Finding Volume of the Solid--washer method The region between the graphs of 
$ y=x^2$  and $y=3x$ is rotated around the line $x=3$
What is the volume of the resulting solid? 
I drew the picture, and I saw that I should be using the washer method. Since the region was being flipped at $x = 3$, I would have to make sure I am subtracting the two functions from 3 and then subtract those two, but I could not seem to get the right answer.
 A: Using washers, the interval of integration will be along the $y$-axis from $y = 0$ to $y = 9$.  The inner radius as a function of the $y$-value will be $r(y) = 3 - \sqrt{y}$, and the outer radius will be $R(y) = 3 - y/3$; thus the differential volume of a washer with thickness $dy$ is $$dV = \pi(R(y)^2 - r(y)^2) \, dy = \pi((3 - y/3)^2 - (3-\sqrt{y})^2) \, dy,$$ and the total volume is given by the integral $$V = \pi \int_{y=0}^9 (3 - y/3)^2 - (3-\sqrt{y})^2 \, dy.$$
Using cylindrical shells, the interval of integration will be along the $x$-axis from $x = 0$ to $x = 3$.  The height of the shell is given by the difference of the functions $y = x^2$ and $y = 3x$, so $h(x) = 3x - x^2$ (since $3x \ge x^2$ on $x \in [0,3]$).  The circumference of a representative shell is simply $2\pi(3-x)$, thus its differential volume is $$dV = 2\pi(3-x)(3x-x^2) \, dx,$$ and the total volume is $$V = \int_{x=0}^3 2\pi(3-x)(3x-x^2) \, dx.$$  Both calculations give the same result.
A: Volume by washers can be used in this case if you stack your washers parallel to the $x$-axis, centered at $x=3$.  Identify the outer radius $R$ (as a function of $y$, since every washer is determined by how high it's sitting in the stack) and the inner radius (again, as a function of $y$); the volume of each thin washer is $\pi(R^2 - r^2)\,\Delta y$ and the total volume is $\pi\int_0^9(R^2 - r^2)\,dy$.
