Let V be the bound region between the surfaces $x^2 = 2z, x^2 +y^2 =8z, z=6$ ($V$ is the inner part of $x^2 =z$) Find the volume of the region V. I can't seem to set the bounds for the integral. so far I've found that (1) and (2) intersect along the line $\pm \sqrt{3}x$ and solved the integral
\begin{gather*} 2 \cdot \int_{0}^{\sqrt{12}} \int_{-\sqrt{3}x}^{\sqrt{3}x} (6-\frac{x^2}{2}) dydx \end{gather*} but the answer I've got is wrong/incomplete.
From $x^2\le 2z$ and $z\le 6$ you get $0\le z \le 6$. Now fix $z\in [0,6]$ and consider the domain $D_z$ defined as $D_{z}:=\{x^2\le 2z, x^2+y^2\le 8z\}$; the volume is then $$V=\int_0^6 \iint_{D_z} \mathrm dy \mathrm dx \mathrm dz.$$ To find the bounds of $x$ and $y$ you can draw a picture in $\Bbb R^2$. From $x^2\le 2z$ you find $-\sqrt {2z}\le x \le \sqrt{2z}$; for $y$ using the other inequality you get $-\sqrt{8z-x^2}\le y \le\sqrt{8z-x^2}$.