4t), (2+2t)
2
+(1+4t)
2
� 5, 2, 4 dt
=
5235e
4
4 �
6285e
2
4 +
9 5sinh
�1 14
3
25 �
9 5sinh
�1 4
3
25 +
5ln 5
2 +
14 41
5 �
4 5
5 �2
=
5235e
4
4 �
6285e
2
4 �
18 5ln 3
25 +
9 5ln (14+ 205 )
25 +
5ln 5
2 +
14 41 �4 5
5 �2
�Cxsin yds=�
2
1
ln tsin (e
�t
) (1/t)
2
+(�e
�t
)
2
dt�0.052
C r(t)=2cos t i+2sin t j 0� t� 2�
F(r(t))= 4cos
2
t, 4cos tsin t r
/
(t)= �2sin t, 2cos t
W =�
C
F� dr=�2�
0
(�8cos
2
tsin t+8cos
2
tsin t)dt=0
27. The part of the astroid that lies in the quadrant is parametrized by , , .
Now and , so
 .
Therefore .
28. We parametrize the line as r ,
 . Using a CAS, we calculate
The first answer is the one given by Maple. The two answers are equivalent by Equation 7.6.3 .
29. A calculator or CAS gives .
30. (a) We parametrize the circle as , . So
 , , and
 .
(b)
From the graph, we see that all of the vectors in the field are perpendicular to the path. This indicates
that the field does no work on the particle, since the field never pulls the particle in the direction in
 9
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.2 Line Integrals
C F�T=0 �CF� dr=0
x=2cos t y=2sin t �
�
2 � t�
�
2
ds=
dx
dt
2
+
dy
dt
2
dt= (�2sin t)
2
+(2cos t)
2
dt=2dt m=�Ckds=2k�
� /2
�� /2
dt=2k(� )
x=
1
2� k �Cxkds=
1
2� �
� /2
�� /2
(2cos t)2dt=
1
2� 4sin t
� /2
�� /2
=
4
� y=
1
2� k �Cykds=
1
2� �
� /2
�� /2
(2sin t)2dt=0
(x,y)=
4
� ,0
x=rcos t y=rsin t 0� t�
�
2
ds=
dx
dt
2
+
dy
dt
2
dt= (�rsin t)
2
+(rcos t)
2
dt=r dt
m=�C(x+y)ds=�
� /2
0
(rcos t+rsin t)r dt=r
2
sin t�cos t
� /2
0
=2r
2
x =
1
2r
2
�Cx(x+y)ds=
1
2r
2
�� /2
0
(r
2
cos
2
t+r
2
cos tsin t)r dt=
r
2
t
2 +
sin 2t
4 �
cos 2t
4
� /2
0
=
r(� +2)
8
y =
1
2r
2
�Cy(x+y)ds=
1
2r
2
�� /2
0
(r
2
sin tcos t+r
2
sin
2
t)r dt
=
r
2 �
cos 2t
4 +
t
2 �
sin 2t
4
� /2
0
=
r(� +2)
8
(x, y)=
r(� +2)
8 ,
r(� +2)
8 .
x=
1
m �Cx� (x, y, z)ds y=
1
m �Cy� (x, y, z)ds z=
1
m �Cz� (x, y, z)ds m=�C� (x, y, z)ds
m=�Ckds=k�
2�
0
4sin
2
t+4cos
2
t+9 dt=k 13�2�
0
dt=2� k 13 x=
1
2� k 13
�2�
0
k2 13 sin t dt=0
y=
1
2� k 13
�2�
0
k2 13 cos t dt=0 z=
1
2� k 13
�2�
0
(k 13 )(3t)dt=3
2� (2�
2
)=3�
which it is going. In other words, at any point along , , and so certainly .
31. We use the parametrization , , . Then
 , so ,
 , .
Hence .
32. We use the parametrization , , . Then
 , so
 ,
 , and
 .
Therefore
33. (a) , , where .
(b) , ,
 , . Hence
 10
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.2 Line Integrals
(x, y, z)=(0, 0, 3� )
m =�C(x
2
+y
2
+z
2
)ds=�2�
0
(t
2
+1) (1)
2
+(�sin t)
2
+(cos t)
2
dt=�2�
0
(t
2
+1) 2 dt
= 2
8
3 �
3
+2�
x=
1
2
8
3 �
3
+2�
�2�
0
2 (t
3
+t)dt
4�
4
+2�
2
8
3 �
3
+2�
3� (2�
2
+1)
4�
2
+3
y=
3
2 2� (4�
2
+3)
�2�
0
( 2 cos t)(t
2
+1)dt=0
z=
3
2 2� (4�
2
+3)
�2�
0
( 2sin t)(t
2
+1)dt=0 (x, y, z)=
3� (2�
2
+1)
4�
2
+3
, 0, 0
� (x, y)=k(1�y) x=cos t y=sin t ds=dt 0� t���
Ix = �Cy
2
� (x, y)ds=��
0
sin
2
t dt=k��
0
(sin
2
t�sin
3
t)dt
= 12 k�
�
0
(1�cos 2t)dt�k��
0
(1�cos
2
t)sin t dt u=t,du=�tdt�3pt
= k
�
2 +�
�1
1
(1�u
2
)du =k
�
2 �
4
3
Iy = �Cx
2
� (x, y)ds=k��
0
cos
2
t (1�sin t)dt=
k
2 �
�
0
(1+cos 2t)dt�k��
0
cos
2
tsin t dt
= k
�
2 �
2
3
x=2sin t y=2cos t z=3t 0� t� 2� � (x, y, z)=k
ds= (2cos t)
2
+(�2sin t)
2
+3
2
= 4(cos
2
t+sin
2
t)+9 = 13
Ix =�C(y
2
+z
2
)� (x, y, z)ds=�2�
0
(4cos
2
t+9t
2
)(k) 13 dt= 13 k 4
1
2 t+
1
4 sin 2t +3t
3 2�
0
 .
34.
 ,
 = = ,
 , and
 . Hence .
35.
From Example 3, , , , and , 
 [Let in the second integral]
 , using the same substitution as above.
36.
The wire is given as , , , with . Then
 and
 11
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.2 Line Integrals
= 13 k(4� +24�
3
)=4 13� k(1+6�
2
)
Iy =�C(x
2
+z
2
)� (x, y, z)ds=�2�
0
4sin
2
t+9t
2( )(k) 13 dt= 13 k 4 12 t�
1
4 sin 2t +3t
3 2�
0
= 13 k(4� +24�
3
)=4 13� k(1+6�
2
)
Iz =�C(x
2
+y
2
)� (x, y, z)ds=�2�
0
(4sin
2
t+4cos
2
t)(k) 13 dt=4 13 k�2�
0
dt=8� 13 k
W =�CF� dr=�
2�
0
t�sin t, 3�cos t � 1�cos t, sin t dt
=�2�
0
(t�tcos t�sin t+sin tcos t+3sin t�sin tcos t)dt
=�2�
0
(t�tcos t+2sin t)dt=
1
2 t
2
�(tsin t+cos t)�2cos t
2�
0
=2�
2
x=x y=x
2
�1� x� 2
W =�2
�1
xsin x
2
, x
2
� 1, 2x dx=�2
�1
(xsin x
2
+2x
3
)dx= �
1
2 cos x
2
+
1
2 x
4 2
�1
=
1
2 (15+cos 1�cos 4)
r(t)= 1+2t, 4t, 2t 0� t� 1
W =�CF� d r=�
1
0
6t, 1+4t, 1+6t � 2, 4, 2 dt=�1
0
(12t+4(1+4t)+2(1+6t))dt
=�1
0
(40t+6)dt= 20t
2
+6t
1
0
=26
r(t)=2 i+t j+5t k 0� t� 1
W = �CF� dr=�
1
0
K 2, t, 5t
(4+26t
2
)
3/2
� 0, 1, 5 dt=K�1
0
26t
(4+26t
2
)
3/2
dt
37.
 [by integrating by parts in the second
term]
38.
 , , ,
39.
 , ,
40.
 , . Therefore
 12
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.2 Line Integrals
= K �(4+26t
2
)
�1/2 1
0
=K
1
2 �
1
30
F=185k
x=20cos t y=20sin t z=
90
6� t=
15
� t 0� t� 6��
W = �CF� dr=�
6�
0
0, 0, 185 � �20sin t, 20cos t,
15
� dt=(185)
15
� �
6�
0
dt=(185)(90)
� 1.67�10
4
ft�lb
m t m=185�
9
6� t=185�
3
2� t F= 185�
3
2� t k
x=20cos t y=20sin t z=
90
6� t=
15
� t 0� t� 6�
W = �CF� dr=�
6�
0
0, 0, 185�
3
2� t � �20sin t, 20cos t,
15
� dt=
15
� �
6�
0
185�
3
2� t dt
= 15
� 185t�
3
4� t
2 6�
0
=90 185�
9
2 �1.62�10
4
ft�lb
r(t)= cos t, sin t 0� t� 2� F= a, b
W =�CF� d r =�
2�
0
a, b � �sin t, cos t dt
=�2�
0
(�asin t+bcos t)dt= acos t+bsin t
2�
0
=a+0�a+0=0
F(x, y)=k x= kx, ky
W =�CF� d r =�
2�
0
kcos t, ksin t � �sin t, cos t dt
=�2�
0
(�ksin tcos t+ksin tcos t)dt=�2�
0
0dt=0
xy �
z=h(x, y)
41.
