Volume between cylinder and plane Problem: Find the volume bounded by $z = y^2, x =0, y =0, z =9-x$.
My working:
$z$ goes from $y^2$ to $9-x$ so these are the limits of integration.
Work out the points of intersection of $9-x$ and $y^2$. When $y=0$, $9-x=0$ and $x=9$. So $x$ goes from 0 to 9. When $x=0$, $y^2 = 9$ so $y=3$ (take the positive one). So $y$ goes from 0 to 9.
Then evaluate 
\begin{align}
\int_{x=0}^{x=9} \int_{y=0}^{y=9} \int_{z=y^2}^{z=9-x} dz dy dz &= \int_{x=0}^{x=9} \int_{y=0}^{y=9} y^2 - 9 + x dy dx 
\\
&= \int_{x=0}^{x=9} 18+3x dx
\\
&= \frac{567}{2}
\end{align}
My textbook says the answer is $\frac{324}{5}$. What have I done wrong?

Alternately the set up is:
$\displaystyle \int_{0}^3 \int_{0}^{9-y^2} \int_{y^2}^{9-x} 1dzdxdy = \dfrac{324}{5}$ (already checked)
