<s>[INST] Volume of an ellipsoid using cylindrical polar coordinates I've been working on a question about finding the volume of an ellipsoid
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1.$$
This is fine if I consider rescaling the axes to give a sphere, but I wanted to try to solve the problem specifically using polar coordinates, $(\rho, \Phi, z)$ in a triple integral. My thoughts (to find the limits of integration) were as follows: 


*

*$z$ varies between $-c$ and $c$; by fixing $z$, we may consider the ellipse given by 
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} =1- \frac{z^2}{c^2}$$

*$\Phi$ varies between $0$ and $2\pi$; by further fixing the angle $\Phi$, we have that $\rho$ varies between $0$ and $\sqrt{(1-\frac{z^2}{c^2})(a^2cos^2\Phi + b^2sin^2\Phi)}$, the horrible square root just being $\sqrt {x^2+y^2}$ in a different form.


However, evaluating $\iiint \rho\, dz\,d\Phi\, d \rho$ with these limits gives me $\frac{2\pi}{3}c(a^2+b^2)$ which is clearly wrong. I'm guessing my mistake is with my limits in the $\rho$ integral and would appreciate some guidance. 
 [/INST] From the equation you should have
$$\rho^2=x^2+y^2=b^2-\frac{b^2}{c^2}z^2\left(1-\frac{b^2}{a^2}x^2\right)$$
thus $\rho$ varies from $0$ to
$$\sqrt{b^2-\frac{b^2}{c^2}z^2+\left(1-\frac{b^2}{a^2}\cos^2\theta\right)}$$
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