Calculus Stewart ET 5e 0534393217;16. Vector Calculus; 16.4 Green's Theorem
1. (a)
C : x=t
1 2
dx=dt , y=0 dx=0dt , y=t
dy=0dt , 0 t dy=dt , 0 t
2. 3. 2. 3.
C : x=2 C : x=2 t
3
dx= dt , y=3 dx=0dt , y=3 t
dy=0dt , 0 t dy= dt , 0 t
C : x=0
4
Thus xy dx+x dy
C C +C +C +C
1 2 3 4
2
3
xy dx+x dy
2 0 3 0 2 0 3
2
3
= 0dt+ 8dt+ =0+24 18+0=6 (b) xy dx+x dy =
C D 2 0 2 2 3
9(2 t)dt+ 0dt
0
x
(x )
3
y
(xy ) dA=
0 0
2
2 3
(3x 2xy)dydx
2
= (9x 9x)dx=24 18=6
2 2
2. (a) x=cos t , y=sin t , 0 t (b)
C
2 . Then
C
ydx xdy= [sin t( sin t) cos t(cos t)]dt=
0 0 2
dt= 2 .
ydx xdy=
D
x
( x)
y
(y) dA= 2
D
dA= 2A(D)= 2 (1) = 2
3. (a)
1
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.4 Green's Theorem
C : x=t
1 2
dx=dt , y=0 dx=0dt , y=t
dy=0dt , 0 t dy=dt , 0 t
1 2. 1.
C : x=1 C : x=1 t
3
dx= dt , y=2 2t
dy= 2dt , 0 t
Thus xydx+x y dy =
C C +C +C
1 2 3
2 3
xydx+x y dy
1 0 2 0 3 1 0 2 0 2 3
2 3
= 0dt+ t dt+ =0+ (b) xydx+x y dy =
C D 1 2 3
(1 t)(2 2t) 2(1 t) (2 2t) dt + 2 3 8 6 (1 t) + (1 t) 3 3
1 0
1 4 t 4
=4
10 2 = 3 3
x
(x y )
2 3
1 2x
y
y=2x y=0
(xy) dA=
0 0 1 0 5 2
(2xy x)dydx 4 3 2 2 = 3 3
3
C :x=0
1
dx=0dt=
0
, y=1 t dy= dt , 4. (a) 1 0 t C :x=t dx=dt
2
1 4 xy xy 2
dx= (8x 2x )dx=
, y=0 dy=0dt , 0 t 1 C :x=1 t
3
dx= dt
2 2
, y=1 (1 t) =2t t dy=(2 2t)dt ,0 t 1 Thus
2
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.4 Green's Theorem
xdx+ydy =
C C +C +C
1 2 3
xdx+ydy
1 0 1 0 1 1 1 0 1 0 2
= (0dt+(1 t)( dt))+ (t dt+0dt)+ ((1 t)( dt)+(2t t )(2 2t)dt) 1 2 t t 2 1 1 = + + 2 2 = 1 2 1 4 3 5 2 t + t 2t + t t 0 2 0 2 2 1 5 2+ 1 =0 2 2 +
(b)
C
xdx+ydy=
D
x
(y)
y
(x) dA=
D
0dA=0 ,0
7
5. We can parametrize C as x=cos
2
, y=sin
2
2 . Then the line integral is sin
6
Pdx+Q dy= cos
C 0
4
sin
5
( sin )d + ( cos
0
2
)cos d =
29 , according to a CAS. 1024
The double integral is Q x P y
1 1 x
dA=
1 1 x
2
( 7x y 5x y )dydx=
6 6
4 4
D
29 , 1024
verifying Green's Theorem in this case. 6. Since y=x along the first part of C and y=x along the second part, the line integral is
1 2 0 1
Pdx+Q dy =
C 0
x sin x+x sin (x )(2x) dx+ (x sin x+x sin x)dx
4
2
2
2
2
