Please check, whether my solution is correct.
Task: Find the volume of the solid bounded by $z=x^2 + y^2$, $x+z=0$
We need to calculate $\iint \limits_D - x - (x^2 + y^2 ) dx dy$, where $D = \{ (x,y): x^2 + y^2 = -x\} = \{(x,y): (x+\frac{1}{2})^2 + y^2 = (\frac{1}{2})^2 \}$.
Easiest solution: change coordinates to polar.
$\psi \in (\frac{\pi}{2}, \frac{3 \pi}{2})$
$ (x+\frac{1}{2})^2 + y^2 = (\frac{1}{2})^2$
$ (r \cos(\psi)+\frac{1}{2})^2 + (r \sin(\psi))^2 = (\frac{1}{2})^2$
$r(r + \cos(\psi)) = 0$, so
$r \in (0, -\cos(\psi))$
$\iint \limits_D - x - (x^2 + y^2 ) dx dy = \int \limits_{\frac{\pi}{2}}^{\frac{3\pi}{2}} \int \limits_0^{-cos \psi } r [-r \cos \psi - r^2 \cos^2 \psi - r^2 \sin^2 \psi ] dr d\psi= ...$