# Find the volume of the solid revolved around y-5

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 $5$.

$\int_{0}^5((5-x)^2−(3)^2)dx$

• Your limits should be from 0 to 2, and you want a factor of $\pi$ in the integral. – 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 $(2,2)$.
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).
• 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 $2$? – Csci319 Jun 20 '15 at 16:25