How can I setup a triple integral for ellipsoid volume?

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?

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.
In cylinderical coordinates: $$2\int_{\theta=0}^{2\pi}\int_{r=0}^1\int_{z=0}^{c\sqrt{1-r^2}}|J|dz~dr~d\theta$$