I am trying to setup this integral but I am having trouble figuring out the bounds.

Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the line $y = 5$.

$y = x$

$y = 2$

$x = 0$

I thought you would subtract $5-x$ and $5-2$ and the lower bound would be zero while upper bound is $5$.


  • 2
    $\begingroup$ Your limits should be from 0 to 2, and you want a factor of $\pi$ in the integral. $\endgroup$ – user84413 Jun 20 '15 at 0:25

To find out the bounds you want to find the points of intersection by which the lines you are given encloses the region. So you are given the lines $y=x, y=2, x=0$, graphically, we can see that the region is enclosed by the points $(0,0),(0,2)$ and $(2,2)$.

enter image description here

After finding the enclosed region, we can then revolve the region around $y=5$, and depending on which method for finding the volume you use, we integrate with respect to $x$ or $y$. (In this case, the bounds are unchanged regardless of whether you integrate with respect to $x$ or to $y$ since the solid lies in between $0$ to $2$ for either $x$ or $y$, but note that this is not always the case).

  • $\begingroup$ So would my integral look like this? $\int_{0}^2pi((5-x)^2−(3)^2)dx$ because the points of intersection are at $0$ and $2$? $\endgroup$ – Csci319 Jun 20 '15 at 16:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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