Suppose the body $D$ is bounded by the $xy-$plane, and the surfaces $z=2y\ $ and $y=4-x^2$.
On a side note, I think the second surface is called a parabolic cylinder, is that right?
What is the volume $V$ of this body?
I calculated it and I got $V = 512/15$
Is that correct?
Let $S_1:=\{(x,y,z)\in \mathbb{R}^3:z=0\}$, $S_2:=\{(x,y,z)\in \mathbb{R}^3:z=2y\}$ and $S_3:=\{(x,y,z)\in \mathbb{R}^3:y=4-x^2\}$. Now define $\tilde S_k$ the compact and convex region of $S_k$ bounded by the other surfaces for $k\in\{1,2,3\}$, then
$$ \tilde S_1=\{(x,y,0)\in \mathbb{R}^3:|x|\leqslant 2\,\land\, 0\leqslant y\leqslant 4-x^2\}\\ \tilde S_2=\{(x,y,2y)\in \mathbb{R}^3:|x|\leqslant 2\,\land\, 0\leqslant y\leqslant 4-x^2\}\\ \tilde S_3=\{(x,4-x^2,z)\in \mathbb{R}^3:|x|\leqslant 2\,\land\, 0\leqslant z\leqslant 8-2x^2\} $$ Thus the region is $$ R=\{(x,y,z)\in \mathbb{R}^3:|x|\leqslant 2\,\land\, 0\leqslant y\leqslant 4-x^2\,\land\, 0\leqslant z\leqslant 2y\} $$ Therefore $$ \int_{R}dxdydz=\int_{-2}^2\int_{0}^{4-x^2}\int_{0}^{2y}dzdydx=\int_{-2}^2\int_{0}^{4-x^2}2ydydx\\ =\int_{-2}^2(4-x^2)^2 dx=\frac{512}{15}. $$