Triple integral x^2 over an ellipsoid I'm having difficulties with an excercise, and having trouble finding resources on the internet that would help me tackle the problem. 
The exercise: 

Calculate $$\iiint x^2\ dV$$
  over the ellipsoid $(\frac{x}{a})^2+(\frac{y}{b})^2+(\frac{z}{c})^2=1$

I'm not entirely sure how to do the variable change into spherical cordinates. Could someone guide me through the process?
 A: 
$$x=a\rho\cos\phi\sin\theta,y=b\rho\sin\phi\sin\theta,z=c\rho\cos\theta$$
value of the Jacobian: $$|I|=abc\rho^2\sin \theta$$

$$\iiint x^2\ dV$$
$$={8}\int_{\theta=0}^{\theta={\pi\over2}}\int_{\phi=0}^{\phi={\pi\over2}}\int_{\rho=0}^{\rho=1}(a\rho\cos\phi\sin\theta)^2abc\,\rho^2\sin \theta \,d\rho\, d\phi\, d\theta$$
$$={8}a^3bc\int_{\theta=0}^{\theta={\pi\over2}}\int_{\phi=0}^{\phi={\pi\over2}}\int_{\rho=0}^{\rho=1}(\rho^4\cos^2\phi\sin^3\theta)\, \,d\rho\, d\phi\, d\theta$$$$={8}a^3bc\int_{\theta=0}^{\theta={\pi\over2}}\sin^3\theta\, d\theta \int_{\phi=0}^{\phi={\pi\over2}}\cos^2\phi \,d\phi \int_{\rho=0}^{\rho=1}\rho^4\, \,d\rho\, $$
$$={8\over5}a^3bc\int_{\theta=0}^{\theta={\pi\over2}}\sin^3\theta\, d\theta [\int_{\phi=0}^{\phi={\pi\over2}}\cos^2\phi \,d\phi]= {8\over5}a^3bc\int_{\theta=0}^{\theta={\pi\over2}}\sin^3\theta\, d\theta[{1\over2}(\phi+\sin \phi\cos \phi)]_{0}^{{\pi\over2}}$$
$$={8\pi\over20}a^3bc\int_{\theta=0}^{\theta={\pi\over2}}\sin^3\theta\, d\theta$$
$$={8\pi\over20}a^3bc[{1\over12}(\cos 3\theta-9\cos \theta)]_{0}^{{\pi\over2}}$$
$$={8\pi\over20}a^3bc[{8\over12}]={4\pi\over15}a^3bc$$
