Finding The Volume Of a Shape That Is Given By the Formula $3x^2 + 2y^2 + z^2 \leq 6$

Question

how can I find the volume of the shape that is given by the formula

$3x^2 + 2y^2 + z^2 \leq 6$ ?

Thanks!

Answer 1

Dividing both sides by 6, this is an ellipsoid in standard form with semi-axes of length $\sqrt{6}$, $\sqrt{3}$, and $\sqrt{2}$. It is well-known that the volume of an ellipsoid $$E: \frac{x^2}{a^2} + \frac{c^2}{b^2} + \frac{z^2}{c^2} = 1$$ has volume $\frac43 \pi abc$, so in this case your answer is $\boxed{6}$.

If you are interested in actually carrying out the integration, then consider rewriting the equation in the above form and then performing the following change of coordinates: $$\begin{cases} x = \sqrt{2}r\cos \theta \sin \phi \\ y = \sqrt{3}r \sin \theta \sin \phi \\ z = \sqrt{6} \cos \phi \end{cases} $$ The new bounds of integration will take $r$ from $0$ to $1$, $\theta$ from $0$ to $2 \pi$, and $\phi$ from $0$ to $\pi$. The resulting Jacobian from the change of variables will be $abc = 6$. The remaining steps are pretty straightforward, so I'll leave that to you.