- #1

Storm Butler

- 78

- 0

## Homework Statement

Calculate the moments of Inertia I[itex]_{1}[/itex], I[itex]_{2}[/itex], I[itex]_{3}[/itex] for a homogenous sphere

## Homework Equations

I[itex]_{jk}[/itex]=[itex]\int[/itex]x[itex]^{2}_{l}[/itex][itex]\delta[/itex][itex]_{ik}[/itex]-x[itex]_{i}[/itex]x[itex]_{k}[/itex]dV

## The Attempt at a Solution

For I[itex]_{x}[/itex] 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

[itex]\rho[/itex][itex]\int[/itex][itex]\int[/itex][itex]\int[/itex](r[itex]^{2}[/itex]-rsin([itex]\vartheta[/itex])cos([itex]\phi[/itex]))r[itex]^{2}[/itex]sin([itex]\vartheta[/itex])d[itex]\phi[/itex]d[itex]\vartheta[/itex]dr

with 0[itex]\leq[/itex]r[itex]\leq[/itex]R, 0[itex]\leq[/itex][itex]\phi[/itex][itex]\leq[/itex]2[itex]\pi[/itex], and 0[itex]\leq[/itex][itex]\vartheta[/itex][itex]\leq[/itex][itex]pi[/itex]

when i solve this integral i get I=[itex]\rho[/itex][itex]\frac{4}{5}[/itex][itex]\pi[/itex]R[itex]^{2}[/itex] and then setting [itex]\rho[/itex]= [itex]\frac{M}{4/3\piR^{3}

}[/itex]

so after simplifying i end up with I=[itex]\frac{3}{5}[/itex]MR[itex]^{2}[/itex]

But the answer i believe is [itex]\frac{2}{5}[/itex]MR[itex]^{2}[/itex] , so i dont really know where is went wrong.