Let . To parametrize the staircase, let
 , , , 
42.
This time is a function of : . So let . To parametrize
the staircase, let , , , . Therefore
43. (a) , , and let . Then
(b) Yes. and
44. Consider the base of the fence in the plane, centered at the origin, with the height given by
 . The fence can be graphed using the parametric equations
 13
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.2 Line Integrals
x=10cos u y=10sin u
z =v 4+0.01((10cos u)
2
�(10sin u)
2
)
=v(4+cos
2
u�sin
2
u)
=v(4+cos 2u) 0� u� 2� 0� v� 1
�Ch(x, y)ds C x=10cos t y=10sin t
0� t� 2�
�Ch(x, y)ds = �
2�
0
4+0.01((10cos t)
2
�(10sin t)
2
) (�10sin t)
2
+(10cos t)
2
dt
= �2�
0
(4+cos 2t) 100 dt=10 4t+
1
2 sin 2t
2�
0
= 10(8� )=80�m
2
160�
2
100
2 160�
100 =1.6��5.03
�CF� dr=�CF�T ds
C 7 �s=2 F�T
(x
*
i , y
*
i ) C F�T
C (x
*
i , y
*
i )
�CF�T ds�	
7
i =1 F(x
*
i , y
*
i )�T (x
*
i , y
*
i ) �s=[2+2+2+2+1+1+1](2)=22.
B
B=|B|T T C x=rcos� y=rsin�
 , ,
 , , .
The area of the fence is where , the base of the fence, is given by , 
, . Then
If we paint both sides of the fence, the total surface area to cover is m , and since 1 L of paint
covers m , we require L of paint.
45. The work done in moving the object is . We can approximate this integral by
dividing into segments of equal length and approximating , that is, the tangential
component of force, at a
point on each segment. Since is composed of straight line segments, is the scalar
projection of each force vector onto . If we choose to be the point on the segment closest to
the origin, then the work done is
 Thus, we estimate the work
done to be approximately 22 J.
46. Use the orientation pictured in the figure. Then since is tangent to any circle that lies in the
plane perpendicular
to the wire, where is the unit tangent to the circle : , . Thus
 14
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.2 Line Integrals
B=|B| �sin� , cos�
�
C
B� dr=�2�
0
|B| �sin� , cos� � �rsin� , rcos� d� =�2�
0
|B|r d� =2� r |B|. |B|
r �
C
B� dr=�
0
I.
|B|=
�
0
I
2� r
 . Then
 (Note that here is the
magnitude of the field at a distance from the wire’s center). But by Ampere’s Law 
Hence .
 15
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.2 Line Integrals
C � f f
�
C
� f � dr f
C 50�10=40.
C r(t)=(t
2
+1) i+(t
3
+t) j 0� t� 1 r
/
(t)=2t i+(3t
2
+1) j
3t
2
+1�0 r
/
(t)�0 C � f f
�C� f � dr= f (r(1))� f (r(0))= f (2, 2)� f (1, 0)=9�3=6
�(6x+5y)/�y=5=�(5x+4y)/�x F R
2
�
F f � f =F f x(x, y)=6x+5y
f y(x, y)=5x+4y f x(x, y)=6x+5y f (x,y)=3x
2
+5xy+g(y)
y f y(x, y)=5x+g
/
(y) 5x+4y=5x+g
/
(y) g
/
(y)=4y
g(y)=2y
2
+K K f (x, y)=3x
2
+5xy+2y
2
+K F
�(x
3
+4xy)/�y=4x �(4xy�y
3
)/�x=4y. F
�(xe
y
)/�y=xe
y
,�(ye
x
)/�x=ye
x
. F
�(e
y
)/�y=e
y
=�(xe
y
)/�x F R
2
F
f � f =F f x(x, y)=e
y
f (x, y)=xe
y
+g(y) f y(x, y)=xe
y
+g
/
(y)
f y(x, y)=xe
y
g
/
(y)=0	 g(y)=K f (x, y)=xe
y
+K F.
�(2xcos y�ycos x)/�y=�2xsin y�cos x=�(�x
2
sin y�sin x)/�x F R
2
F
f � f =F f x(x, y)=2xcos y�ycos x
f (x, y)=x
2
cos y�ysin x+g(y) f y(x, y)=�x
2
sin y�sin x+g
/
(y) f y(x, y)=�x
2
sin y�sin x
g
/
(y)=0	 g(y)=K f (x, y)=x
2
cos y�ysin x+K F
�(1+2xy+ln x)/�y=2x=�(x
2
)/�x F { (x, y) | x>0 } �
F f � f =F
f x(x, y)=1+2xy+ln x f (x, y)=x+x
2
y+xln x�x+g(y) f y(x, y)=x
2
+g
/
(y) f y(x, y)=x
2
g
/
(y)=0	 g(y)=K f (x,y)=x
2
y+xln x+K F
1. appears to be a smooth curve, and since is continuous, we know is differentiable. Then
Theorem 2 says that the valueof is simply the difference of the values of at the terminal
and initial points of . From the graph, this is 
2. is represented by the vector function , , so .
Since , we have , thus is a smooth curve. is continuous, and hence is
differentiable, so by Theorem 2 we have .
3. and the domain of is which is open and simply connected, so by
Theorem 6 is conservative. Thus, there exists a function such that , that is, 
and . But implies and differentiating both sides
of this equation with respect to gives . Thus so and
 where is a constant. Hence is a potential function for .
4. , Since these are not equal, is not conservative.
5. Since these are not equal, is not conservative.
6. and the domain of is . Hence is conservative so there exists a
function such that . Then implies and . But
 so . Then is a potential function for 
7. and the domain of is . Hence is
conservative so there exists a function such that . Then implies
 and . But so
 . Then is a potential function for .
8. and the domain of is which is open and simply
connected. Hence is conservative, so there exists a function such that . Then
 implies and . But so
 . Then is a potential function for .
 1
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.3 The Fundamental Theorem for Line Integrals
�(ye
x
+sin y)/�y=e
x
+cos y=�(e
x
+xcos y)/�x F R
2
F
f � f =F f x(x, y)=ye
x
+sin y f (x, y)=ye
x
+xsin y+g(y)
f y(x, y)=e
x
+xcos y+g
/
(y) f y(x, y)=e
x
+xcos y g(y)=K f (x, y)=ye
x
+xsin y+K
F
�(xycosh xy+sinh xy)
�y =x
2
ysinh xy+xcosh xy+xcosh xy=x
2
ysinh xy+2xcosh xy=
�(x
2
cosh xy)
�x
F R
2
F f � f =F
f x(x, y)=xycosh xy+sinh xy f (x, y)=xsinh xy+g(y)	 f y(x, y)=x
2
cosh xy+g
/
(y)
f y(x, y)=x
2
cosh xy g(y)=K f (x, y)=xsinh xy+K F
F �
�
�y 2xy=2x=
�
�x (x
2
) R
2
� F
F �CF� dr
C �CF� dr
f � f =F f x(x, y)=2xy f y(x, y)=x
2
f x(x, y) x f (x, y)=x
2
y+g(y)
y f y(x, y)=x
2
+g
/
(y) x
2
+g
/
(y)=x
2
	 g
/
(y)=0	 g(y)=K
f (x, y)=x
2
y+K (1, 2) (3, 2)
�
C
F� dr= f (3, 2)� f (1, 2)=18�2=16
f x(x, y)=y f (x, y)=xy+g(y) f y(x, y)=x+g
/
(y) f y(x, y)=x+2y
g
/
(y)=2y	 g(y)=y
2
+K K=0 f (x, y)=xy+y
2
�CF� dr= f (2, 1)� f (0, 1)=3�1=2
f x(x, y)=x
3
y
4
f (x, y)=
1
4 x
4
y
4
+g(y) f y(x, y)=x
4
y
3
+g
/
(y) f y(x, y)=x
4
y
3
g
/
(y)=0	 g(y)=K K=0 f (x, y)=
1
4 x
4
y
4
9. and the domain of is . Hence is conservative so
there exists a function such that . Then implies 
and . But so and is a
potential function for .
10. and
the domain of is . Thus is conservative, so there exists a function such that . Then
 implies . But
 so and is a potential function for .
11. (a) has continuous first order partial derivatives and on , which is
open and simply connected. Thus, is conservative by Theorem 6. Then we know that the line
integral of is independent of path; in particular, the value of depends only on the endpoints
of . Since all three curves have the same initial and terminal points, will have the same
value for each curve.
(b) We first find a potential function , so that . We know and .
Integrating with respect to , we have . Differentiating both sides with
respect to gives , so we must have , a constant.
Thus . All three curves start at and end at , so by Theorem 2,
 for each curve.
12. (a) implies and . But so
 . We can take , so .
(b) .
13. (a) implies and . But so
 , a constant. We can take , so .
 2
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.3 The Fundamental Theorem for Line Integrals
C r(0)=(0, 1) r(1)=(1, 2)
�CF� dr= f (1, 2)� f (0, 1)=4�0=4
f x(x, y)=y
2
/(1+x
2
) f (x, y)=y
2
arctanx+g(y)	 f y(x, y)=2yarctanx+g
/
(y)
f y(x, y)=2yarctanx g
/
(y)=0	 g(y)=K K=0 f (x, y)=y
2
arctanx.
