Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 17.4 GREEN’S THEOREM ET SECTION 16.4 ¤ 281
(c) 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
23. We orient the quarter-circular region as shown in the figure.
A = 1 4 πa2 so x = 1
x 2 dy and y = − 1
πa 2 /2
πa 2 /2
C
C
y 2 dx.
Here C = C 1 + C 2 + C 3 where C 1 : x = t, y =0, 0 ≤ t ≤ a;
C 2 : x = a cos t, y = a sin t, 0 ≤ t ≤ π 2 ;and
C 3 : x =0, y = a − t, 0 ≤ t ≤ a. Then
C x2 dy = C 1
x 2 dy + C 2
x 2 dy + C 3
x 2 dy = a
0 0 dt + π/2
0
(a cos t) 2 (a cos t) dt + a
0 0 dt
so x = 1
πa 2 /2
= π/2
a 3 cos 3 tdt= a 3 π/2
(1 − sin 2 t)costdt= a 3 sin t − 1 0 0 3 sin3 t π/2
= 2 0 3 a3
x 2 dy = 4a
C 3π .
C y2 dx = C 1
y 2 dx + C 2
y 2 dx + C 3
y 2 dx = a
0 0 dt + π/2
0
(a sin t) 2 (−a sin t) dt + a
0 0 dt
so y = − 1
πa 2 /2
= π/2
(−a 3 sin 3 t) dt = −a 3 π/2
(1 − cos 2 t)sintdt= −a 3 1
0 0 3 cos3 t − cos t π/2
= − 2 0 3 a3 ,
y 2 dx = 4a
4a
C 3π . Thus (x, y) = 3π , 4a
3π
25. By Green’s Theorem, − 1 3 ρ C y3 dx = − 1 3 ρ D (−3y2 ) dA = D y2 ρdA = I x and
1
ρ 3 C x3 dy = 1 ρ 3 D (3x2 ) dA = D x2 ρdA= I y .
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. ThusP = −y/(x 2 + y 2 ) and Q = x/(x 2 + y 2 ) have continuous partial derivatives on
this open region containing D and we can apply Green’s Theorem. But by Exercise 17.3.33(a) [ ET 16.3.33(a)],
∂P/∂y = ∂Q/∂x,so F · dr = 0 dA =0.
C D
29. Using the firstpartof(5),wehavethat dx dy = A(R) = ∂h ∂h
xdy.Butx = g(u, v),anddy = du +
R ∂R
∂u ∂v dv,
and we orient ∂S by taking the positive direction to be that which corresponds, under the mapping, to the positive direction
along ∂R,so
∂R
∂h
xdy= g(u, v)
∂S ∂u
= ± S
= ± S
= ± S
∂
∂u
∂h
du +
∂v dv = g(u, v) ∂h
∂S ∂u
g(u, v)
∂h
g(u, v)
∂h
∂v
−
∂
∂v
∂g ∂h
+ g(u, v) ∂2 h
− ∂g ∂h
− g(u, v) ∂2 h
∂u ∂v ∂u ∂v ∂v ∂u ∂v ∂u
∂x ∂y
− ∂x ∂y
∂u ∂v ∂v ∂u
du + g(u, v)
∂h
∂v dv
∂u dA [using Green’s Theorem in the uv-plane]
dA [using the Chain Rule]
dA [by the equality of mixed partials] = ±
S
∂(x,y)
du dv
∂(u,v)
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
Therefore A(R) = dx dy =
R
S
∂(x, y)
∂(u, v) du dv.
∂ (x, y)
∂(u, v) .