Volume between two "orthogonal" paraboloids 
Find the volume enclosed by the paraboloids $z=x^2+y^2$ and $x=y^2+z^2$.

I'm honestly not even sure where to start.  I feel like the integral should be taken with respect to cylindrical coordinates.  But in that case, I'm not sure what the bounds on the angle should be, which makes me think this is the wrong approach.
All of the other two paraboloid problems seem to have an upward opening and downward opening paraboloid, rather than a vertically opening and horizontally opening paraboloid.
 A: Let's write your equations in the form 
\begin{align}
z &= x^2+y^2 \\
z &= \sqrt{x-y^2}.
\end{align}
If we graph these together, we see something like so:

The intersection between these two has the form
$$\sqrt{x-y^2}=x^2+y^2$$
or, after squaring both sides and transferring everything to one side,
$$x - y^2 - (x^2 + y^2)^2 = 0.$$
The curve you see in the $xy$-plane in the picture above is exactly this curve and it looks like a circle. Indeed, it's the circle $r=\cos(\theta)$ as you can verify by setting $x=\cos^2(\theta)$ and $y=\cos(\theta)\sin(\theta)$ into the above equation and simplifying with a few trig identities.
This all suggests that we evaluate the following polar integral:
$$
\int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}\int _0^{\cos
   (t)}\left(\sqrt{r \cos (t)-r^2 \sin ^2(t)}-r^2\right)
   r\,dr\,dt.
$$
Fortunately, you can use the symmetry of the figure to simplify the integral to 
$$
2 \int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}\int _0^{\cos
   (t)}\left(r \cos (t)-r^2\right) r\,dr\,dt.
$$
The answer is then easily computed to be $\pi/16$.
