Solução_Calculo_Stewart_6e

F.
TX.10 SECTION 17.5 CURL AND DIVERGENCE ET SECTION 16.5 ¤ 285
∂
grad(div F) −∇ 2 2 P 1
F =
∂x + ∂2 Q 1
2 ∂x∂y + ∂2 R 1 ∂ 2 P 1
i +
∂x∂z ∂y∂x + ∂2 Q 1
∂y + ∂2 R 1 ∂ 2 P 1
j +
2 ∂y∂z ∂z∂x + ∂2 Q 1
∂z∂y + ∂2 R 1
k
∂z 2
∂ 2 P 1
−
∂x + ∂2 P 1
2 ∂y + ∂2 P 1 ∂ 2 Q 1
i +
2 ∂z 2 ∂x + ∂2 Q 1
2 ∂y + ∂2 Q 1
j
2 ∂z 2
∂ 2 R 1
+
∂x + ∂2 R 1
2 ∂y + ∂2 R 1
k
2 ∂z 2
∂ 2 Q 1
=
∂x∂y + ∂2 R 1
∂x∂z − ∂2 P 1
∂y − ∂2 P 1 ∂ 2 P 1
i +
2 ∂z 2 ∂y∂x + ∂2 R 1
∂y∂z − ∂2 Q 1
∂x − ∂2 Q 1
j
2 ∂z 2
∂ 2 P 1
+
∂z∂x + ∂2 Q 1
∂z∂y − ∂2 R 1
∂x − ∂2 R 2
k
2 ∂y 2
Then applying Clairaut’s Theorem to reverse the order of differentiation in the second partial derivatives as needed and
comparing, we have curl curl F =graddivF −∇ 2 F as desired.
31. (a) ∇r = ∇ x 2 + y 2 + z 2 =
(b) ∇ × r =
∂
∂x
1
(c) ∇ = ∇
r
i j k
∂
∂y
∂
∂z
x y z
1
x2 + y 2 + z 2
x
x2 + y 2 + z 2 i +
1
−
2 x
=
2 + y 2 + z (2x)
2 i −
x 2 + y 2 + z 2
= − x i + y j + z k
(x 2 + y 2 + z 2 ) 3/2 = − r
r 3
y
x2 + y 2 + z 2 j +
z
x2 + y 2 + z k = x i + y j + z k
2 x2 + y 2 + z = r 2 r
∂
=
∂y (z) − ∂ ∂
∂z (y) i +
∂z (x) − ∂ ∂
∂x (z) j +
∂x (y) − ∂
∂y (x) k = 0
1
2 x 2 + y 2 + z 2 (2y)
x 2 + y 2 + z 2
(d) ∇ ln r = ∇ ln(x 2 + y 2 + z 2 ) 1/2 = 1 2 ∇ ln(x2 + y 2 + z 2 )
=
x
x 2 + y 2 + z 2 i +
y
x 2 + y 2 + z 2 j +
j −
1
2 x 2 + y 2 + z 2 (2z)
x 2 + y 2 + z 2
z
x 2 + y 2 + z k = x i + y j + z k
2 x 2 + y 2 + z = r
2 r 2
33. By (13), C f(∇g) · n ds = D div(f∇g) dA = D [f div(∇g)+∇g · ∇f] dA by Exercise 25. But div(∇g) =∇2 g.
Hence D f∇2 gdA= f(∇g) · n ds − ∇g · ∇f dA.
C D
k
35. Let f(x, y) =1.Then∇f = 0 and Green’s first identity (see Exercise 33) says
D ∇2 gdA= (∇g) · n ds − 0 · ∇gdA ⇒ C D D ∇2 gdA= ∇g · n ds. Butg is harmonic on D,so
C
∇ 2 g =0 ⇒ ∇g · n ds =0and D C C ngds= (∇g · n) ds =0.
C
37. (a) We know that ω = v/d, and from the diagram sin θ = d/r ⇒ v = dω =(sinθ)rω = |w × r|. Butv is perpendicular
to both w and r,sothatv = w × r.