Q:

Problem with finding the limits for a triple integral

I have the function $$x^2+y^2-z=0 \ , (x^2+y^2)^2=x^2-y^2$$ and i must calculate the volume. So i am using cylindrical coordinates but my problem is i cannot find the right limits or maybe i am making another mistake. So i have found that $$0 \le Z \le r^2 \\0 \le r \le \sqrt{cos2\phi} \\ 0 \le\phi \le \frac{\pi}{2}$$ I am totally not sure about limit of integration for $\phi$, i observe the $\sqrt{cos2\phi}$ should be positive. and it is positive only in the first and fourth quadrant of the unit circle so i decide to take this limit. For the other two i solved the equations for the function by substituting $x=rcos \phi$ and $y=rsin \phi$ . I am getting wrong result that is not near to the right answer that must be equal to $\frac{\pi}{8}$. I want to ask for help with setting up the limit. Thank you in advance.

A:

The integration set is bounded by the lemniscate of Bernoulli. In the polar coordinates, it is
$$
r^2=\cos2\phi.
$$
Clearly, $\cos2\phi\ge 0$ $\Leftrightarrow$ $\phi\in[-\pi/4,\pi/4]\cup [3\pi/4,5\pi/4]$. By the symmetry, it is enough to consider the first interval only and multiply by 2. Thus, the volume is
$$
\iiint1\,dxdydz=2\int_{-\pi/4}^{\pi/4}\int_0^{\sqrt{\cos2\phi}}\int_0^{r^2} r\,dzdrd\phi=[...]=\frac{\pi}{8}.
$$

