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?

Your limits of integration don't make sense.  The region of integration is given by the set $$R = \{(x,y,z) \in \mathbb R^3 \mid (y^2 \le z \le 9-x) \cap (0 \le x \le 9) \cap (0 \le y \le 3)\}.$$  The projection of $R$ onto the $xz$-plane is simply the triangle $x \ge 0$, $z \ge 0$, $x + z \le 9$.  Over this triangle, the curve $y = \sqrt{z}$ is the boundary, so the integral is $$\int_{x=0}^9 \int_{z=0}^{9-x} \int_{y=0}^\sqrt{z} 1 \, dy \, dz \, dx = \frac{324}{5}.$$  You may also project $R$ onto the $yz$-plane, in which case we would have $$\int_{y=0}^3 \int_{z=y^2}^9 \int_{x=0}^{9-z} 1 \, dx \, dz \, dy = \frac{324}{5}.$$  Projecting onto the $xy$-plane is trickier, but yields the integral $$\int_{y=0}^3 \int_{x=0}^{9-y^2} \int_{z=y^2}^{9-x} 1 \, dz \, dx \, dy = \frac{324}{5}.$$