C r(0)=(0, 0) r(1)=(1, 2)
�CF� d r= f (1, 2)� f (0, 0)=4arctan1�0=4�
�
4 =�
f x(x, y, z)=yz f (x, y, z)=xyz+g(y, z) f y(x, y, z)=xz+gy(y, z) f y(x, y, z)=xz
gy(y, z)=0	 g(y, z)=h(z) f (x, y, z)=xyz+h(z) f z(x, y, z)=xy+h
/
(z) f z(x, y, z)=xy+2z
h
/
(z)=2z	 h(z)=z
2
+K f (x, y, z)=xyz+z
2
K=0
�CF� dr= f (4, 6, 3)� f (1, 0, �2)=81�4=77
f x(x, y, z)=2xz+y
2
f (x, y, z)=x
2
z+xy
2
+g(y, z) f y(x, y, z)=2xy+gy(y, z)
f y(x, y, z)=2xy gy(y, z)=0	 g(y, z)=h(z) f (x, y, z)=x
2
z+xy
2
+h(z) f z(x, y, z)=x
2
+h
/
(z)
f z(x, y, z)=x
2
+3z
2
h
/
(z)=3z
2
	 h(z)=z
3
+K f (x, y, z)=x
2
z+xy
2
+z
3
K=0
t=0 (0, 1, �1) t=1 (1, 2, 1)
�
C
F� dr= f (1, 2, 1)� f (0, 1, �1)=6�(�1)=7
f x(x, y, z)=y
2
cos z f (x, y, z)=xy
2
cos z+g(y, z) f y(x, y, z)=2xycos z+gy(y, z)
f y(x, y, z)=2xycos z gy(y, z)=0	 g(y, z)=h(z) f (x, y, z)=xy
2
cos z+h(z)
f z(x, y, z)=�xy
2
sin z+h
/
(z) f z(x, y, z)=�xy
2
sin z h
/
(z)=0	 h(z)=K f (x, y, z)=xy
2
cos z
K=0
r(0)= 0, 0, 0 r(� )= �
2
, 0,� �
C
F� dr= f (�
2
, 0,� )� f (0, 0, 0)=0�0=0
f x(x, y, z)=e
y
f (x, y, z)=xe
y
+g(y, z) f y(x, y, z)=xe
y
+gy(y, z) f y(x, y, z)=xe
y
gy(y, z)=0	 g(y, z)=h(z) f (x, y, z)=xe
y
+h(z) f z(x, y, z)=0+h
/
(z) f z(x, y, z)=(z+1)e
z
(b) The initial point of is and the terminal point is , so
 .
14. (a) implies . But
 so . We can take , so 
(b) The initial point of is and the terminal point is , so
 .
15. (a) implies and so . But 
so . Thus and . But 
, so . Hence (taking ).
(b) .
16. (a) implies and so . But
 so . Thus and .
But , so . Hence (taking ).
(b) corresponds to the point and corresponds to , so
 .
17. (a) implies and so . But
 so . Thus and
 . But , so . Hence 
(taking ).
(b) , so .
18. (a) implies and so . But 
so . Thus and . But 
, so
 3
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.3 The Fundamental Theorem for Line Integrals
h
/
(z)=(z+1)e
z
	 h(z)=ze
z
+K f (x, y, z)=xe
y
+ze
z
K=0
r(0)= 0, 0, 0 r(1)= 1, 1, 1 �CF� dr= f (1, 1, 1)� f (0, 0, 0)=2e�0=2e
F(x, y)=tan y i+xsec
2
y j f (x, y)=xtan y F � f =F
F
�
C
tan ydx+xsec
2
ydy=�
C
F� d r= f 2,
�
4 � f (1, 0)=2tan
�
4 �tan 0=2
F(x, y)=(1�ye
�x
) i+e
�x
j f (x, y)=x+ye
�x
F � f =F
F
�
C
(1�ye
�x
)dx+e
�x
dy=�
C
F� d r= f (1, 2)� f (0, 1)=(1+2e
�1
)�1=2/e
F(x, y)=2y
3/2
i+3x y j W =�CF� d r �(2y
3/2
)/�y=3 y=�(3x y )/�x
f � f =F f x(x, y)=2y
3/2
	 f (x, y)=2xy
3/2
+g(y)	 f y(x, y)=3xy
1/2
+g
/
(y)
f y(x, y)=3x y g
/
(y)=0 g(y)=K K=0	 f (x, y)=2xy
3/2
W =�
C
F� d r= f (2, 4)� f (1, 1)=2(2)(8)�2(1)=30
F(x, y)= y
2
x
2
i�
2y
x j W =�CF� dr
�
�y
y
2
x
2
= 2y
x
2
=
�
�x �
2y
x
f � f =F f x=y
2
/x
2
	 f (x, y)=�y
2
/x+g(y)	 f y=�2y/x+g
/
(y)	 g
/
(y)=0
f (x, y)=�y
2
/x F
W =�CF� dr= f (4, �2)� f (1, 1)=�[(�2)
2
/4]+(1/1)=0
F C
�CF� dr=0 C
C C C
 (using integration by parts). Hence (taking ).
(b) , so .
19. Here . Then is a potential function for , that is, 
so is conservative and thus its line integral is independent of path. Hence
 .
20. Here . Then is a potential function for , that is, so
 is conservative and thus its line integral is independent of path. Hence
 .
21. , . Since , there exists a
function such that . In fact, .
But so or . We can take . Thus
 .
22. , . Since , there exists a
function such that . In fact, , so
we can take as a potential function for . Thus
 .
23. We know that if the vector field (call it ) is conservative, then around any closed path,
 . But take to be some circle centered at the origin, oriented counterclockwise. All of the
field vectors along oppose motion along , so the integral around will be negative. Therefore the
field is not conservative.
24.
 4
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.3 The Fundamental Theorem for Line Integrals
F
�
�y (2xy+sin y)=2x+cos y
�
�x (x
2
+xcos y)=2x+cos y
F
F
(2, 1)
�F� dr�0
�
�y
x�2y
1+x
2
+y
2
=(x�2y) �y
(1+x
2
+y
2
)
3/2
� 2
1+x
2
+y
2
= �2�2x
2
�xy
(1+x
2
+y
2
)
3/2
�
�x
x�2
1+x
2
+y
2
=(x�2) �x
(1+x
2
+y
2
)
3/2
+ 1
1+x
2
+y
2
= 1+y
2
+2x
(1+x
2
+y
2
)
3/2
� f (x, y)=cos (x�2y) i�2cos (x�2y) j
�
C1
F� dr=�
C1
� f � dr= f (r(b))� f (r(a)) C
1
t=a t=b
f (0, 0)=sin 0=0 f (� ,� )=sin (� �2� )=0 C
1
(0, 0) (� ,� ) r(t)=� t i+� t j 0� t� 1
�
C2
F� dr= f (r(b))� f (r(a)) f (0, 0)=sin 0=0 f
�
2 ,0 =1
C
2
r(t)=
�
2 t i 0� t� 1 (0, 0)
�
2 , 0
From the graph, it appears that is conservative, since around all closed paths, the number and size
of the field vectors pointing in directions similar to that of the path seem to be roughly the same as the
number and size of the vectors pointing in the opposite direction. To check, we calculate
 , 
Thus is conservative, by Theorem 6.
25. From the graph, it appears that is not conservative. For example, any closed curve containing
the point seems to have many field vectors pointing counterclockwise along it, and none
pointing clockwise. So along this path the integral . To confirm our guess, we calculate
 ,
 .
These are not equal, so the field is not conservative, by Theorem 5.
26.
(a) We use Theorem 2: where starts at and ends at .
So because and , one possible curve is the straight line from
 to ; that is, , .
(b) From (a), . So because and , one possible
curve is , , the straight line from to .
 5
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.3 The Fundamental Theorem for Line Integrals
F f F=� f P= f x Q= f y
R= f z P Q R
�P/�y= f xy= f yx=�Q/�x �P/�z= f xz= f zx=�R/�x �Q/�z= f yz= f zy=�R/�y
F(x, y, z)=y i+x j+xyzk �P/�z=0 �R/�x=yz
F F
D={ (x, y) | x>0, y>0 }=
D D D
D D D
D �
D={ (x, y) | x�0 } xy � y �
D
y � D D
D �
D={ (x, y) | 1<x
2
+y
2
<4 }= (0, 0) 1
2
D
D
D � x
2
+y
2
=(1.5)
2
D
D D �
D={ (x, y) | x
2
+y
2
� 1 4� x
2
+y
2
� 9 }= x
2
+y
2
=1
x
2
+y
2
=4 x
2
+y
2
=9
D (0, 2) D
D (0, 0) (0, 2.5) D
D
D � x
2
+y
2
=9 D
D
P=� y
x
2
+y
2
�P
�y =
y
2
�x
2
(x
2
+y
2
)
2
Q= x
x
2
+y
2
�Q
�x =
y
2
�x
2
(x
2
+y
2
)
2
�P
�y =
�Q
�x
C
1
x=cos t y=sin t 0� t�� C
2
x=cos t y=sin t t=2� t=�
27. Since is conservative, there exists a function such that , that is, , , and
 . Since , and have continuous first order partial derivatives, Clairaut’s Theorem says that
 , , and .
28. Here . Then using the notation of Exercise 27, while .
Since these aren’t equal, is not conservative. Thus by Theorem 4, the line integral of is not
independent of path.
29. the first quadrant (excluding the axes).
(a) is open because around every point in we can put a disk that lies in .
(b) is connected because the straight line segment joining any two points in lies in .
(c) is simply connected because it’s connected and has no holes.
30. consists of all points in the plane except for those on the axis.
