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

I have $$y=|x|+1, 2x+3y=6\text{ about }x=-20.$$

I need to set up an integral (but not evaluate it) that will give the volume of the solid when the region enclosed by those curves is revolved about the given line. I have to use either the shell or washer's method.

I have been trying to use the shell method (I think) but I am having trouble setting up the integral. How do I figure out the bounds for it? I'm not sure how I get the numbers for "top" and "bottom" either. I solved for y in the $2x+3y=6$ equation and got $y=-(2/3)x+2$ but how do I know where that belongs in the integral?

Thank you for any help

share|cite|improve this question
up vote 1 down vote accepted

Hint: If you plot the two curves, they form a triangle. This is the area that you are revolving, so find the coordinates of the three corners of the triangle. If you use cylindrical shells, the minimum $y$ value is the start of your integration and the maximum $y$ is the end. Then for a given $y$ in that range, you need to find the minimum $x$ and maximum $x$. The minimum will always be along $y=1-x$ (why?) but the maximum will shift from one line to the other part way along, so you probably want to split the integral there.

share|cite|improve this answer
Thanks. I guess it's a good idea to always plot them out first to see what the figure will look like? – Ryan May 20 '11 at 19:14
@Ryan: it often helps. Especially with the absolute value there, one can get confused easily. Did you get the same corners as Christian Blatter? – Ross Millikan May 20 '11 at 21:49

Whatever the suggested methods, the simplest way to find this volume $V$ is to use Guldin's rule: $V$ is equal to the area $A$ of the given triangle $\Delta$ times the circumference of the circle described by the centre of gravity $S$ of $\Delta$. The vertices of $\Delta$ being $(0,1)$, $({3\over5}, {8\over5})$ and $(-3,4)$ we get $A={9\over5}$ and $S=(-{4\over5},{11\over5})$. This implies $$V= {9\over5} \cdot 2\pi (-{4\over5}+20) ={1728\over25}\ \pi\ .$$

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.