# 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?

• In spherical coordinates, $dV=\rho^2\sin\phi\;d\rho\;d\theta\;d\phi$ and $x=\rho\sin\phi\cos\theta$. Now, the ellipsoid is formed by taking $0\leq\rho\leq\left(\frac{1}{a^2}\sin^2\phi\cos^2\theta+\frac{1}{b^2}\sin^2\phi\sin^2\theta+\frac{1}{c^2}\cos^2\phi\right)^{-1/2}$, $0\leq\theta\leq 2\pi$, $0\leq\phi\leq 2\pi$. However the resulting integral seems no easier to compute than in rectangular coordinates. Commented Feb 13, 2017 at 10:26
• I was thinking that you'd first perform a variable change so that you'd get a more easy volume to integrate over, for example (x,y,z) = (au,bv,cw) then from there on switching into spherical coordinates. However that way I end up with a^3 in the answer which doesn't seem right. Commented Feb 13, 2017 at 10:41
• Are you sure that generalized polar coordinates do not work? en.wikipedia.org/wiki/Ellipsoid#Parameterization Commented Feb 13, 2017 at 10:45
• Actually my main issue is that I'm not sure weather my answer is right or wrong. After the variable change and changing into spherical coordinates solving the integrals is easy. I end up with the answer 4*pi*(a^3)*b*c/15. A friend told me that the dimensions have to make sense after the integration. So if we integrate over a volume, the answer should bee in 3 dimensions (not five like my answer). Is my friend crazy? Commented Feb 13, 2017 at 11:11

$$x=a\rho\cos\phi\sin\theta,y=b\rho\sin\phi\sin\theta,z=c\rho\cos\theta$$
value of the Jacobian: $$|I|=abc\rho^2\sin \theta$$
$$\iiint x^2\ dV$$ $$={8}\int_{\theta=0}^{\theta={\pi\over2}}\int_{\phi=0}^{\phi={\pi\over2}}\int_{\rho=0}^{\rho=1}(a\rho\cos\phi\sin\theta)^2abc\,\rho^2\sin \theta \,d\rho\, d\phi\, d\theta$$ $$={8}a^3bc\int_{\theta=0}^{\theta={\pi\over2}}\int_{\phi=0}^{\phi={\pi\over2}}\int_{\rho=0}^{\rho=1}(\rho^4\cos^2\phi\sin^3\theta)\, \,d\rho\, d\phi\, d\theta$$$$={8}a^3bc\int_{\theta=0}^{\theta={\pi\over2}}\sin^3\theta\, d\theta \int_{\phi=0}^{\phi={\pi\over2}}\cos^2\phi \,d\phi \int_{\rho=0}^{\rho=1}\rho^4\, \,d\rho\,$$ $$={8\over5}a^3bc\int_{\theta=0}^{\theta={\pi\over2}}\sin^3\theta\, d\theta [\int_{\phi=0}^{\phi={\pi\over2}}\cos^2\phi \,d\phi]= {8\over5}a^3bc\int_{\theta=0}^{\theta={\pi\over2}}\sin^3\theta\, d\theta[{1\over2}(\phi+\sin \phi\cos \phi)]_{0}^{{\pi\over2}}$$ $$={8\pi\over20}a^3bc\int_{\theta=0}^{\theta={\pi\over2}}\sin^3\theta\, d\theta$$ $$={8\pi\over20}a^3bc[{1\over12}(\cos 3\theta-9\cos \theta)]_{0}^{{\pi\over2}}$$ $$={8\pi\over20}a^3bc[{8\over12}]={4\pi\over15}a^3bc$$