Find the volume between two paraboloids Find the volume of the solid enclosed by the paraboloids $z = 1-x^2-y^2$ and $z = -1 + (x-1)^2 + y^2$.
Using triple integrals, it is known that $V = \iiint_R \mathrm dx\,\mathrm dy\,\mathrm dz$, and I will have to change variables. But I can't just say that $r^2 = x^2 + y^2$, because the second paraboloid has a "$(x-1)^2$" term. How do I change variables without making a mess? 

Since $z=1-x^2-y^2$  is a downward paraboloids and $z=-1+(x-1)^2+y^2$     is upward , the limits for $z$ are:
$$
-1+(x-1)^2+y^2 \le z \le 1-x^2-y^2 
$$
The projection on the $xy$ plane of the curve of intersection between the two paraboloids is the circumference:
$$
x^2+y^2-x-\frac{1}{2}=0
$$
so the limits for $y$ are:
$$-\sqrt{\frac{1}{2}+x-x^2} \le y \le \sqrt{\frac{1}{2}+x-x^2}
$$
that has real solution if $x$ in the limits
$$
\frac{1-\sqrt{3}}{2}\le x \le \frac{1+\sqrt{3}}{2}
$$
so the volume is given by:
$$
\int_{\frac{1-\sqrt{3}}{2}}^{\frac{1+\sqrt{3}}{2}} \int_{-\sqrt{\frac{1}{2}+x-x^2}}^{\sqrt{\frac{1}{2}+x-x^2}}  \int_{[-1+(x-1)^2+y^2]}^{[1-x^2-y^2]} dzdydx
$$
Since the problem has not an axis of symmetry, it is unlikely that the use of cylindrical coordinates gives a simpler integration.
