# Calculating the Inertia Tensor of a Homogeneous Sphere

## Homework Statement

Calculate the moments of Inertia I$_{1}$, I$_{2}$, I$_{3}$ for a homogenous sphere

## Homework Equations

I$_{jk}$=$\int$x$^{2}_{l}$$\delta$$_{ik}$-x$_{i}$x$_{k}$dV

## The Attempt at a Solution

For I$_{x}$ i set up the equation using the above equation in cartesian coordinates and then i switched into polar coordinates and i get the following integral
$\rho$$\int$$\int$$\int$(r$^{2}$-rsin($\vartheta$)cos($\phi$))r$^{2}$sin($\vartheta$)d$\phi$d$\vartheta$dr
with 0$\leq$r$\leq$R, 0$\leq$$\phi$$\leq$2$\pi$, and 0$\leq$$\vartheta$$\leq$$pi$

when i solve this integral i get I=$\rho$$\frac{4}{5}$$\pi$R$^{2}$ and then setting $\rho$= $\frac{M}{4/3\piR^{3} }$
so after simplifying i end up with I=$\frac{3}{5}$MR$^{2}$
But the answer i believe is $\frac{2}{5}$MR$^{2}$ , so i dont really know where is went wrong.

Related Introductory Physics Homework Help News on Phys.org
How did you end up with that integral? What are i an k in your expression? Remember that the principal moments of inertia are the diagonal terms in Iij. Start by making the easiest choice for i=j and then think if you have to calculate more.

as i started to type i response i realized my integrand was essentially r$^{2}$-x. when i should have been r$^{2}$-x$^{2}$. So what i should have is r$^{4}$-r$^{4}$sin($\vartheta$)$^{3}$cos$\phi$$^{2}$

If i integrate this then i get 2/5$\pi$$^{2}$R$^{5}$. which looks better.

although if i then multiply this by rho i get 3/10$\pi$MR$^{2}$...

as i started to type i response i realized my integrand was essentially r$^{2}$-x. when i should have been r$^{2}$-x$^{2}$. So what i should have is r$^{4}$-r$^{4}$sin($\vartheta$)$^{3}$cos$\phi$$^{2}$

If i integrate this then i get 2/5$\pi$$^{2}$R$^{5}$. which looks better.
That's not right either, one sin(θ) should be in the integration measure. Perhaps it would be easier to calculate only the zz-component of I. Why is this enough?