(a) is open.
(b) Points on opposite sides of the axis cannot be joined by a path that lies in , so is not
connected.
(c) is not simply connected because it is not connected.
31. the annular region between the circles with center and radii and
 .
(a) is open.
(b) is connected.
(c) is not simply connected. For example, is simple and closed and lies within but
encloses points that are not in . (Or we can say, has a hole, so is not simply connected.)
32. or the points on or inside the circle , together with
the points on or between the circles and .
(a) is not open because, for instance, no disk with center lies entirely within .
(b) is not connected because, for example, and lie in but cannot be joined by a
path that lies entirely in .
(c) is not simply connected because, for example, is a simple closed curve in but
encloses points that are not in .
33. (a) , and , . Thus .
(b) : , , , : , , to . Then
 6
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.3 The Fundamental Theorem for Line Integrals
�
C1
F� dr=��
0
(�sin t)(�sin t)+(cos t)(cos t)
cos
2
t+sin
2
t
dt=��
0
dt=� �
C2
F� dr=��
2�
dt=��
F
�
C3
F� dr=�2�
0
dt=2� C
3
x
2
+y
2
=1
F R
2
�
F(r)=cr/ |r|
3
r=x i+y j+zk f (r)=�c/ |r| F
� f =F F
P
1
=(x
1
, y
1
, z
1
) P
2
=(x
2
, y
2
, z
2
)
W =�CF� dr= f (P2)� f (P1)=�
c
(x
2
2
+y
2
2
+z
2
2
)
1/2
+ c
(x
2
1
+y
2
1
+z
2
1
)
1/2
=c 1
d
1
� 1
d
2
c=�(mMG)	
W = �mMG
1
1.52
10
8
� 1
1.47
10
8
= �(5.97
10
24
)(1.99
10
30
)(6.67
10
�11
)(�2.2377
10
�10
)�1.77
10
35
J
c=� qQ	
W =� qQ 1
10
�12
� 1
5
10
�13
=(8.985
10
10
)(1)(�1.6
10
�19
)(�10
12
)�1.4
10
4
 and 
Since these aren’t equal, the line integral of isn’t independent of path. (Or notice that
 where is the circle , and apply the contrapositive of Theorem 3). This
doesn’t contradict Theorem 6, since the domain of , which is except the origin, isn’t simply
connected.
34. (a) Here and . Then is a potential function for , that is,
 . (See the discussion of gradient fields in Section 17.1). Hence is conservative and its line
integral is independent of path. Let and .
 .
(b) In this case, 
(c) In this case, 
 J.
 7
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.3 The Fundamental Theorem for Line Integrals
C
1
x=t � dx=dt y=0� dy=0dt 0� t� 2
C
2
x=2� dx=0dt y=t � dy=dt 0� t� 3
C
3
x=2�t � dx=�dt y=3� dy=0dt 0� t� 2
C
4
x=0� dx=0dt y=3�t � dy=�dt 0� t� 3
�
C
xy
2
dx+x
3
dy �
C1+C2+C3+C4
xy
2
dx+x
3
dy
=�
0
2
0dt+�
0
3
8dt+�
0
2
�9(2�t)dt+�
0
3
0dt
=0+24�18+0=6
�
C
xy
2
dx+x
3
dy =��
D
�
�x (x
3
)�
�
�y (xy
2
) dA=�
0
2
�
0
3
(3x
2
�2xy)dydx
=�
0
2
(9x
2
�9x)dx=24�18=6
x=cos t y=sin t 0� t� 2� �
C
ydx�xdy=�
0
2�
[sin t(�sin t)�cos t(cos t)]dt=� �
0
2�
dt=�2�
�
C
ydx�xdy=��
D
�
�x (�x)�
�
�y (y) dA=�2��D
dA=�2A(D)=�2� (1)
2
=�2�
1. (a)
 : , , .
 : , , .
 : , , .
 : , , .
Thus
(b)
2. (a) , , . Then .
(b)
3. (a)
 1
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.4 Green’s Theorem
C
1
x=t � dx=dt y=0� dy=0dt 0� t� 1C
1
x=t � dx=dt y=0� dy=0dt 0� t� 1
C
2
x=1� dx=0dt y=t � dy=dt 0� t� 2C
2
x=1� dx=0dt y=t � dy=dt 0� t� 2
C
3
x=1�t � dx=�dt y=2�2t � dy=�2dt 0� t� 1C
3
x=1�t � dx=�dt y=2�2t � dy=�2dt 0� t� 1
�
C
xydx+x
2
y
3
dy = �
C1+C2+C3
xydx+x
2
y
3
dy
=�
0
1
0dt+�
0
2
t
3
dt+�
0
1
�(1�t)(2�2t)�2(1�t)
2
(2�2t)
3
dt
=0+
1
4 t
4 2
0
+
2
3 (1�t)
3
+
8
3 (1�t)
6 1
0
=4�
10
3 =
2
3
�
C
xydx+x
2
y
3
dy =��
D
�
�x (x
2
y
3
)�
�
�y (xy) dA=�0
1
�
0
2x
(2xy
3
�x)dydx
=�
0
1 1
2 xy
4
�xy
y=2x
y=0
dx=�
0
1
(8x
5
�2x
2
)dx=
4
3 �
2
3 =
2
3
C
1
:x=0� dx=0dty=1�t�
dy=�dt
0� t� 1
C
2
:x=t � dx=dt
y=0�
dy=0dt
0� t� 1
C
3
:x=1�t � dx=�dt
y=1�(1�t)
2
=2t�t
2
�
dy=(2�2t)dt
0� t� 1
 : , , : , , 
 : , , . : , , .
 : , , . : , , .
ThusThus
(b)
4. (a)
,
 ,
,
 ,
,
,
Thus 2
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.4 Green’s Theorem
�
C
xdx+ydy = �
C1+C2+C3
xdx+ydy
=�
0
1
(0dt+(1�t)(�dt))+�
0
1
(t dt+0dt)+�
0
1
((1�t)(�dt)+(2t�t
2
)(2�2t)dt)
=
1
2 t
2
�t
1
0
+
1
2 t
2 1
0
+
1
2 t
4
�2t
3
+
5
2 t
2
�t
1
0
=�
1
2 +
1
2 +
1
2 �2+
5
2 �1 =0
�
C
xdx+ydy=��
D
�
�x (y)�
�
�y (x) dA=��D
0dA=0
C x=cos� y=sin� 0��� 2�
�
C
Pdx+Qdy=�
0
2�
cos
4
� sin
5
� (�sin� )d� +�
0
2�
(�cos
7
� sin
6
� )cos� d� =�
29�
1024
��
D
�Q
�x �
�P
�y dA=��1
1
�
� 1�x
2
1�x
2
(�7x
6
y
6
�5x
4
y
4
)dydx=�
29�
1024
y=x
2
C y=x
�
C
Pdx+Qdy = �
0
1
x
4
sin x+x
2
sin (x
2
)(2x) dx+�
1
0
(x
2
sin x+x
2
sin x)dx
= �16cos 1�23sin 1+28
(b)
5. We can parametrize as , , . Then the line integral is
 , according to a CAS.
The double integral is
 ,
verifying Green’s Theorem in this case.
6.
Since along the first part of and along the second part, the line integral is
according to a CAS. The double integral is
 3
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.4 Green’s Theorem
��
R
�Q
�x �
�P
�y dA=�0
1
�
x
2
x
(2xsin y�2ysin x)dydx=�16cos 1�23sin 1+28
D C [0, 1]� [0, 1]
�
C
e
y
dx+2xe
y
dy =��
D
�
�x (2xe
y
)�
�
�y (e
y
) dA=�
0
1
�
0
1
(2e
y
�e
y
)dydx
=�
0
1
dx�
0
1
e
y
dy=(1)(e
1
�e
0
)=e�1
D C { (x, y) | 0� x� 1, 3x� y� 3 }
�
C
x
2
y
2
dx+4xy
3
dy =��
D
�
�x (4xy
3
)�
�
�y (x
2
y
2
) dA=�
0
1
�
3x
3
(4y
3
�2x
2
y)dydx
=�
0
1
y
4
�x
2
y
2 y=3
y=3xdx=�0
1
(81�9x
2
�72x
4
)dx=81�3�
72
5 =
318
5
�
C
(y+e
x
)dx+(2x+cos y
2
)dy =��
D
�
�x (2x+cos y
2
)�
�
�y (y+e
x
) dA
=�
0
1
�
y
2
y
(2�1)dxdy=�
0
1
(y
1/2
�y
2
)dy=
1
3
�
C
xe
�2x
dx+(x
4
+2x
2
y
2
)dy =��
D
�
�x (x
4
+2x
2
y
2
)�
�
�y (xe
�2x
) dA=��
D
(4x
3
+4xy
2
�0)dA
=4��
D
x(x
2
+y
2
)dA=4�
0
2�
�
1
2
(rcos� )(r
2
)rdr d�
=4�
0
2�
cos� d� �
1
2
r
4
dr=4 sin�
2�
0
1
5 r
5 2
1
=0
7. The region enclosed by is , so
8. The region enclosed by is given by , so
9.
10.
11.
