Volume of a solid generated by revolving a region about vertical line

Question

Given the region bounded by these functions: $y = x, y = 0, y = 7$; how do I find the volume of that region revolving around the vertical line of $x = 8$? I've had trouble finding the answer to this specific question and a few other similar ones, my biggest issue is setting up the integral, my first thought was this: $\pi\int_0^7[(8-x)^2-1^2]dx =\cdots {490\pi\over3}$, but it was not correct. Please help!

Answer 1

Using the disk/washer method, the differential volume, as a function of $y$ is

$ dV = \pi ( 8^2 - (8 - y)^2) dy = \pi ( 16 y - y^2 ) dy $

And you integrate this between $ y = 0 $ and $ y = 7 $, this will a volume of

$ V = \dfrac{833 \pi}{3} $

Using the shell method, the differential volume is

$ dV = 2 \pi (8 - x) (7 - x) dx $

And you integrate this between $ x= 0 $ and $ x= 7 $

$ V = 2 \pi \displaystyle \int_{0}^{7} (56 - 15 x + x^2) dx = \dfrac{833\pi}{3}$