Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

320 0 CHAPTER 16 VECTOR CALCULUS
d(d . F) n 2p-[(EPH+fiQ1+fiR1)·+(8 2 P1 +8 2 Ch 8 2 R1)· (8 2 P1. 8 2 Q1 8 2 1
R1)k]
gra IV - V - OX2 OXOY OXOZ
oyox oy2 + [)y[)z J + , OZOX + Ozoy + OZ 2
_ [(8 2 P1 + 8 2 P1 + B 2 P1) i + (8 2 Q1 + 8 2 Q1 + 8 2 Q1) j
8x 2 8y 2 8z 2 8x 2 8y2 8z 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 = grad div F - \l 2 F as desired.
31. (a)\lr =V.Jx2+y2 + z2= x i + y j + z k = xi + yj+zk =~
. .Jx2+y2+z2 .Jx2+y2+z2 .Jx2+y2+z2 .Jx2+y2 + z2 r
j
k
(b)\lxr= :x :V :z =l:V(z) - !(v)]i + [:z(x) - :x(z)]j + [:x(y)-:V(x)]k = O
X y Z
(c) \l (.!) = \l ( 1 ) .
r .J x2 + y2 + z2 ·
1
1
1
-
(2x) ·
(2y)
(2z)
2 .J x2 + y2 + z2 . 2 .J x2 + y2 + z2 . 2 V x2 + y2 + z2
--~~~~~---1 - J - k
x2 + y2 + z2 x2 + y2 + z2 x2 + y2 + z~
x i+ y j +z k r
(x2 + y2 + z2)3/2 =-r3
(d) \lin r = \lln(x2 + y2 + z2)1/2 = ~ \lln(x2 + y2 + z2)
x . y • z k x i + y j +z k r
--::----::-----::- 1 + J + = = -
x2 + y2 + z2 x2 + y2 + z2 x2 + y2 + z2 x2 + y2 + z2 7.2
33. By (13), fc f(\lg) · n ds = ffv div(f\lg) dA = JJ'o!f div(\lg) + \lg · \l f] dA by Exercise 25. But div(\lg) = \l 2 g.
Hence ffv f\l 2 gdA = fc f(Vg) · nds - ffv \lg · \lf dA.
35. Let f(x, y) = 1. Then V f = 0 and Green's first identity (see Exercise 33) says
ffv \l 2 gdA = f c (Vg) · n ds - ffv 0 · \lgdA =>
ffv \l 2 gdA = J~ \lg · n ds. But g is harmonic on D, so
V 2 g = 0 =>
fc\lg· nds = OandfcDngds = fc(\lg· n)ds =O.
37. (a) We know thatw = vjd, and from the diagram sin (;I= d/r => v = d!JJ = (sinO)rw = lw x rl. But vis perpendicular
to both wand r, so that v = w x r.
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