I frequent Stack Overflow, but I am new here. I was given this problem:
Use the method of cylindrical shells to find the volume $V$ generated by rotating the region bounded by the given curves about the specified axis. $x = 2y^2$, $y \geq 0$, $x = 2$; about $y = 2$
My setup for the problem is $2\pi \displaystyle\int_0^2(2-y)(2y^2) dy$. I derived this from $V=2\pi rh$ where $r = 2-y$ and $h=x=2y^2$. I've solved this to get $\frac{16\pi}{3}$ but that answer seems to be incorrect.
Disclaimer: While I appreciate answers, I don't just want "The answer is: ...." I want to know WHY my setup is incorrect. Also, if someone could format this for me that would be very helpful, as I am unfamiliar with MathJax. Thank you!
You almost got it. The correct setup is
$$2\pi \int_0^{\color{red}1}(2-y)\underbrace{(2-2y^2)}_{\text{"height"}~h} dy$$
The "height" of the cylinder is from the parabolic boundary $x = 2y^2$ on the "left" to the $x=2$ on the "right". Note the upper integration limit is $1$, again due to $x = 2y^2$ meeting $x=2$ giving $x = 2 = 2y^2 \implies y = 1$ (and also $y$ is restricted to be positive).