Calculating the moment of inertia with respect to z-axis (volume integral)

Question

Question states: Consider a body with a surface defined by $2(x^{2}+y^{2})+4z^{2}=1$. Calculate the moment of inertia with respect to the z-axis, i.e. $I_{z}=\int\int\int_{V}(x^{2}+y^{2})dxdydz$.

I applied the generalized spherical coordinates: s=$\frac{1}{\sqrt{2}}r\cos(\psi) \sin(\theta)$ i +$\frac{1}{\sqrt{2}}r\sin(\psi) \sin(\theta)$ j +$\frac{1}{2}r\cos(\theta)$ k.

What are the boundaries of r, $\theta$ and $\psi$ in that case?

The volume integral than states: $I_{z}=\int\int\int_{V}(r^{2}\sin^{2}(\theta))\ dr\ d\theta \ d\psi$.

So, the rest to do is to calculate the triple integral with adequate boundaries?

Answer 1

Hint: By Divergence Theorem $$\int_{V}\nabla\cdot \mathbf{F}=\int_{\partial V}\mathbf{F}\cdot d\mathbf{S}$$. Now you can consider the vector field $\mathbf{F}=(x^3/3+y^2x,0,0)$. At this point we have $$\nabla\cdot \mathbf{F}=x^2+y^2,$$ hence your moment of inertia is $$I_z=\int_{\partial V}(x^3/3+y^2x,0,0)\cdot d\mathbf{S}.$$

Answer 2

The body is an ellipsoid $2x^2+2y^2+4z^2=1$ where the $z$-axis is an axis of rotational symmetry. Using cylinder coordinates $x=r\cos\phi$, $y=r\sin\phi$, $x^2+y^2=r^2$ and Jacobian $r$ the integral is $$ I_z=\int_{-1/2}^{1/2} dz \int_{2(x^2+y^2)\le 1-4z^2} r dr \int_0^{2\pi} d\phi r^2 $$ $$ =2\pi \int_{-1/2}^{1/2} dz \int_0^{\sqrt{(1-4z^2)/2}} r^3 dr $$ $$ =2\pi \int_{-1/2}^{1/2} dz \frac{1}{64}(2-8z^2)^2 $$ $$ =\pi/15 . $$