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.
The polar equation relative to its center of an ellipse in standard position is $$\rho = {a b \over \sqrt{(b\cos\Phi)^2+(a\sin\Phi)^2}}.$$ Each horizontal slice of the ellipsoid must be scaled by a factor of $\sqrt{1-z^2/c^2}$, therefore $\rho$ ranges from $0$ to $${a b \sqrt{1-z^2/c^2} \over \sqrt{(b\cos\Phi)^2+(a\sin\Phi)^2}}.$$
However, instead of slogging through a triply-iterated integral, it seems to me a bit easier and certainly less error-prone to take advantage of the coordinate system by integrating $\frac12\rho\,d\Phi$ directly for each slice, and then integrating over $z$.