Solução_Calculo_Stewart_6e

F.
SECTION 16.9TX.10
CHANGE OF VARIABLES IN MULTIPLE INTEGRALS ET SECTION 15.9 ¤ 255
(b) The wedge in question is the shaded area rotated from θ = θ 1 to θ = θ 2.
Letting
V ij = volume of the region bounded by the sphere of radius ρ i
and the cone with angle φ j (θ = θ 1 to θ 2 )
and letting V be the volume of the wedge, we have
V =(V 22 − V 21 ) − (V 12 − V 11 )
= 1 (θ 3 2 − θ 1 ) ρ 3 2(1 − cos φ 2 ) − ρ 3 2(1 − cos φ 1 ) − ρ 3 1(1 − cos φ 2 )+ρ 3 1(1 − cos φ 1 )
= 1 3 (θ2 − θ1) ρ 3 2 − ρ1 3 (1 − cos φ2 ) − ρ 3 2 − ρ1 3 (1 − cos φ1 ) = 1 3 (θ2 − θ1)
ρ 3 2 − ρ 3 1 (cos φ1 − cos φ 2 )
θ2
θ 1
ρ2 sin φ 2
ρ 1 sin φ 1
r cot φ1
r cot φ 2
Or: Show that V =
rdzdrdθ.
(c) By the Mean Value Theorem with f(ρ) =ρ 3 there exists some ˜ρ with ρ 1 ≤ ˜ρ ≤ ρ 2 such that
f(ρ 2 ) − f(ρ 1 )=f 0 (˜ρ)(ρ 2 − ρ 1 ) or ρ 3 1 − ρ 3 2 =3˜ρ 2 ∆ρ. Similarly there exists φ with φ 1 ≤ ˜φ ≤ φ 2
such that cos φ 2 − cos φ 1 = − sin ˜φ
∆φ. Substituting into the result from (b) gives
∆V =(˜ρ 2 ∆ρ)(θ 2 − θ 1 )(sin ˜φ) ∆φ =˜ρ 2 sin ˜φ ∆ρ ∆φ ∆θ.
16.9 Change of Variables in Multiple Integrals ET 15.9
1. x =5u − v, y = u +3v.
∂(x, y)
The Jacobian is
∂(u, v) = ∂x/∂u ∂x/∂v
∂y/∂u ∂y/∂v = 5 −1
=5(3)− (−1)(1) = 16.
1 3 3. x = e −r sin θ, y = e r cos θ.
∂(x, y)
∂(r, θ) = ∂x/∂r ∂x/∂θ
∂y/∂r ∂y/∂θ = −e −r sin θ
e r cos θ
e −r cos θ
−e r sin θ = e−r e r sin 2 θ − e −r e r cos 2 θ =sin 2 θ − cos 2 θ or − cos 2θ
5. x = u/v, y = v/w, z = w/u.
∂x/∂u ∂x/∂v ∂x/∂w
∂(x, y, z)
1/v −u/v 2 0
∂(u, v, w) = ∂y/∂u ∂y/∂v ∂y/∂w
=
0 1/w −v/w 2
∂z/∂u ∂z/∂v ∂z/∂w
−w/u 2 0 1/u
= 1 1/w −v/w 2
v 0 1/u − − u 0 −v/w 2
v 2 −w/u 2 1/u +0 0 1/w
−w/u 2 0
= 1 1
v uw − 0 + u
0 − v
+0= 1
v 2 u 2 w uvw − 1
uvw =0