Set up: Volume by Integration Having difficulty setting up the equation.
Find Volume bound by 
$$y = x; y = 0; x = 2; x = 4$$
and rotated about $x = 1$
The issue I'm having is that I get a final answer of $\frac{16\pi}{3}$ ... which clearly is incorrect.
 A: Draw a picture. We use cylindrical shells. Take a vertical strip of width "$dx$" going from $x$ to $x+dx$, and rotate about $x=1$. We get a cylindrical shell of height $y$ and radius $x-1$, so of approximate volume $2\pi y(x-1)\,dx$. "Add up" (integrate) from $x=2$ to $x=4$. Since $y=x$, the volume is
$$\int_2^4 2\pi x(x-1)\,dx.$$
A: Well I'm glad to see you realize that a negative volume doesn't quite make sense.
With this said let's try to analyze this problem. Note that we have
$$
V = \int_0^4 A(y) \, \mathrm{d} y
$$
where $A(y)$ stands for the area of a disk at a given $y$. We have that the outer radius of rotation will be
$$
r_{out}(y) = 3
$$
and then the inner will be
$$
r_{in}(y) = \begin{cases} 1 & 0 \le y \le 2 \\ y - 1 & 2 < y \le 4 \end{cases}
$$
so finally our area function will be
$$
A(y) = \pi \left( r_{out}^2(y) - r_{in}^2 (y) \right)
$$
and now we can integrate:
$$
\begin{eqnarray}
V & = & \pi \int_0^4 r_{out}^2(y) - r_{in}^2 (y) \, \mathrm{d} y \\
& = & \pi \left( \int_0^2 3^2 - 1^2 \, \mathrm{d} y + \int_2^4 3^2 - (y-1)^2 \, \mathrm{d} y\right) \\
& = & \pi \left( 16 + \frac{28}{3} \right) = \frac{76}{3} \pi
\end{eqnarray}
$$
EDIT: Sorry, forgot to square my radius at first, and the second time I didn't square it properly >.<
Note the following plot shows the region which we want to rotate. You can more easily see then that the disk created by rotating around $x=1$ will have an area of $3^2 \pi - \pi$ when $y \in [0,2]$ and will have an area of $3^2\pi - (y-1)^2 \pi$ when $y \in [2,4]$. Finally to find the volume we integrate over all the possible $y$'s (from $0 \to 4$) in order to add up all the areas from the disks to get the volume of the region.

