Use triple integrals to find the volume of... The solid enclosed by the parabaloid $$x=y^2+z^2$$ and the plane $$x=6$$
I wanted to make sure that I'm setting up the correct integral before I start to integrate it. $$\int_{-\sqrt{6}}^{\sqrt{6}}\int_{-\sqrt{6-y^2}}^{\sqrt{6-y^2}}\int_{y^2+z^2}^6dxdydz$$
If this is incorrect, would someone be able to help guide me in the right direction?
 A: In any case, you can always validate your result by using the formula for a solid of revolution, which in your case gives:
$$V = \pi \int f(x)^2 dx = \pi \int_0^6 \sqrt{x}^2 dx = 18\pi$$
A: 
I wanted to make sure that I'm setting up the correct integral before I start to integrate it. 
  $$\int_{-\sqrt{6}}^{\sqrt{6}}\int_{-\sqrt{6-y^2}}^{\sqrt{6-y^2}}\int_{y^2+z^2}^6dxdydz$$  

The integration looks almost correct, except, the order of the integrated variable should be
$$\int_{-\sqrt{6}}^{\sqrt{6}}\int_{-\sqrt{6-y^2}}^{\sqrt{6-y^2}}\int_{y^2+z^2}^6 dx \color{red}{dz dy},$$
or 
$$\int_{-\sqrt{6}}^{\sqrt{6}}\int_{\color{red}{-\sqrt{6-z^2}}}^{\color{red}{\sqrt{6-z^2}}}\int_{y^2+z^2}^6dxdydz.$$  

This is before we learned about cylindrical coordinates, so I'm not allowed to use them yet unfortunately.

Of course the integral can be computed w/o any help from any spherization of the coordinate system. 
$$
V = \int_{-\sqrt{6}}^{\sqrt{6}}\int_{-\sqrt{6-z^2}}^{\sqrt{6-z^2}} (6-y^2-z^2) dydz
\\
= \int_{-\sqrt{6}}^{\sqrt{6}}\frac{4}{3}(6-z^2)\sqrt{6-z^2}dz
\\
= \frac{8}{3}\int_{0}^{\sqrt{6}}(6-z^2)\sqrt{6-z^2}dz,
$$
which will yields the same answer with nbubis's answer if you use the trigonometric substitution you learned from Cal II: let $z = 6\sin \theta$ for $0\leq \theta\leq \pi/2$.