 4
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.4 Green’s Theorem
�
C
y
3
dx�x
3
dy =��
D
�
�x (�x
3
)�
�
�y (y
3
) dA=��
D
(�3x
2
�3y
2
)dA=�
0
2�
�
0
2
(�3r
2
)rdr d�
=�3�
0
2�
d� �
0
2
r
3
dr=�3(2� )(4)=�24�
�
C
sin ydx+xcos ydy=��
D
�
�x (xcos y)�
�
�y (sin y) dA=��D
(cos y�cos y)dA=��
D
0dA=0
F(x, y)= x +y
3
, x
2
+ y D C
{ (x, y) | 0� x�� , 0� y� sin x }.C �C
�
C
F� dr =� �
�C
x +y
3( )dx+ x2+ y( )dy=���
D
�
�x x
2
+ y( )� ��y x +y
3( ) dA
=��
0
�
�
0
sin x
(2x�3y
2
)dydx=��
0
�
2xy�y
3 y=sin x
y=0 dx
=��
0
�
(2xsin x�sin
3
x)dx=��
0
�
(2xsin x�(1�cos
2
x)sin x)dx
=� 2sin x�2xcos x+cos x�
1
3 cos
3
x
�
0
=� 2� �2+
2
3 =
4
3 �2�
F(x, y)= y
2
cos x, x
2
+2ysin x D C
{ (x, y) | 0� x� 2, 0� y� 3x } C �C
�
C
F� dr =� �
�C
(y
2
cos x)dx+(x
2
+2ysin x)dy
=���
D
�
�x (x
2
+2ysin x)�
�
�y (y
2
cos x) dA
=���
D
(2x+2ycos x�2ycos x)dA=��
0
2
�
0
3x
2xdydx
12.
13. and the region enclosed by is given by
 is traversed clockwise, so gives the positive orientation.
14. and the region enclosed by is given by
 . is traversed clockwise, so gives the positive orientation.
 5
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.4 Green’s Theorem
=��
0
2
2x y
y=3x
y=0 dx=��0
2
6x
2
dx= �2x
3 2
0
=�16
F(x, y)= e
x
+x
2
y, e
y
�xy
2
D C x
2
+y
2
� 25
C �C
�
C
F� dr =� �
�C
(e
x
+x
2
y)dx+(e
y
�xy
2
)dy
=���
D
�
�x (e
y
�xy
2
)�
�
�y (e
x
+x
2
y) dA=���
D
(�y
2
�x
2
)dA
=��
D
(x
2
+y
2
)dA=�
0
2�
�
0
5
(r
2
)rdr d� =�
0
2�
d� �
0
5
r
3
dr=2�
1
4 r
4 5
0
=
625
2 �
F x, y( )= y�ln (x2+y2), 2tan �1 yx D C
1 (2, 3).C
�CF� d r =�
C
(y�ln (x
2
+y
2
))dx+ 2tan
�1 y
x dy
=��
D
�
�x 2tan
�1 y
x �
�
�y (y�ln (x
2
+y
2
)) dA
=��
D
2 �yx
�2
1+(y/x)
2
� 1� 2y
x
2
+y
2
dA=��
D
� 2y
x
2
+y
2
�1+ 2y
x
2
+y
2
dA
=���
D
dA=�( D)=��
W =�
C
F� dr=�
C
x(x+y)dx+xy
2
dy=��
D
(y
2
�x)dydx C
D C
W = �
0
1
�
0
1�x
(y
2
�x)dydx=�
0
1 1
3 y
3
�xy
y=1�x
y=0
dx=�
0
1 1
3 (1�x)
3
�x(1�x) dx
15. and the region enclosed by is the disk .
 is traversed clockwise, so gives the positive orientation.
16. and the region enclosed by is the disk with radius
 centered at is oriented positively, so
 area of 
17. By Green’s Theorem, where is the path described
in the question and is the triangle bounded by . So
 6
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.4 Green’s Theorem
= �
1
12 (1�x)
4
�
1
2 x
2
+
1
3 x
3 1
0
= �
1
2 +
1
3 � �
1
12 =�
1
12
W =�
C
F� dr=�
C
xdx+(x
3
+3xy
2
)dy=��
D
(3x
2
+3y
2
�0)dA D
C
W =3�
0
2
�
0
�
r
2
� rd� dr=3�
1
4 r
4 2
0
=12�
C
1
(0, 0) (2� , 0) 0� t� 2�
C
2
(2� , 0) (0, 0) C
2
x=2� �t y=0 0� t� 2� .
C=C
1
	C
2
�C �C
A=� �
�C
ydx =�
C1
ydx+�
C2
ydx=�
0
2�
(1�cos t)(1�cos t)dt+�
0
2�
0(�dt)
=�
0
2�
(1�2cos t+cos
2
t)dt+0= t�2sin t+
1
2 t+
1
4 sin 2t
2�
0
=3�
A= �
C
xdy=�
0
2�
(5cos t�cos 5t)(5cos t�5cos 5t)dt
=�
0
2�
(25cos
2
t�30cos tcos 5t+5cos
2
5t)dt
= 25
1
2 t+
1
4 sin 2t �30
1
8 sin 4t+
1
12 sin 6t +5
1
2 t+
1
20 sin 10t
2�
0
=30�
x=(1�t)x
1
+tx
2
y=(1�t)y
1
+ty
2
0� t� 1 dx=(x
2
�x
1
)dt dy=(y
2
�y
1
)dt
�
C
xdy�ydx = �
0
1
[(1�t)x
1
+tx
2
](y
2
�y
1
)dt+[(1�t)y
1
+ty
2
](x
2
�x
1
)dt
18. By Green’s Theorem, , where is the
semicircular region bounded by . Converting to polar coordinates, we have
 .
19. Let be the arch of the cycloid from to , which corresponds to , and let
 be the segment from to , so is given by , , Then
 is traversed
clockwise, so is oriented positively. Thus encloses the area under one arch of the cycloid and
from (5) we have
20.
21. (a) Using Equation 17.2.8 , we write parametric equations of the line segment as ,
 , . Then and , so
 7
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.4 Green’s Theorem
= �
0
1
(x
1
(y
2
�y
1
)�y
1
(x
2
�x
1
)+t [(y
2
�y
1
)(x
2
�x
1
)�(x
2
�x
1
)(y
2
�y
1
)])dt
= �
0
1
(x
1
y
2
�x
2
y
1
)dt=x
1
y
2
�x
2
y
1
C=C
1
	C
2
	 � � � 	Cn Ci
(xi, yi) (xi+1, yi+1) i=1 2 .. . n�1 Cn (xn, yn) (x1, y1)
1
2 �C
xdy�ydx=��
D
dA D C
=A(D)=��
D
dA=
1
2 �C
xdy�ydx
=
12 �
C1
xdy�ydx+�
C2
xdy�ydx+...+ �
Cn�1
xdy�ydx+�
Cn
xdy�ydx
A(D)=[(x
1
y
2
�x
2
y
1
)+(x
2
y
3
�x
3
y
2
)+...+(xn�1yn�xnyn�1)+(xny1�x1yn)].
A =
1
2 [(0� 1�2� 0)+(2� 3�1� 1)+(1� 2�0� 3)+(0� 1�(�1)� 2)+(�1� 0�0� 1)]
=
1
2 (0+5+2+2)=
9
2
1
2A �C
x
2
dy=
1
2A ��D
2xdA=
1
A ��D
xdA=x
�
1
2A �C
y
2
dx=�
1
2A ��D
(�2y)dA=
1
A ��D
ydA=y
A=
1
2 (1)(1)=
1
2 C=C1+C2+C3 C1 x=x y=0 0� x� 1
C
2
x=x y=1�x x=1 x=0 C
3
x=0 y=1 y=0
x=
1
2A �C
x
2
dy=�
C1
x
2
dy+�
C2
x
2
dy+�
C3
x
2
dy=0+�
1
0
(x
2
)(�dx)+0=
1
3 .
(b) We apply Green’s Theorem to the path , where is the line segment that
joins to for , , , , and is the line segment that joins to 
. From (5), , where is the polygon bounded by . Therefore
 area of polygon
To evaluate these integrals we use the formula from (a) to get
(c)
22. By Green’s Theorem, and
 .
23. Here and , where : , , ;
 : , , to ; and : , to . Then
 Similarly,
 8
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.4 Green’s Theorem
y=�
1
2A �C
y
2
dx=�
C1
y
2
dx+�
C2
y
2
dx+�
C3
y
2
dx=0+�
1
0
(1�x)
2
(�dx)+0=
1
3 .
(x, y)=
1
3 ,
1
3 .
A=
� a
2
2 x=
1
� a
2
�
C
x
2
dy y=� 1
� a
2
�
C
y
2
dx.
x =
1
� a
2
�
C1+C2
x
2
dy
= 1
� a
2
0+�
0
�
(a
2
cos
2
t)(acos t)dt =0
y=� 1
� a
2
�
�a
a
0dx+�
0
�
(a
2
sin
2
t)(�asin t)dt =
a
� �0
�
sin
3
tdt=
a
� �cos t+
1
3 (cos
3
t)
�
0
=
4a
3�
(x, y)= 0,
4a
3�
�
1
3 � �C
y
3
dx=�
1
3 ���D
(�3y
2
)dA=��
D
y
2
�dA=Ix
1
3 � �C
x
3
dy=
1
3 ���D
(3x
2
)dA=��
D
x
2
�dA=Iy
Therefore
24. so and 
Orienting the semicircular region as in the figure,
and
 .
Thus .
25. By Green’s Theorem, and
 .
