Math 522
Homework 9 Solutions
1. To solve this integral we will use a change of variables to n-dimensional spherical
coordinates and consider the limit as the “radius” variable goes to infinity. The
variables in our new coordinate system will be r, φ1 , φ2 , . . . , φn−1 where 1 ≤ r < ∞,
φ1 , φ2 , . . . , φn−2 ∈ [0, π], and φn−1 ∈ [0, 2π). We introduce the new variables by
defining f by the following.
f : x1 = r cos(φ1 )
f : x2 = r sin(φ1 ) cos(φ2 )
f : x3 = r sin(φ1 ) sin(φ2 ) cos(φ3 )
..
.
f : xn−1 = r sin(φ1 ) sin(φ2 ) · · · sin(φn−2 ) cos(φn−1 )
f : xn = r sin(φ1 ) sin(φ2 ) · · · sin(φn−1 ).
We then see that |x| = r and det ∇f = rn−1 sinn−2 (φ1 ) sinn−3 sin(φ2 ) · · · sin(φn−2 )
so
Z
F
1
dx = lim
t→∞
|x|k
Z
2π
Z
π
Z
π
Z
π
Z
t
···
φn−1 =0 φn−2 =0
Z t
n−1−k
= cn lim
t→∞
r
φn−3 =0
φ1 =0
r=1
1
| det ∇f |drdφ1 · · · dφn−1 .
rk
dr
1
n−k
cn t
1
−
t→∞ n − k
n−k
= lim
where cn is a constant. This converges when n − k < 0 so the given integral is
finite when k > n.
Page 368
6. Graph of F in the x1 , x2 -plane:
Graph of F in the u1 , u2 -plane:
Solving for each variable we find that
x21
q
+ = u21 + u22 ,
s q
1
2
2
x1 =
u1 + u2 + u1 , and
2
s q
1
2
2
u1 + u2 − u1 .
x2 =
2
x22
Now
f −1 : u1 = x21 − x22 ,
u2 = 2x1 x2
so
det ∇f −1 = 4x21 + 4x22 .
Therefore
det ∇f =
4x21
1
.
+ 4x22
From the change of variables we have
Z
x21
+
x22 dV2 (x)
Z
4
Z
4
Z
2
=
2
F
Z
1
2
=
2
1
p
u2 + u22
p1
du1 du2
4 u21 + u22
1
du1 du2
4
1
=
2
9. The given inequalities tell us that ρ ∈ [0, 1], ϕ ∈ [0, 2π) and θ ∈ [0, π4 ]. Finding
the Jacobian of the given change of variables we get ρ2 sin θ. Thus,
Z
Z
2π
Z
π/4
Z
x3 dV3 (x) =
0
F
π
2
π
=
8
Z
=
0
π/4
1
ρ3 sin θ cos θ dρ dθ dϕ
0
sin θ cos θ dθ
0