Find the volume of the ellipsoid
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} \leq 1$$
where $a,b,c \in \mathbb R$.
Attempt:
Ok well, I figured spherical coordinates would probably be the best approach.
In spherical, $(x,y,z) \mapsto (sin(\phi)cos(\theta),sin(\phi)sin(\theta),cos(\phi))$ where $\phi \in [0,\pi]$ and $\theta \in [0,2\pi]$.
So would my integral look something like:
$$\int_0^{2\pi}\int_0^{\pi}\int_0^{1} \rho ^2sin(\phi) \space dp \space d\phi \space d\theta$$
Where $\rho = \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = \frac{sin^2(\phi)cos^2(\theta)}{a^2} + \frac{sin^2(\phi)sin^2(\theta)}{b^2} + \frac{cos^2(\phi)}{c^2}$ … ?
If not, could someone guide me through the process of setting up the integral?
Best Answer
If you make the change of variables $(x,y,z)=(au,bv,cw)$ the Jacobian is just $abc$ (assuming they are positive) and the region becomes $$u^2+v^2+w^2\le 1$$ Therefore the new integral is $abc$ times an integral for the volume of a unit sphere, hence the answer is $${4\over 3}\pi abc$$ Note that when $a=b=c=r$ we get the usual formula for volume of a sphere.