Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

"Use the method of cylindrical shells to find the volume V of the solid obtained by rotating the region bounded by the given curves about the x-axis."

I have two equations : x= 2y^(2)-y^(3) and x= 0. I am supposed to rotate about x axis. I'm stuck on what region I am looking at for rotation. I got x=0 being the y axis, and the other equation to look something like a sin(x) curve. The area rotate is not very intuitive for me.

share|cite|improve this question
Maybe with cylindrical shells you are accustomed to $y=f(x)$ rotate about $y$ axis. If it will make you feel more comfortable, rotate region bounded by $y=2x^2-x^3$ and the $y$-axis, about the $y$-axis. – André Nicolas Oct 11 '11 at 22:52
up vote 0 down vote accepted

Your cylinders lie along the $x$ axis. At a given $y$, the minimum $x$ is $0$ and the maximum is $2y^2-y^3$. The upper plot in this Alpha page shows it well (but note the axes are switched as André Nicolas suggests). You are rotating the region above the horizontal axis around the vertical axis. This would give $\int_0^2 (2y^2-y^3) 2\pi y\;dy$ where the height of the shell is $x=2y^2-y^3$, the radius is $y$, and the thickness of the shell is $dy$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.