Finding the volume bounded by surface $y^2=4ax$ and the planes $x+z=a$ and $z=0$ The problem is stated below:

Let $V$ be volume bounded by surface $y^2=4ax$ and the planes $x+z=a$ and $z=0$.
Express $V$ as a multiple integral, whose limits should be clearly stated. Hence calculate $V$.

Progress
I want to find out the limits of the multiple integral that's needed to calculate $V$.
I'm guessing that:
$x=a-z$, $x=(y^2)/4a$
$$y= \pm \sqrt{4ax},\quad z=0,\quad z=a-x $$
but it seems like I've used the upper plane equation twice?
Also, it would really help if we could compare our answer volumes to check that this is right from the start!
Thanks everyone!!! :)
 A: Sometimes a picture is worth a thousand words. Eventually, a math student should try plot graphs in their minds with out the use of software (although, I've always preferred Play Doh  because after solving the problem I could always make little fishies and green froggies).
Geometrically, Your problem should look like this...

We see that $x=a$ when the upper plane intersect the $x-y$ plane. We also see that when $x=t$
$$y^2=4at$$
$$y=\pm2\sqrt{at}$$
So the integral becomes
$$\begin{array}{lll}
\int^a_0\int_{-2\sqrt{ax}}^{2\sqrt{ax}}(a-x)dydx&=&\int^a_0(a-x)(4\sqrt{ax})dx\\
&=&\int^a_0(4a^\frac{3}{2}x^\frac{1}{2} - 4a^\frac{1}{2}x^\frac{3}{2})dx\\
&=&(\frac{2}{3}\cdot 4a^\frac{3}{2}a^\frac{3}{2} - \frac{2}{5}\cdot 4a^\frac{1}{2}a^\frac{5}{2})\\
&=&8a^3(\frac{1}{3}-\frac{1}{5})\\
&=&8a^3(\frac{2}{15})\\
&=&\frac{16}{15}a^3\\
\end{array}$$
A: OK so first of all we must turn "bounded by" into inequalities. If you picture the three surfaces, you realize the only bounded region is:
$$\left\lbrace y^2\leq 4ax,z\geq0,x+z\leq a\right\rbrace=\left\lbrace y\in[-\sqrt{4ax},\sqrt{4ax}],x,z\geq0,x\leq a-z\right\rbrace.$$
The limits for $y$ are explicit and depend on $x$, so we put the $dy$ integral inside. With that, $x,z$ don't depend on $y$, so we will have one from 0 to $a$, and the other in such a way that the sum is less than $a$. I wrote the equation in terms of $x$ to suggest my approach, but swapping the integrals shouldn't give any change.
How do we see that is the region? Well we have two planes and a 'parabolc prism'. The region outside the prism will be unbounded, which for example gives $x\geq0$ and the $y$ inequality. Under the $z=0$ plane, the region can go to infinity in $x$, since there is no upper bound for it from any of the surfaces. So $z\geq0$. The other bit, $x\leq a-z$, is just the bounding of the plane, which has to be an upper bound since the other plane is a lower one and the 'prism' doesn't touch $z$ in any way.
Hope I've been clear.
