integrate $2\sqrt{1-x^2/4-y^2/9}$

Question

I am trying to find the volume of the ellipsoid $E$ given by $x^2/4 + y^2/9 + z^2 \leq 1$ by computing $\iiint_E dV$

I ended up with $$\int_{-2}^{2}\int_{-3\sqrt{1-x^2/4}}^{3\sqrt{1-x^2/4}}\int_{-\sqrt{1-x^2/4-y^2/9}}^{\sqrt{1-x^2/4-y^2/9}}1dzdydx$$

After doing the inner integral I get $$\int_{-2}^{2}\int_{-3\sqrt{1-x^2/4}}^{3\sqrt{1-x^2/4}}2\sqrt{1-x^2/4-y^2/9}\space dydx$$

I am not sure about how to figure out the next one

The hint is to use trig substitution, though I am not sure if it is talking about this step or the next one

Answer 1

HINT:

If you introduce new coordinates according to

$$ (x,y,z) \rightarrow (au,bv,cw),$$

where $a=2, b=3, c = 1$ then the problem changes to determining the volume of a sphere of unit radius given by the equation

$$S: u^2 + v^2 +w^2 \leq 1.$$

The Jacobian associated with such a transformation is given by

$$ J = \left|\frac{\partial(x,y,z)}{\partial (u,v,w)} \right| =abc. $$

The problem can now be stated as

$$ abc \iiint dV. $$

You can see this method explained in detail on youtube.

Answer 2

$$\int_{-2}^{2}\int_{-3\sqrt{1-x^2/4}}^{3\sqrt{1-x^2/4}}2\sqrt{1-x^2/4-y^2/9}\space dydx$$ is the same as $$\int_{-2}^{2}\int_{-\sqrt{9-9x^2/4}}^{\sqrt{9-9x^2/4}}2\sqrt{1-x^2/4-y^2/9}\space dydx$$ then do trig sub so $\frac{y^2}{9}=1-\frac{x^2}{4}\sin^2{\theta}$

then $y=\sqrt{9-\frac{9x^2}{4}}\sin{\theta}$, $dy=\sqrt{9-\frac{9x^2}{4}}\cos{\theta}d\theta$

then you get $$\int_{-2}^{2}\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}6(1-\frac{x^2}{4})\cos^2{\theta}d\theta dx=\int_{-2}^{2}\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}3(1-\frac{x^2}{4})(1+\cos{2\theta})d\theta dx$$

then you can keep going to end up with $8\pi$

(this was the given solution)