26. By symmetry the moments of inertia about any two diameters are equal. Centering the disk at the
origin, the moment of inertia about a diameter equals
 9
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.4 Green’s Theorem
Iy =
1
3 � �C
x
3
dy=
1
3 � �0
2�
(a
4
cos
4
t)dt=
1
3 a
4
� �
0
2� 3
8 +
1
2 cos 2t+
1
8 cos 4t dt
= 13 a
4
� �
3(2� )
8 =
1
4 � a
4
�
C
D P=�y/(x
2
+y
2
) Q=x/(x
2
+y
2
)
D
�P/�y=�Q/�x �
C
F� dr=��
D
0dA=0
D D={ (x, y) | f
1
(y)� x� f
2
(y), c� y� d } f
1
f
2
��
D
�Q
�x dA=�c
d
�
f 1(y)
f 2(y) �Q
�x dxdy=�c
d
Q( f
2
(y), y)�Q( f
1
(y), y) dy
�
C
Qdy= �
C1+C2+C3+C4
Qdy.
�
C1
Qdy=�
d
c
Q( f
1
(y), y)dy �
C2
Qdy=�
C4
Qdy=0 �
C3
Qdy=�
c
d
Q( f
2
(y), y)dy
�
C
Qdy=�
c
d
Q( f
2
(y), y)�Q( f
1
(y), y) dy=��
D
(�Q/�x)dA.
��
R
dxdy=A(R)=�
�R
xdy x=g(u, v)
dy=
�h
�u du+
�h
�v dv �S
�R
27. Since is a simple closed path which doesn’t pass through or enclose the origin, there exists an
open region that doesn’t contain the origin but does contain . Thus and 
have continuous partial derivatives on this open region containing and we can apply Green’s
Theorem. But by Exercise 17.3.33(a) , , so .
28. We express as a type II region: where and are
continuous functions. Then by
the Fundamental Theorem of Calculus. But referring to the figure, Then
 , , and . Hence
29. Using the first part of (5), we have that . But , and
 , and we orient by taking the positive direction to be that which corresponds,
under the mapping, to the positive direction along , so
 10
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.4 Green’s Theorem
�
�R
xdy = �
�S
g(u, v)
�h
�u du+
�h
�v dv =��S
g(u, v)
�h
�u du+g(u, v)
�h
�v dv
= 
��S
�
�u g(u, v)
�h
�v �
�
�v g(u, v)
�h
�u dA uv �
= 
��
S
�g
�u
�h
�v +g(u, v)
�
2
h
�u�v �
�g
�v
�h
�u �g(u, v)
�
2
h
�v�u dA
= 
��
S
�x
�u
�y
�v �
�x
�v
�y
�u dA =
��S
�(x, y)
�(u, v) dudv
�S
A(R)
�(x, y)
�(u, v) A(R)=��R
dxdy=��
S
�(x, y)
�(u, v) dudv.
 [using Green’s Theorem in the 
plane]
 [by the equality of mixed partials] 
The sign is chosen to be positive if the orientation that we gave to corresponds to the usual positive
orientation, and it is negative otherwise. In either case, since is positive, the sign chosen must be
the same as the sign of . Therefore 
 11
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.4 Green’s Theorem
F =��F=
i
�/�x
xyz
j
�/�y
0
k
�/�z
�x
2
y
=(�x
2
�0) i�(�2xy�xy) j+(0�xz)k
=�x
2
i+3xy j�xzk
F=��F=
�
�x (xyz)+
�
�y (0)+
�
�z (�x
2
y)=yz+0+0=yz
curlF =��F=
i
�/�x
x
2
yz
j
�/�y
xy
2
z
k
�/�z
xyz
2
=(xz
2
�xy
2
) i�(yz
2
�x
2
y) j+(y
2
z�x
2
z)k
=x(z
2
�y
2
) i+y(x
2
�z
2
) j+z(y
2
�x
2
)k
F=��F=
�
�x (x
2
yz)+
�
�y (xy
2
z)+
�
�z (xyz
2
)=2xyz+2xyz+2xyz=6xyz
curlF =��F=
i
�/�x
1
j
�/�y
x+yz
k
�/�z
xy� z
=(x�y) i�(y�0) j+(1�0)k
=(x�y) i�y j+k
F=��F=
�
�x (1)+
�
�y (x+yz)+
�
�z (xy� z )=z�
1
2 z
curlF =��F=
i
�/�x
0
j
�/�y
cos xz
k
�/�z
�sin xy
=(�xcos xy+xsin xz) i�(�ycos xy�0) j+(�zsin xz�0)k
=x(sin xz�cos xy) i+ycos xy j�zsin xzk
F=��F=
�
�x (0)+
�
�y (cos xz)+
�
�z (�sin xy)=0+0+0=0
1. (a)
curl
(b) div
2. (a)
(b) div
3. (a)
(b) div
4. (a)
(b) div
 1
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.5 Curl and Divergence
F=��F=
i
�/�x
e
x
sin y
j
�/�y
e
x
cos y
k
�/�z
z
=(0�0)i�(0�0) j+(e
x
cos y�e
x
cos y)k=0
F=��F=
�
�x (e
x
sin y)+
�
�y (e
x
cos y)+
�
�z (z)=e
x
sin y�e
x
sin y+1=1
F =��F=
i
�/�x
x
x
2
+y
2
+z
2
j
�/�y
y
x
2
+y
2
+z
2
k
�/�z
z
x
2
+y
2
+z
2
= 1
(x
2
+y
2
+z
2
)
2
[(�2yz+2yz) i�(�2xz+2xz) j+(�2xy+2xy)k]=0
F =��F=
�
�x
x
x
2
+y
2
+z
2
+
�
�y
y
x
2
+y
2
+z
2
+
�
�z
z
x
2
+y
2
+z
2
= x
2
+y
2
+z
2
�2x
2
(x
2
+y
2
+z
2
)
2
+ x
2
+y
2
+z
2
�2y
2
(x
2
+y
2
+z
2
)
2
+ x
2
+y
2
+z
2
�2z
2
(x
2
+y
2
+z
2
)
2
= x
2
+y
2
+z
2
(x
2
+y
2
+z
2
)
2
= 1
x
2
+y
2
+z
2
curlF =��F=
i
�/�x
ln x
j
�/�y
ln (xy)
k
�/�z
ln (xyz)
=
xz
xyz �0 i�
yz
xyz �0 j+
y
xy �0 k
=
1
y , �
1
x ,
1
x
F=��F=
�
�x (ln x)+
�
�y (ln (xy))+
�
�z (ln (xyz))=
1
x +
x
xy +
xy
xyz =
1
x +
1
y +
1
z
5. (a) curl
(b) div
6. (a)
curl
(b)
div
7. (a)
(b) div
8. (a)
 2
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.5 Curl and Divergence
curlF =��F=
i
�/�x
xe
�y
j
�/�y
xz
k
�/�z
ze
y
=(ze
y
�x) i�(0�0) j+(z�xe
�y
(�1))k
= ze
y
�x, 0, z+xe
�y
F=��F=
�
�x (xe
�y
)+
�
�y (xz)+
�
�z (ze
y
)=e
�y
+0+e
y
=e
y
+e
�y
F=P i+Q j+Rk R=0 x �
F 0 P=0
�P
�x =
�P
�y =
�P
�z =
�R
�x =
�R
�y =
�R
�z =0 Q y
�Q
�y <0 Q x � z �
�Q
�x =
�Q
�z =0
divF=
�P
�x +
�Q
�y +
�R
�z =0+
�Q
�y +0<0
curlF=
�R
�y �
�Q
�z i+
�P
�z �
�R
�x j+
�Q
�x �
�P
�y k=(0�0) i+(0�0) j+(0�0)k=0
F=P i+Q j+Rk R=0 P Q z
�
�R
�x =
�R
�y =
�R
�z =
�P
�z =
�Q
�z =0 x x � F
y �
�P
�x >0
�Q
�x =0y
y � x �
�Q
�y >0
�P
�y =0
divF=
�P
�x +
�Q
�y +
�R
�z =
�P
�x +
�Q
�y +0>0
curlF=
�R
�y �
�Q
�z i+
�P
�z �
�R
�x j+
�Q
�x �
�P
�y k=(0�0) i+(0�0) j+(0�0)k=0
F=P i+Q j+Rk R=0 y �
F 0 Q=0
�Q
�x =
�Q
�y =
�Q
�z =
�R
�x =
�R
�y =
�R
�z =0 P y
�P
�y >0 P x � z �
�P
�x =
�P
�z =0
(b) div
9. If the vector field is , then we know . In addition, the component of each
vector of is , so , hence . decreases as increases, so
 , but doesn’t change in the or directions, so .
(a)
(b)
10. If the vector field is , then we know . In addition, and don’t vary in the 
direction, so . As increases, the component of each vector of 
increases while the component remains constant, so and . Similarly, as increases,
the component of each vector increases while the component remains constant, so and
 .
(a)
(b)
11. If the vector field is , then we know . In addition, the component of each
vector of is , so , hence . increases as increases, so
 , but doesn’t change in the or directions, so .
