Solução_Calculo_Stewart_6e

F.
TX.10
PROBLEMS PLUS ¤ 267
11.
x
y
0 0
And ∂u
∂θ = −∂u
∂u
ρ sin φ sin θ + ρ sin φ cos θ,while
∂x ∂y
Therefore
∂ 2 u
∂θ 2 = −2 ∂2 u
∂y ∂x ρ2 sin 2 φ cos θ sin θ + ∂2 u
∂x 2 ρ2 sin 2 φ sin 2 θ
∂ 2 u
∂ρ + 2 ∂u
2 ρ ∂ρ + cot φ
ρ 2
+ ∂2 u
∂y 2 ρ2 sin 2 φ cos 2 θ − ∂u
∂u
ρ sin φ cos θ − ρ sin φ sin θ
∂x ∂y
∂u
∂φ + 1 ∂ 2 u
ρ 2 ∂φ 2 + 1
ρ 2 sin 2 φ
∂ 2 u
∂θ 2
= ∂2 u
∂x 2
(sin 2 φ cos 2 θ)+(cos 2 φ cos 2 θ)+sin 2 θ
+ ∂2 u (sin 2 φ sin 2 θ)+(cos 2 φ sin 2 θ)+cos 2 θ + ∂2 u cos 2 φ +sin 2 φ
∂y 2 ∂z 2
+ ∂u
2sin 2 φ cos θ +cos 2 φ cos θ − sin 2 φ cos θ − cos θ
∂x
ρ sin φ
+ ∂u
2sin 2 φ sin θ +cos 2 φ sin θ − sin 2 φ sin θ − sin θ
∂y
ρ sin φ
But 2sin 2 φ cos θ +cos 2 φ cos θ − sin 2 φ cos θ − cos θ =(sin 2 φ +cos 2 φ − 1) cos θ =0andsimilarlythecoefficient of
∂u/∂y is 0. Alsosin 2 φ cos 2 θ +cos 2 φ cos 2 θ +sin 2 θ =cos 2 θ (sin 2 φ +cos 2 φ)+sin 2 θ =1,andsimilarlythe
coefficient of ∂ 2 u/∂y 2 is 1. So Laplace’s Equation in spherical coordinates is as stated.
z f (t) dt dz dy = f (t) dV ,where
0 E
E = {(t, z, y) | 0 ≤ t ≤ z, 0 ≤ z ≤ y, 0 ≤ y ≤ x}.
If we let D be the projection of E on the yt-plane then
D = {(y, t) | 0 ≤ t ≤ x, t ≤ y ≤ x}. And we see from the diagram
that E = {(t, z, y) | t ≤ z ≤ y, t ≤ y ≤ x, 0 ≤ t ≤ x}. So
x
y
0 0
z
0 f(t) dt dz dy = x
0
= x
0
x
t
y
t
f(t) dz dy dt = x
x
0
1
2 y2 − ty f(t) y = x
y = t
(y − t) f(t) dy dt
t
1
2 x2 − tx − 1 2 t2 + t 2 f(t) dt
dt = x
0
= x
1
0 2 x2 − tx + 1 t2 f(t) dt = x
1
2 0 2 x2 − 2tx + t 2 f(t) dt
x (x − 0 t)2 f(t) dt
= 1 2