= 16cos 1 23sin 1+28 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
1 x
dA=
0 x
(2xsin y 2ysin x)dydx= 16cos 1 23sin 1+28
2
7. The region D enclosed by C is [0, 1] [0, 1] , so e dx+2xe dy =
C D 1 0 1 0 y y
x
y
(2xe )
1
y
y
0
(e ) dA=
0 0
y
1 1
(2e e )dydx
y
y
= dx e dy=(1)(e e )=e 1 8. The region D enclosed by C is given by { (x, y) | 0 x 1, 3x y 3 } , so x y dx+4xy dy =
C D 1 2 2 3
x
4
(4xy )
3
y
1 0
(x y ) dA=
2 4
2 2
1 3 0 3x
(4y 2x y)dydx 72 318 = 5 5
3
2
=
0
y x y
2 2 y=3
y=3x
dx= (81 9x 72x )dx=81 3
9. (y+e
C x
)dx+(2x+cos y )dy =
D 1 y
2
2
x
(2x+cos y )
1 0 1/2
2
y
2
(y+e 1 3
x
) dA
=
0 y
(2 1)dxdy= (y
y )dy=
10. xe
C 2x
dx+(x +2x y )dy =
D
4
2 2
x
2
(x +2x y )
2 2 2
4
2 2
y
(xe
2x
) dA=
D 2
(4x +4xy 0)dA
3
2
=4
D 2
x(x +y )dA=4
0 1 2 1 4
(rcos )(r )rdr d
2 0
=4 cos d
0
r dr=4 sin
1 5 r 5
2 1
=0
11.
4
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.4 Green's Theorem
y dx x dy =
C D 2
3
3
x
2 0
( x)
3
3
y
(y ) dA=
D
3
( 3x 3y )dA=
0 0
2
2
2
2
( 3r )rdr d
2
= 3 d
0
r dr= 3(2 )(4)= 24
12. sin ydx+xcos ydy=
C 3 2 D
x
(xcos y)
y
(sin y) dA=
D
(cos y cos y)dA=
D
0dA=0
13. F(x, y)= x +y , x + y and the region D enclosed by C is given by { (x, y) | 0 x , 0 y sin x }.C is traversed clockwise, so C gives the positive orientation. F dr =
C
(
0 0
x +y dx+ x + y dy=
D 2
3
) (
2
)
C sin x
( x2+ y ) x
y=0
( y
x +y
3
)
dA
= =
0
(2x 3y )dydx=
0 3
2xy y
3 y=sin x
dx
2
(2xsin x sin x)dx=
0
(2xsin x (1 cos x)sin x)dx 1 3 cos x 3
= =
2sin x 2xcos x+cos x 2
2
0
2+
2 3
=
2
4 2 3
14. F(x, y)= y cos x, x +2ysin x and the region D enclosed by C is given by { (x, y) | 0 x 2, 0 y 3x } . C is traversed clockwise, so C gives the positive orientation. F dr =
C C
(y cos x)dx+(x +2ysin x)dy (x +2ysin x)
2
2
2
=
D
x
y
(y cos x) dA
2 3x
2
=
D
(2x+2ycos x 2ycos x)dA=
0 0
2xdydx
5
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.4 Green's Theorem
2
=
0
2x y
y=3x y=0
2 0 y 2
dx=
6x dx= 2x
2
3 2
0
= 16
15. F(x, y)= e +x y, e xy and the region D enclosed by C is the disk x +y C is traversed clockwise, so C gives the positive orientation. F dr =
C C
x
2
2
2
25 .
