# Moment of Inertia for Ellipsoid

• ledphones

## Homework Statement

a)Evaluate ∫∫∫E dV, where E is the solid enclosed by the ellipsoid x^2/a^2+y^2/b^2+z^2/c^2 =1. Use the transformation x=au, y=bv, z=cw.

b)If the solid in the above has density k find the moment of inertia about the z-axis.

∅=phi

## The Attempt at a Solution

I got a correct 4/3∏abc

i thought b) would be 0 based on the following.

Jacobian is abc and the solid enclosed would be u^2+v^2+w^2≤1

x^2*y=(a^2*u^2)(bv)

V=∫∫∫(a^2bu^2v)(abc)dudvdw
=∫0→2∏ ∫0→∏ ∫0→1 (a^3b^2c)(ρ^2sin^2∅cos^2θ)(ρsin∅sinθ)ρ^2sin∅dρd∅dθ
=a^3b^2c∫∫∫(ρ^5sin^4∅cos^2θsinθ)dρd∅dθ
=a^3b^2c∫0→2∏ cos^2θsinθdθ∫0→∏ sin^4∅d∅ ∫0→1 ρ^5dρ
=0 since the 1st integral is 0

## Homework Statement

a)Evaluate ∫∫∫E dV, where E is the solid enclosed by the ellipsoid x^2/a^2+y^2/b^2+z^2/c^2 =1. Use the transformation x=au, y=bv, z=cw.

b)If the solid in the above has density k find the moment of inertia about the z-axis.

∅=phi

## The Attempt at a Solution

I got a correct 4/3∏abc

i thought b) would be 0 based on the following.

Jacobian is abc and the solid enclosed would be u^2+v^2+w^2≤1

x^2*y=(a^2*u^2)(bv)

The integrand for a second moment is always positive. Your problem is in the last step above. The moment arm is the distance squared from the z axis: r2=x2+y2, not x2y.