I need to calculate the moment of inertia about $z$ axis of domain :
$$E=\{(p,\theta,\phi)\,|\,0\le p\le2,0\le\theta\le2\pi,\frac{\pi}{3}\le\phi\le\frac{\pi}{2}\}$$
Which, if I'm right, is domain between cone $z=\sqrt{(x^2+y^2)/3}$ and under half sphere $z=\sqrt{4-x^2-y^2}$.
I know that :
$$I_z=\int\int\int_E(x^2+y^2)\rho(x,y,z)dV$$
Further, with constant density $\rho(x,y,z)=k$.
$$I_z=k\int\int\int_E(x^2+y^2)dV$$
I'm just not sure how to transpose this to spherical coordinates…
$$I_z=k\int_0^2\int_0^{2\pi}\int_{\pi/3}^{\pi/2}\rho^2\sin^2(\phi)\cos^2(\theta)+\rho^2\sin^2(\phi)\sin^2(\theta)\rho^2\sin(\phi)\,\,\rho d\theta d\phi$$
$$I_z=k\int_0^2\int_0^{2\pi}\int_{\pi/3}^{\pi/2}\rho^2\sin^2(\phi)\rho^2\sin(\phi)\,\,\rho d\theta d\phi$$
$$I_z=k\int_0^2\int_0^{2\pi}\int_{\pi/3}^{\pi/2}\rho^4\sin^3(\phi)\,\,\rho d\theta d\phi$$
Is that correct?
Thanks !
Best Answer
You have an extra $\rho$ in it, otherwise it is correct.
$$I_z=k\int_0^2\int_0^{2\pi}\int_{\pi/3}^{\pi/2}\rho^2\sin^2(\phi)\rho^2\sin(\phi)\,\, d\phi d\theta d\rho$$
Or that $\rho$ should be $d\rho$, if that is what you meant.