Washer Method about a Vertical Line question The question I was given states, "Find the volume of the solid revolving the region bounded by the graphs of $x=0$, $y=x(x-1)$ and $y=0$ about $x=-1$."
I know that I need to take the integral in terms of $y$ because the solid is revolving around $x=-1$; however, I don't know how to solve for $y$ in the equation $y=x(x-1)$ to do that. 
Any help would be greatly appreciated :) thanks!
 A: I prefer to take a more systematic approach to these problems than to try to memorize such things the disk, shell, or washer methods. To that end, let's step back and look at the basic definitions. The area and centroid are given by
$$
A=\int\!\!\!\int dy~dx=\int y(x)~dx\\
R_x=\frac{\int\!\!\!\int x~dy~dx}{\int\!\!\!\int dy~dx}=\frac{1}{A}\int x~y(x)~dx\\
R_y=\frac{\int\!\!\!\int y~dy~dx}{\int\!\!\!\int dy~dx}=\frac{1}{2A}\int y^2(x)~dx\\
$$
And finally, Pappus's $2^{nd}$ Centroid Theorem states that the volume of a planar area of revolution is the product of the area $A$ and the length of the path traced by its centroid $R$, i.e., $V=2πRA$. In the present problem, we have an axis of rotation that is offset from the $y$-axis. In that case we can say that
$$V=2\pi (R_x-x_0) A=2\pi \left(-x_0 A+ \int_0^{x_{max}} x~y(x)~dx\right)$$
where $x_{max}$ is the point on positive $x$-axis where $y=0$, i.e., $x_{max}=1$ as seen in the figure below. Also note that $x_0=-1$.
The area and centroid can be readily computed (note that area will be negative; you should take the absolute value).
You should be able to take it from here. I found that $V=\pi/2$ which I verifed numerically.