 3
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.5 Curl and Divergence
divF=
�P
�x +
�Q
�y +
�R
�z =0+0+0=0
curlF =
�R
�y �
�Q
�z i+
�P
�z �
�R
�x j+
�Q
�x �
�P
�y k
=(0�0)i+(0�0) j+ 0�
�P
�y k=�
�P
�y k
�P
�y >0 �
�P
�y k z �
curl f =�� f f
grad f
divF
curl(grad f )
gradF F
grad(divF)
div(grad f )
grad(div f ) f
curl(curlF)
div(divF) divF
(grad f )� (divF) divF
div(curl(grad f ))
F=��F=
i
�/�x
yz
j
�/�y
xz
k
�/�z
xy
=(x�x) i�(y�y) j+(z�z)k=0
F R
3
F f F=� f f x(x, y, z)=yz
f (x, y, z)=xyz+g(y, z) f y(x, y, z)=xz+gy(y, z) f y(x, y, z)=xz g(y, z)=h(z)
f (x, y, z)=xyz+h(z) f z(x, y, z)=xy+h
/
(z) f z(x, y, z)=xy h(z)=K
F f (x, y, z)=xyz+K
F=��F=
i
�/�x
3z
2
j
�/�y
cos y
k
�/�z
2xz
=(0�0) i�(2z�6z) j+(0�0)k=4z j�0
F
(a)
(b)
Since , is a vector pointing in the negative direction.
12. (a) is meaningless because is a scalar field.
(b) is a vector field.
(c) is a scalar field.
(d) is a vector field.
(e) is meaningless because is not a scalar field.
(f) is a vector field.
(g) is a scalar field.
(h) is meaningless because is a scalar field.
(i) is a vector field.
(j) is meaningless because is a scalar field.
(k) is meaningless because is a scalar field.
(l) is a scalar field.
13. curl 
and is defined on all of with component functions which have continuous partial derivatives, so
by Theorem 4, is conservative. Thus, there exists a function such that . Then 
implies and . But , so and
 . Thus but so , a constant. Hence a
potential function for is .
14. curl ,
so is not conservative.
 4
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.5 Curl and Divergence
F=��F=
i
�/�x
2xy
j
�/�y
x
2
+2yz
k
�/�z
y
2
=(2y�2y) i�(0�0) j+(2x�2x)k=0 F R
3
F
f � f =F f x(x, y, z)=2xy f (x, y, z)=x
2
y+g(y, z)
f y(x, y, z)=x
2
+gy(y, z) f y(x, y, z)=x
2
+2yz g(y, z)=y
2
z+h(z) f (x, y, z)=x
2
y+y
2
z+h(z)
f z(x, y, z)=y
2
+h
/
(z) f z(x, y, z)=y
2
h(z)=K f (x, y, z)=x
2
y+y
2
z+K
F=��F=
i
�/�x
e
z
j
�/�y
1
k
�/�z
xe
z
=(0�0) i�(e
z
�e
z
) j+(0�0)k=0 F R
3
F
f � f =F f x(x, y, z)=e
z
f (x, y, z)=xe
z
+g(y, z)�
f y(x, y, z)=gy(y, z). f y(x, y, z)=1 g(y, z)=y+h(z) f (x, y, z)=xe
z
+y+h(z)
f z(x, y, z)=xe
z
+h
/
(z) f z(x, y, z)=xe
z
h(z)=K F
f (x, y, z)=xe
z
+y+K
F=��F=
i
�/�x
ye
�x
j
�/�y
e
�x
k
�/�z
2z
=(0�0) i�(0�0) j+(�e
�x
�e
�x
)k=�2e
�x
k�0
F
curlF =��F=
i
�/�x
ycos xy
j
�/�y
xcos xy
k
�/�z
�sin z
=(0�0) i�(0�0) j+[(�xysin xy+cos xy)�(�xysin xy+cos xy)]k=0
F R
3
F
f � f =F f x(x, y, z)=ycos xy
f (x, y, z)=sin xy+g(y, z)� f y(x, y, z)=xcos xy+gy(y, z) f y(x, y, z)=xcos xy g(y, z)=h(z)
f (x, y, z)=sin xy+h(z) f z(x, y, z)=h
/
(z) f z(x, y, z)=�sin z h(z)=cos z+K
15. curl , is defined on all of 
, and the partial derivatives of the component functions are continuous, so is conservative. Thus
there exists a function such that . Then implies and
 . But , so and . Thus
 but so and .
16. curl and is defined on all of 
with component functions that have continuous partial deriatives, so is conservative. Thus there
exists a function such that . Then implies 
 But , so and . Thus
 but , so , a constant. Hence a potential function for is
 .
17. curl ,
so is not conservative.
18.
 is defined on all of , and the partial derivatives of the component functions are continuous, so 
is conservative. Thus there exists a function such that . Then implies
 . But , so and
 . Thus but so and a potential
 5
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.5 Curl and Divergence
F f (x, y, z)=sin xy+cos z+K
G div(curlG)=y
2
+z
2
+x
2
�0
G div(curlG)=xz�0
curlF=
i
�/�x
f (x)
j
�/�y
g(y)
k
�/�z
h(z)
=(0�0)i+(0�0) j+(0�0)k=0
F= f (x) i+g(y) j+h(z)k
divF=
�( f (y, z))
�x +
�(g(x, z))
�y +
�(h(x, y))
�z =0 F
div(F+G) =
�(P
1
+P
2
)
�x +
�(Q
1
+Q
2
)
�y +
�(R
1
+R
2
)
�z
=
�P
1
�x +
�Q
1
�y +
�R
1
�z +
�P
2
�x +
�Q
2
�y +
�R
3
�z =divF+divG
curlF+curlG=
�R
1
�y �
�Q
1
�z i+
�P
1
�z �
�R
1
�x j+
�Q
1
�x �
�P
1
�y k
+
�R
2
�y �
�Q
2
�z i+
�P
2
�z �
�R
2
�x j+
�Q
2
�x �
�P
2
�y k
=
�(R
1
+R
2
)
�y �
�(Q
1
+Q
2
)
�z i+
�(P
1
+P
2
)
�z �
�(R
1
+R
2
)
�x j
+
�(Q
1
+Q
2
)
�x �
�(P
1
+P
2
)
�y k=curl F+G( )
div fF( ) =
�( fP
1
)
�x +
�( fQ
1
)
�y +
�( fR
1
)
�z
= f
�P
1
�x +P1
� f
�x + f
�Q
1
�y +Q1
� f
�y + f
�R
1
�z +R1
� f
�z
function for is .
19. No. Assume there is such a . Then , which contradicts Theorem 11.
20. No. Assume there is such a . Then which contradicts Theorem 11.
21. .
Hence is irrotational.
22. so is incompressible.
23.
24.
25.
 6
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.5 Curl and Divergence
= f
�P
1
�x +
�Q
1
�y +
�R
1
�z + P1, Q1, R1 �
� f
�x ,
� f
�y ,
� f
�z = f divF+F� � f
curl fF( ) =
�( fR
1
)
�y �
�( fQ
1
)
�z i+
�( fP
1
)
�z �
�( fR
1
)
�x j+
�( fQ
1
)
�x �
�( fP
1
)
�y k
= f
�R
1
�y +R1
� f
�y � f
�Q
1
�z �Q1
� f
�z i+ f
�P
1
�z +P1
� f
�z � f
�R
1
�x �R1
� f
�x j
+ f
�Q
1
�x +Q1
� f
�x � f
�P
1
�y �P1
� f
�y k
= f
�R
1
�y �
�Q
1
�z i+ f
�P
1
�z �
�R
1
�x j+ f
�Q
1
�x �
�P
1
�y k
+ R
1
� f
�y �Q1
� f
�z i+ P1
� f
�z �R1
� f
�x j+ Q1
� f
�x �P1
� f
�y k
= f curlF+ � f( )�F
div(F)�G =�� (F�G)=
�/�x
P
1
P
2
�/�y
Q
1
Q
2
�/�z
R
1
R
2
=
�
�x
Q
1
Q
2
R
1
R
2
�
�
�y
P
1
P
2
R
1
R
2
+
�
�z
P
1
P
2
Q
1
Q
2
= Q
1
�R
2
�x +R2
�Q
1
�x �Q2
�R
1
�x �R1
�Q
2
�x
� P
1
�R
2
�y +R2
�P
1
�y �P2
�R
1
�y �R1
�P
2
�y
+ P
1
�Q
2
�z +Q2
�P
1
�z �P2
�Q
1
�z �Q1
�P
2
�z
= P
2
�R
1
�y �
�Q
1
�z +Q2
�P
1
�z �
�R
1
�x +R2
�Q
1
�x �
�P
1
�y
� P
1
�R
2
�y �
�Q
2
�z +Q1
�P
2
�z �
�R
2
�x +R1
�Q
2
�x �
�P
2
�y
=G� curlF�F� curlG
26.27.
 7
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.5 Curl and Divergence
div(� f ��g)=�g� curl(� f )�� f � curl(�g)=0
curl curlF =�� (��F)=
i
�/�x
�R
1
/�y��Q
1
/�z
j
�/�y
�P
1
/�z��R
1
/�x
k
�/�z
�Q
1
/�x��P
1
/�y
=
�
2
Q
1
�y�x �
�
2
P
1
�y
2
�
�
2
P
1
�z
2
+
�
2
R
1
�z�x i+
�
2
R
1
�z�y �
�
2
Q
1
�z
2
�
�
2
Q
1
�x
2
+
�
2
P
1
�x�y j
+
�
2
P
1
�x�z �
�
2
R
1
�x
2
�
�
2
R
1
�y
2
+
�
2
Q
1
�y�z k
graddivF��
2
F
�
2
F
grad divF��
2
F =
�
2
P
1
�x
2
+
�
2
Q
1
�x�y +
�
2
R
1
�x�z i+
�
2
P
1
�y�x +
�
2
Q
1
�y
2
+
�
2
R
1
�y�z j
+
�
2
P
1
�z�x +
�
2
Q
1
�z�y +
�
2
R
1
�z
2
k
�
�
2
P
1
�x
2
+
�
2
P
1
�y
2
+
�
2
P
1
�z
2
i+
�
2
Q
1
�x
2
+
�
2
Q
1
�y
2
+
�
2
Q
1
�z
2
j
+
�
2
R
1
�x
2
+
�
2
R
1
�y
2
+
�
2
R
1
�z
2
k
=
�
2
Q
1
�x�y +
�
2
R
1
�x�z �
�
2
P
1
�y
2
�
�
2
P
1
�z
2
i+
�
2
P
1
�y�x +
�
2
R
1
�y�z �
�
2
Q
1
�x
2
�
�
2
Q
1
�z
2
j
+
�
2
P
1
�z�x +
�
2
Q
1
�z�y �
�
2
R
1
�x
2
�
�
2
R
2
�y
2
k
28. (by Theorem 3)
29.
