I've been reviewing the shell method for my Calculus II final, but I suppose I need a little refresher.
For example, I am given the equations $y=2x+15$ and $y=x^2$ and told to revolve the enclosed region around the line $x=5$. To find the limits of integration, I set the first two equations equal to each other and solved for $x$, in which I got $x=-3, x=5$.
Then, I determined that the shell radius would be simply $x$, and the shell height would be $2x+15-x^2$. Finally, I set up the integral using all of this information as follows:
$$\int_{-3}^5 x(2x+15-x^2) = \frac{2048\pi}{12}$$
However, the answer is apparently $\frac{2048\pi}{3}$. I feel like I might have used the incorrect shell radius, but I'm not sure. Any suggestions?
Best Answer
You're right; your shell radius is incorrect. For instance, when $x = 5$, the radius of your shell should be $r = 0$. When $x = 2$, the radius of your shell should be $r = 3$. In general, the radius is $r = 5 - x$. So we find that the volume is: $$ 2\pi\int_{-3}^5 (5 - x)(2x + 15 - x^2) \, dx = \frac{2048\pi}{3} $$ as desired.