My book doesn't have the answer. Can anyone help me with volumes of solids by slicing? How do I find the volume in the first octant bounded by the surfaces $x^2=y+2z$ and $x=2$ use the slice in the figure to compute the volume? 

 A: While you can do it with triple integrals, the slice given hinted at a much simpler answer:
Note that $x^2=y+2z$ has only linear terms in $y$ and $z$, so the constant-$x$ slice in the picture is a triangle with base along $z$-direction of length $x^2/2$ and height along $y$-direction of length $x^2$.  So integrating the slices,
$$
\operatorname{Volume}=\int_0^2 \frac12\cdot\frac12x^2\cdot x^2\,\mathrm{d}x
$$
which you should know how to do.
A: The variable $x$ varies between $0$ and $2$. The variable $y$ varies from $0$ to $x^2$, and $z$ varies from $0$ to $\dfrac{x^2-y}{2}$.
You want 
$$V=\iiint_D 1\, dV = \int_0^2 \int_0^{x^2} \int_0^{\frac{x^2-y}{2}} 1 \, dz\,dy\,dx$$
Note this is only one of six different ways to set up the integral. For example:
$$V=\int_0^4 \int_{\sqrt{y}}^2 \int_0^{\frac{x^2-y}{2}} 1 \, dz\,dx\,dy$$
or
$$V=\int_0^2 \int_{\sqrt{2z}}^2 \int_0^{x^2-2z} 1 \, dy\,dx\,dz$$
A: Look at the diagram carefully. Why make it complicated?
$$ dV = x.y. dz = z^2.z.dz = z^3 dz $$
can you take it further?
A: Try the following set:

$$\{(x,y,z)\in\mathbb{R}^3; 0\leq x\leq2,\ 0\leq z\leq \frac{x^2}{2}\, 0\leq y\leq 2z-x^2\}$$

