Find the volume of the region which lies inside both $x^2+y^2=r^2$ and $y^2+z^2=r^2$ Find the region inside two cylinders $x^2+z^2 \le r^2$ and $y^2+z^2 \le r^2$. 
I attempted this question by combining the two inequalities $x^2+2z^2+y^2 \le 2r^2$ and I got the bounds for $z$ which is $-\sqrt{{(2r^2-x^2-y^2)}/{2}} \le z \le \sqrt{{(2r^2-x^2-y^2)}/{2}}$. 
And I found the bound on $y$ to be $-\sqrt{(2r^2-x^2)} \le y \le \sqrt{(2r^2-x^2)}$, the bound on x is $0 \le x \le r$.
I am not sure if the above attempt is correct, if not, can anyone provide the correct way to find the bounds on z and y?

If you visualize it carefully, you might be able to see that for fixed $y$ the cross-section is a square. Then:
$$|x| \le \sqrt{r^2 - y^2},$$
$$|z| \le \sqrt{r^2 - y^2}.$$
Then, the cross-section is a square of area $4 (r^2 - y^2)$ and the volume should be:
$$\int_{-r}^r dy 4 (r^2 - y^2) = 16 r^3/3.$$