(e +x y)dx+(e xy )dy
y 2 x 2 2 2
x
2
y
2
=
D 2
x
2
(e xy )
2 5 2
y
(e +x y) dA=
D 2 0 5 0 3
( y x )dA 1 4 r 4
5 0
=
D
(x +y )dA=
0 0
(r )rdr d = d
r dr=2
=
625 2
y and the region D enclosed by C is the disk with radius x 1 centered at (2, 3).C is oriented positively, so 16. F ( x, y ) = y ln (x +y ), 2tan
2 2 1 C
F d r = (y ln (x +y ))dx+ 2tan
C
2
2
1
y x y
dy (y ln (x +y )) dA dA= 2y
D 2 2
=
D
x 2
2tan yx
1
y x 1
2 2
=
2y x +y
2 2
D
1+(y/x)
x +y
2
2
1+
2y x +y
2 2
dA
=
D
dA= ( area of D)=
17. By Green's Theorem, W = F dr= x(x+y)dx+xy dy=
C 11 x 1 0 C y=1 x y=0 1 0 D
2
(y x)dydx where C is the path described
2
in the question and D is the triangle bounded by C . So W=
0 0
(y x)dydx=
2
1 3 y xy 3
dx=
1 3 (1 x) x(1 x) dx 3
6
Stewart Calculus ET 5e 0534393217;16. Calculus; Vector 16.4 Green's Theorem
=
1 4 1 2 1 3 (1 x) x+ x 12 2 3
1 0
=
1 1 + 2 3
3 2
1 12
=
1 12
2 2
18. By Green's Theorem, W = F dr= xdx+(x +3xy )dy=
C 2 C 2 0 D
(3x +3y 0)dA , where D is the
semicircular region bounded by C . Converting to polar coordinates, we have W =3
0 0
r rd dr=3
2
1 4 r 4
=12 . 2 , and let
19. Let C be the arch of the cycloid from (0, 0) to (2 , 0) , which corresponds to 0 t
1
C be the segment from (2 , 0) to (0, 0) , so C is given by x=2
2 2
t , y=0 , 0 t
2 . Then
C=C
1
C is traversed
2
clockwise, so C is oriented positively. Thus C encloses the area under one arch of the cycloid and from (5) we have
2 2 0 2
A=
C
ydx = ydx+ ydx= (1 cos t)(1 cos t)dt+ 0( dt)
C 2 0
1
C
2
0
= (1 2cos t+cos t)dt+0= t 2sin t+
1 1 t+ sin 2t 2 4
2 0
=3
20.
2
A=
C
xdy= (5cos t cos 5t)(5cos t 5cos 5t)dt
0 2 0
= (25cos t 30cos tcos 5t+5cos 5t)dt = 25 =30 21. (a) Using Equation 17.2.8 , we write parametric equations of the line segment as x=(1 t)x +tx ,
1 2
2
2
1 1 t+ sin 2t 2 4
30
1 1 sin 4t+ sin 6t +5 8 12
1 1 t+ sin 10t 2 20
2 0
y=(1 t)y +ty , 0 t
1 2 1
1 . Then dx=(x x )dt and dy=(y y )dt , so
2 1 2 1
xdy ydx =
C 0
[(1 t)x +tx ](y y )dt+[(1 t)y +ty ](x x )dt
1 2 2 1 1 2 2 1
7
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.4 Green's Theorem
1
=
0 1
(x (y y ) y (x x )+t [(y y )(x x ) (x x )(y y )])dt
1 2 1 1 2 1 2 1 2 1 2 1 2 1
=
0
(x y x y )dt=x y x y
1 2 2 1 1 2
2 1
(b) We apply Green's Theorem to the path C=C
i i i+1 i+1
1
C
2 n
C , where C is the line segment that
n i n n 1 1
joins (x , y ) to (x , y ) for i=1 , 2 , ... , n 1 , and C is the line segment that joins (x , y ) to (x , y ) . From (5), 1 xdy ydx= 2C dA , where D is the polygon bounded by C . Therefore
D
area of polygon =A(D)=
D
dA=
1 xdy ydx 2C xdy ydx+ xdy ydx
n 1
=
1 2
xdy ydx+ xdy ydx+...+
C
1
C
2
C
C
n
To evaluate these integrals we use the formula from (a) to get A(D)=[(x y x y )+(x y x y )+...+(x y x y )+(x y x y )].
1 2 2 1 2 3 3 2 n 1 n n n 1 n 1 1 n
(c) A= 1 [(0 1 2 0)+(2 3 1 1)+(1 2 0 3)+(0 1 ( 1) 2)+( 1 0 0 1)] 2 1 9 = (0+5+2+2)= 2 2 1 2A
2
22. By Green's Theorem, 1 2A y dx=
C 2
x dy=
C
1 2A
2xdA=
D
1 A
xdA=x and
D
1 2A
( 2y)dA=
D
1 A
ydA=y .
