8—Multivariable Calculus 191b
a
A Moment of Inertia
The moment of inertia about an axis is
∫
r^2 ⊥dm. Here,r⊥is the perpendicular distance to the axis.
What is the moment of inertia of a uniform sheet of massM in the shape of a right triangle of sides
aandb? Take the moment about the right angled vertex. The area mass density,σ=dm/dAis
2 M/ab. The moment of inertia is then
∫
(x^2 +y^2 )σdA=
∫a0dx
∫b(a−x)/a0dyσ(x^2 +y^2 ) =
∫a0dxσ
[
x^2 y+y^3 / 3
]b(a−x)/a0=∫a0dxσ
[
x^2
b
a
(a−x) +
1
3
(
b
a
) 3
(a−x)^3
]
=σ
[
b
a
(
a^4
3
−
a^4
4
)
+
1
3
(
b^3
a^3
a^4
4
)]
=
1
12
σ
(
ba^3 +ab^3
)
=
M
6
(
a^2 +b^2
)
The dimensions are correct. For another check take the case wherea= 0, reducing this toMb^2 / 6. But
wait, this now looks like a thin rod, and I remember that the moment of inertia of a thin rod about its
end isMb^2 / 3. What went wrong? Nothing. Look again more closely. Show why this limiting answer
ought to be less thanMb^2 / 3.
Volume of a Sphere
What is the volume of a sphere of radiusR? The most obvious approach would be to use spherical
coordinates. See problem8.16for that. I’ll use cylindrical coordinates instead. The element of volume
isdV =rdrdφdz, and the integrals can be done a couple of ways.
∫
d^3 r=
∫R
0rdr
∫ 2 π0dφ
∫+√R (^2) −r 2
−
√
R^2 −r^2
dz=
∫+R
−Rdz
∫ 2 π0dφ
∫√R (^2) −z 2
0
rdr (8.21)
You can finish these now, see problem8.17.
A Surface Charge Density
An example that appears in electrostatics: The surface charge density,dq/dA, on a sphere of radius
Risσ(θ,φ) =σ 0 sin^2 θcos^2 φ. What is the total charge on the sphere?
The element of area isR^2 sinθdθdφ, so the total charge is
∫
σdA,
Q=
∫π0sinθdθR^2
∫ 2 π0dφσ 0 sin^2 θcos^2 φ=R^2
∫+1
− 1dcosθσ 0
(
1 −cos^2 θ
)∫^2 π
0dφcos^2 φ
The mean value ofcos^2 is 1 / 2. so theφintegral givesπ. For the rest, it is
σ 0 πR^2
[
cosθ−
1
3
cos^3 θ
]+1
− 1=
4
3
σ 0 πR^2
Limits of Integration
Sometimes the trickiest part of multiple integrals is determining the limits of integration. Especially