# Calculating triple integral over an ellipsoid

I'm trying to calculate $$\iiint \frac{dx\,dy\,dz}{\sqrt{1 - \left(\frac{x^2}{9}+\frac{y^2}{16}+\frac{z^2}{25}\right)}}$$ over the ellipsoid $\frac{x^2}{9}+\frac{y^2}{16}+\frac{z^2}{25}=1$. I couldn't find a solution. Can anyone help me?

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Is there a change of coordinates that'll map your ellipsoid to the unit ball? Because if you can replace your integral over a solid ellipsoid with an integral over the unit ball, there's a certain further change of coordinates that'll make things crystal clear. –  Branimir Ćaćić Jan 4 '13 at 23:52
I added the homework tag, which is how you should indicate that a question is homework in the future. –  Alex Becker Jan 5 '13 at 0:01
I can't get anywhere with changing the coordinates it's getting complicated or I'm doing something wrong. –  LuckyStrike Jan 5 '13 at 0:16

First make the change of variables $(x,y,z) = (3u,4v,5w)$. The Jacobian of this transformation is a diagonal matrix $[3, 4, 5]$ with determinant $3\cdot 4 \cdot 5 = 60$. Thus we have $$\iiint_E \frac{dx dy dz}{\sqrt{1-(x^2/9+y^2/16+z^2/25)}} = \iiint_B \frac{60du dv dw}{\sqrt{1-(u^2+v^2+w^2)}}.$$ Here $E$ is the original ellipsoid, and $B$ is the unit ball. Then we change to spherical coordinates $$=\int_{0}^{2\pi}\int_0^{\pi}\int_0^1 \frac{60\rho^2\sin \phi d\rho d\phi d\theta }{\sqrt{1-\rho^2}}.$$ Since the bounds are constants we may evaluate the integrals in any order we prefer. Integrating in $\phi$ gives a factor of $2$, and integrating in $\theta$ a factor of $2\pi$ so $$=240\pi\int_0^1 \frac{\rho^2}{\sqrt{1-\rho^2}}d\rho.$$ This is a standard integral which can be calculated either by parts (differentiate $\rho^2\to2\rho$, integrate $\frac{1}{\sqrt{1-\rho^2}}\to\arcsin \rho$) or trig substitution and it comes out to $$=240\pi \cdot \pi/4 = 60 \pi^2.$$