Q:

Find the volume of the region using cylindrical shells.

A region is bounded by the line $y=x+6$ and the parabola $y=x^2$ rotated about the x-axis. I had a similar problem rotated about the $x=3$ line but now this is rotated on the x-axis. I am having trouble identifying what the height of the cylindrical shell is. I believe that the radius is $y$, thickness is $dy$ and the limits of integration go from $0$ to $9$. I drew the graph out and drew a line parallel to the axis of rotation, but the height seems to go from $y=x^2$ to $y=x+6$ and it also goes from one end of $y=x^2$ to the other when $y < 4$. Did I draw this picture wrong since I cannot determine the height of the shell?

A:

The method of shells is non-optimal here, "washers" are much easier. And a picture, and being able to point to parts of it, would be very useful. 
We will be integrating with respect to $y$. Note that the geometry  from $x=-2$ to $x=2$, is different from the geometry for $x$ between $2$ and $3$. Or, to put it in terms of $y$, the situation from $y=0$ to $y=4$ is different than the situation from $y=4$ to $y=9$.
So we will have to break up the integral into two parts, $y=0$ to $4$ and $y=4$ to $9$, and add.
First Integral: We are going from $y=0$ to $y=4$. Maybe draw a horizontal line at height $y=4$. 
Draw a thin horizontal strip of width $dy$ at height $y$, where $0\le y\le 4$. The length of the strip is $2x$, which is $2\sqrt{y}$. The height (radius, distance from the $x$-axis) is $y$. Thus the volume of this chunk is 
$$\int_{y=0}^4 (2\pi)(2\sqrt{y})(y)\,dy.$$
Second integral: Now we are going $y=0$ to $y=9$. The radius is again $y$. But now the length of the strip is different.  It starts on the line $x=y-6$, and ends on $x=\sqrt{y}$, so has length $\sqrt{y}-(y-6)$. So this part of our volume is
$$\int_{y=4}^9 (2\pi)(y)\left(\sqrt{y}-(y-6)\right)\,dy.$$
Evaluate the two integrals and add. 

