Thomas Calculus 13th [Solutions]

1182 Chapter 16 Integrals and Vector Fields
27.
3 3 2 2
M x cos t, N y sin t dx 3 cos t sin t dt, dy 3 sin t cos t dt Area 1 x dy y dx
2 C
1
2 2 2 2 2 2 2 2 3
2 2 3
4
3 sin t cos t cos t sin t dt 1 3 sin t cos t dt sin 2t dt
2 0 2 0 8 0 16 0
2
sin u du
3 u sin 2u
4
3
16 2 4
0
8
28. C1: M x t, N y 0 dx dt, dy 0; C2: M x (2 t) sin(2 t) 2 t sin t,
N y 1 cos(2 t) 1 cos t dx (cos t 1) dt, dy sin t dt
Area
1 x dy y dx 1 x dy y dx 1 x dy y dx
2 C 2 C 2 C
1 2
1
2
1
2
1
2
(0) dt (2 t sin t)(sin t) (1 cos t) (cos t 1) dt (2 cos t t sin t 2 2 sin t)
dt
2 0 2 0 2 0
1
2
3 sin
2 t t cos t 2 t 2 cos t 3
0
29. (a) M f ( x), N g( y) M 0, N 0 f ( x) dx g( y) dy N M dx dy 0 dx dy 0
y x C
x y
R
R
(b) M ky, N hx M k,
N h ky dx hx dy N M dx dy
y x C
x y
R
( h k) dx dy ( h k)(Area of the region)
R
30.
2 2 2 2
M xy , N x y 2x M 2 xy, N 2xy 2 xy dx x y 2x dy N M dx dy
y x C
x y
R
(2xy 2 2 xy) dx dy 2 dx dy 2 times the area of the square
R
R
31. The integral is 0 for any simple closed plane curve C. The reasoning: By the tangential form of Greens
3
4 3 4 4 3
Theorem, with M 4x y and N x , 4x y dx x dy x 4x y dx dy
C
x y
R
3 3
4x 4x dx dy 0.
R
0
32. The integral is 0 for any simple closed curve C. The reasoning: By the normal form of Greens theorem, with
M
3
x and
3 3 3 3 3
N y , y dy x dx y x dx dy 0.
C
x y
R
0 0
33. Let M x and N 0 M 1 and N 0 M dy N dx M N dx dy x dy
x
y C
x y
C
R
(1 0) dx dy Area of R dx dy x dy ; similarly, M y and
C
N 0 M 1
y
and N
x
0
R
R
M dx N dy N M dy dx y dx (0 1) dy dx y dx dx dy Area of R
C x y
C C
R R R
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