Triple Integration in vector calc NIn the following example I have been asked to find the volume V of the solid bounded by the sphere $ x^2 + y^2 +z^2 = 2 $ and the paraboloid $ x^2 + y^2 = z $ by using triple integration.
I am not quite sure how to set up this triple integration and what each of the integrands should be.
 A: Note $z=x^2+y^2>0$, therefor 
$$V=\int_{-\sqrt{2}}^{\sqrt{2}}\int_{-\sqrt{2-x^2}}^{\sqrt{2-x^2}}\int_{x^2+y^2}^{\sqrt{2-x^2-y^2}}dzdydx$$
Now apply cylinder coordinate.
A: It is comfortable to describe your solid in cylindrical coordinates since both the sphere and the paraboloid have the $z$-axis as an axis of symmetry (both surfaces are surfaces of revolution around the $z$-axis). 
In cylindrical coordinates, the sphere is described by $\rho^2 + z^2 = 2$ and the paraboloid is described by $\rho^2 = z$. The solid bounded between them is described by $\rho^2 \leq z \leq \sqrt{2 - \rho^2}$ and so
$$ V = \int_{0}^1 \int_{0}^{2\pi} \int_{\rho^2}^{\sqrt{2 - \rho^2}} \rho \, dz \, d\theta \, d\rho = 2\pi \int_0^{1} \rho \left( \sqrt{2 - \rho^2} - \rho^2 \right) \, d\rho = 
2\pi \left( \frac{1}{2} \int_1^2 \sqrt{u} \, du - \int_0^{1} \rho^3 \right)
=  2\pi \left( \frac{2^{\frac{3}{2}} - 1}{3} - \frac{1}{4}\right).$$
where we used the substitution $u = 2 - \rho^2, \, du = -2\rho$ for the first integral.
