Find the volume interior to $y^2+z^2=2$ and exterior to $x^2-y^2-z^2=2$. I have solved a few exercises of this kind, but all of them had one dependent function of $x$ and $y$ only, but here both of them have $z$ in the equation.
To find the volume using double integrals you must use: $$V=\iint z\ dA$$ Where $z$ is the inferior base. Also, it could be that the best way to solve the exercise is using polar coordinates, in that case you use $x=ρcos\theta $ and $y=ρsin\theta $
Here's a graph of the functions:
The answer is:
$$2 \int\limits_{r=0}^{\sqrt{2}} 2 \pi r \sqrt{r^2+2}\ dr = \frac{8}{3} \left(4-\sqrt{2}\right) \pi$$
Use cylindrical shells around the $x$ axis and integrate in just the positive side ($0 \leq x \leq \sqrt{r^2 + 2}$). Here $r = \sqrt{y^2 + z^2}$.
One shell in red (up to the radius limit, in yellow):
You could use double integrals (over $\theta$ and $r$), but the $\theta$ integral is trivial ($2 \pi$), so I just used that in the above.
$$2 \int\limits_{\theta = 0}^{2 \pi} \int\limits_{r=0}^{\sqrt{2}} r \sqrt{r^2 + 2}\ d\theta\ dr$$