Find the volume of the ellipsoid $\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$ by solving the triple integral after making the transformation $\displaystyle x = au, y = bv$ and $\displaystyle z = cw$

I'm just a bit confused here because this does not seem like a Jacobian transformation. I know what the triple integral would be in cartesian terms, ie, $\displaystyle \displaystyle{\int_{-a}^a \int_{-\sqrt{b^2-\frac{b^2x^2}{a^2}}}^{\sqrt{b^2-\frac{b^2x^2}{a^2}}} \int_{-\sqrt{c^2-c^2\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}\right)}}^ {\sqrt{c^2-c^2\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}\right)}} 1 dzdydx}$

But clearly, we don't want to integrate that, so what does it mean by "after making the transformation $\displaystyle x = au, y = bv$ and $\displaystyle z = cw$"? Do I just simply replace all the x with au and y with bv etc, then what happens to the dz dy and dx?

Thanks heaps!