Calculating the Inertia Tensor of a Homogeneous Sphere

[Storm Butler]

Homework Statement

Calculate the moments of Inertia I_{1}, I_{2}, I_{3} for a homogenous sphere

Homework Equations

I_{jk}=\intx^{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\phid\varthetadr
with 0\leqr\leqR, 0\leq\phi\leq2\pi, and 0\leq\vartheta\leqpi

when i solve this integral i get I=\rho\frac{4}{5}\piR^{2} and then setting \rho= \frac{M}{4/3\piR^{3}<br /> }
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 don't really know where is went wrong.

 

[clamtrox]

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 I_ij. Start by making the easiest choice for i=j and then think if you have to calculate more.

 

[Storm Butler]

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.

 

[Storm Butler]

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

 

[clamtrox]

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?