Now let’s consider and compare with the above.
(Note that is defined on page 1130).
Then applying Clairaut’s Theorem to reverse the order of differentiation in the second partial
derivatives as needed and comparing, we have
 8
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.5 Curl and Divergence
curl curlF=grad divF��
2
F
�� r=
�
�x i+
�
�y j+
�
�z k � (x i+y j+zk)=1+1+1=3
�� (rr) =�� x
2
+y
2
+z
2
(x i+y j+zk)
= x
2
x
2
+y
2
+z
2
+ x
2
+y
2
+z
2
+ y
2
x
2
+y
2
+z
2
+ x
2
+y
2
+z
2
+ z
2
x
2
+y
2
+z
2
+ x
2
+y
2
+z
2
= 1
x
2
+y
2
+z
2
(4x
2
+4y
2
+4z
2
)=4 x
2
+y
2
+z
2
=4r
�� (rr)=div(rr)=rdivr+r� �r=3r+r�
r
r =4r
�
2
r
3
=�
2
(x
2
+y
2
+z
2
)
3/2
=
�
�x
3
2 (x
2
+y
2
+z
2
)
1/2
(2x) +
�
�y
3
2 (x
2
+y
2
+z
2
)
1/2
(2y)
+
�
�z
3
2 (x
2
+y
2
+z
2
)
1/2
(2z)
=3
1
2 (x
2
+y
2
+z
2
)
�1/2
(2x)(x)+(x
2
+y
2
+z
2
)
1/2
+3
1
2 (x
2
+y
2
+z
2
)
�1/2
(2y)(y)+(x
2
+y
2
+z
2
)
1/2
+3
1
2 (x
2
+y
2
+z
2
)
�1/2
(2z)(z)+(x
2
+y
2
+z
2
)
1/2
=3(x
2
+y
2
+z
2
)
�1/2
(4x
2
+4y
2
+4z
2
)=12(x
2
+y
2
+z
2
)
1/2
=12r
�
�x (x
2
+y
2
+z
2
)
3/2
=3x x
2
+y
2
+z
2
��r
3
=3r (x i+y j+zk)=3r r
�
2
r
3
=���r
3
=�� (3r r)=3(4r)=12r
 as desired.
30. (a)
(b)
Another method:
By Exercise 25, .
(c)
Another method: , so
 by part (b).
 9
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.5 Curl and Divergence
�r =� x
2
+y
2
+z
2
= x
x
2
+y
2
+z
2
i+ y
x
2
+y
2
+z
2
j+ z
x
2
+y
2
+z
2
k
= x i+y j+zk
x
2
+y
2
+z
2
=
r
r
�� r =
i
�
�x
x
j
�
�y
y
k
�
�z
z
=
�
�y (z)�
�
�z (y) i+
�
�z (x)�
�
�x (z) j+
�
�x (y)�
�
�y (x) k=0
�
1
r
=� 1
x
2
+y
2
+z
2
=
� 1
2 x
2
+y
2
+z
2
(2x)
x
2
+y
2
+z
2
i�
1
2 x
2
+y
2
+z
2
(2y)
x
2
+y
2
+z
2
j�
1
2 x
2
+y
2
+z
2
(2z)
x
2
+y
2
+z
2
k
=� x i+y j+zk
(x
2
+y
2
+z
2
)
3/2
=� r
r
3
� ln r =� ln x
2
+y
2
+z
2( ) 1/2= 12 � ln x
2
+y
2
+z
2( )
= x
x
2
+y
2
+z
2
i+ y
x
2
+y
2
+z
2
j+ z
x
2
+y
2
+z
2
k= x i+y j+zk
x
2
+y
2
+z
2
= r
r
2
r=x i+y j+zk�r=|r|= x
2
+y
2
+z
2
F= r
r
p
= x
(x
2
+y
2
+z
2
)
p/2
i+ y
(x
2
+y
2
+z
2
)
p/2
j+ z
(x
2
+y
2
+z
2
)
p/2
k
31. (a)
(b)
(c)
(d)
32. , so
 10
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.5 Curl and Divergence
�
�x
x
(x
2
+y
2
+z
2
)
p/2
= (x
2
+y
2
+z
2
)�px
2
(x
2
+y
2
+z
2
)
1+ p/2
= r
2
�px
2
r
p+2
�
�y
y
(x
2
+y
2
+z
2
)
p/2
= r
2
�py
2
r
p+2
�
�z
z
(x
2
+y
2
+z
2
)
p/2
= r
2
�pz
2
r
p+2
divF = ��F= r
2
�px
2
r
p+2
+ r
2
�py
2
r
p+2
+ r
2
�pz
2
r
p+2
= 3r
2
�px
2
�py
2
�pz
2
r
p+2
=
3r
2
�p(x
2
+y
2
+z
2
)
r
p+2
= 3r
2
�pr
2
r
p+2
= 3�p
r
p
p=3 divF=0
	
C
f (�g)� nds=		
D
div ( f�g)dA=		
D
[ f div(�g)+�g� � f ]dA
div(�g)=�
2
g. 		
D
f�
2
gdA= 	
C
f (�g)� nds�		
D
�g� � f dA.
		
D
f�
2
gdA= 	
C
f (�g)� nds�		
D
�g� � f dA
		
D
g�
2
f dA= 	
C
g(� f )� nds�		
D
� f � �gdA.
		
D
( f�
2
g�g�
2
f )dA = 	
C
f (�g)� n�g(� f )� n ds+		
D
(� f � �g��g� � f )dA
= 	
C
f�g�g� f � nds
� =v/d sin� =d/r� v=d� =(sin� )r� =|w� r |. v
w r v=w� r.
v=w� r=
i
0
x
j
0
y
k
�
z
=(0� z�� y) i+(� x�0� z) j+(0� y�x� 0)k=�� y i+� x j
curlv=�� v=
i
�/�x
��y
j
�/�y
�x
k
�/�y
0
Then . Similarly,
 and . Thus
Consequently, if we have .
33. By (13), by Exercise 25. But
 Hence 
34. By Exercise 33, and
 Hence
35. (a) We know that , and from the diagram But is
perpendicular to both and , so that 
(b) From (a), 
(c)
 11
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.5 Curl and Divergence
=
�
�y (0)�
�
�z (� x) i+
�
�z (�� y)�
�
�x (0) j+
�
�x (� x)�
�
�y (�� y) k
=[� �(�� )]k=2� k=2w
H= h
1
, h
2
, h
3
E= E
1
, E
2
, E
3
.
�� (��E) =�� (curlE)=�� �
1
c
�H
�t =�
1
c
i
�/�x
�h
1
/�t
j
�/�y
�h
2
/�t
k
�/�z
�h
3
/�t
=�
1
c
�
2
h
3
�y�t �
�
2
h
2
�z�t i+
�
2
h
1
�z�t �
�
2
h
3
�x�t j+
�
2
h
2
�x�t �
�
2
h
1
�y�t k
=�
1
c
�
�t
�h
3
�y �
�h
2
�z i+
�h
1
�z �
�h
3
�x j+
�h
2
�x �
�h
1
�y k
=�
1
c
�
�t curlH=�
1
c
�
�t
1
c
�E
�t =�
1
c
2
�
2
E
�t
2
�� (��H) =�� (curlH)=��
1
c
�E
�t =
1
c
i
�/�x
�E
1
/�t
j
�/�y
�E
2
/�t
k
�/�z
�E
3
/�t
=
1
c
�
2
E
3
�y�t �
�
2
E
2
�z�t i+
�
2
E
1
�z�t �
�
2
E
3
�x�t j+
�
2
E
2
�x�t �
�
2
E
1
�y�t k
=
1
c
�
�t
�E
3
�y �
�E
2
�z i+
�E
1
�z �
�E
3
�x j+
�E
2
�x �
�E
1
�y k
=
1
c
�
�t curlE=
1
c
�
�t �
1
c
�H
�t =�
1
c
2
�
2
H
�t
2
curl curlE=grad divE��
2
E�
36. Let and 
(a)
[assuming that the partial derivatives are continuous so that the order of differentiation
does not matter]
(b)
[assuming that the partial derivatives are continuous so that the order of differentiation
does not matter]
(c) Using Exercise 29, we have that 
 12
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.5 Curl and Divergence
�
2
E=grad divE�curl curlE=grad0+ 1
c
2
�
2
E
�t
2
= 1
c
2
�
2
E
�t
2
.
�
2
H=grad divH�curl curlH=grad 0+ 1
c
2
�
2
H
�t