D
1 1 (1)(1)= and C=C +C +C , where C : x=x , y=0 , 0 x 1 ; 1 2 3 1 2 2 C : x=x , y=1 x , x=1 to x=0 ; and C : x=0 , y=1 to y=0 . Then 23. Here A=
2 3
1 1 2 2 2 2 2 x= x dy= x dy+ x dy+ x dy=0+ (x )( dx)+0= . Similarly, 2A C 3 1
C
1
0
C
2
C
3
8
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.4 Green's Theorem
1 1 2 2 2 2 2 y= y dx= y dx+ y dx+ y dx=0+ (1 x) ( dx)+0= . 2A C 3 1
C
1
0
C
2
C
3
Therefore (x, y)=
2
1 1 , 3 3
.
a 2 2 1 1 24. A= so x= x dy and y= y dx. 2 2 2 a C a C Orienting the semicircular region as in the figure, x = 1 a = 1 a and
2 2 C +C
1 2
x dy
2
0+ (a cos t)(acos t)dt
0
2
2
=0
y=
1 a
2
a a 0
0dx+ (a sin t)( asin t)dt 4a 3 . 1 3
2
2
=
a
0
sin tdt=
3
a
cos t+
1 3 (cos t) 3
0
=
4a . 3
Thus (x, y)= 0,
25. By Green's Theorem, 1 3 x dy=
C 3
y dx=
C 2
3
1 3
y
( 3y )dA=
D D
2
y dA=I and
x
2
1 3
(3x )dA=
D D
2
x dA=I .
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
I = 1 y 3 = 1 4 a 3
C
1 x dy= 3
3
2
1 4 (a cos t)dt= a 3 0
4 4
2 0
3 1 1 + cos 2t+ cos 4t dt 8 2 8
3(2 ) 1 4 = a 8 4
27. Since C 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 D . Thus P= y/(x +y ) and Q=x/(x +y ) have continuous partial derivatives on this open region containing D and we can apply Green's Theorem. But by Exercise 17.3.33(a) , P/ y= Q/ x , so
C 2 2 2 2
F dr=
D
0dA=0 . y d } where f and f are
1 2
28. We express D as a type II region: D={ (x, y) | f (y) x
1
f (y), c
2
continuous functions. Then
D
Q dA= x c
d f 2(y) f (y)
1
Q dxdy= Q( f (y), y) Q( f (y), y) dy by 2 1 x c Qdy=
C C +C +C +C
1 2 3
d
the Fundamental Theorem of Calculus. But referring to the figure,
c d 1
Qdy. Then
4
Qdy= Q( f (y), y)dy ,
C
1
Qdy= Qdy=0 , and
C
2
Qdy= Q( f (y), y)dy . Hence
C
3
d d
C
4
c
2
Qdy=
C c
Q( f (y), y) Q( f (y), y) dy=
2 1 D
( Q/ x)dA.
29. Using the first part of (5), we have that
R
dxdy=A(R)= xdy . But x=g(u, v) , and
R
h h dy= du+ dv , and we orient S by taking the positive direction to be that which corresponds, u v under the mapping, to the positive direction along R , so
10
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.4 Green's Theorem
xdy =
R S
g(u, v)
h h du+ dv u v g(u, v) h v
2
= g(u, v)
S
h h du+g(u, v) dv u v h u dA [using Green's Theorem in the uv
=
S
u
v
g(u, v)
plane] =
S
g u x u
h h +g(u, v) v u v y v x v y u
g v
h h g(u, v) u v u
2
dA (x, y) dudv (u, v)
=
S
dA [by the equality of mixed partials] =
S
The sign is chosen to be positive if the orientation that we gave to S corresponds to the usual positive orientation, and it is negative otherwise. In either case, since A(R) is positive, the sign chosen must be the same as the sign of (x, y) . Therefore A(R)= (u, v) dxdy=
R S
(x, y) (u, v)
dudv.
11