The volume of the solid from the region bounded by $x=9-y^2$, $y=x-7$, $x=0$ about $y=3$ using cylindrical shells. The volume of the solid from the region bounded by $x=9-y^2$, $y=x-7$, $x=0$ about $y=3$ using cylindrical shells.
I've tried creating two separate regions: 
$V_1=2\pi(3-y)(9-y^2)dy$ from 3 to 1
and
$V_2=2\pi(3-y)(y+7)dy$ from 1 to 0
but the answer this gives isn't correct. I can't seem to find any errors in my calculations. Am I devising the solution incorrectly?

Here again is the graph of the things mentioned in the problem:
$\color{blue}{x=9-y^2}$, $\color{red}{x=y+7}$, $\color{green}{x=0}$, $y=3$ (dashed)

You're correct that the circumference of the shell at $y$ is given by $2\pi(3-y)$, but you should be integrating over all of $-3\leq y\leq 3$, with the height of the shell at $y$ given by
$$\begin{cases} 9-y^2&\text{ if }1\leq y\leq 3,\\ y+7 & \text{ if }-2\leq y\leq 1,\\
9-y^2 & \text{ if }-3\leq y\leq -2.\end{cases}$$
Thus, the volume is
$$V=\int_1^3 (3-y)(9-y^2)\,dy+\int_{-2}^1(3-y)(y+7)\,dy+\int_{-3}^2(3-y)(9-y^2)\,dy$$
which you can check gives the right answer.
