The region R enclosed by the curves y=5x−6 and y=−x2+12x−16 is rotated about the x-axis. Find the volume of the resulting solid.

Thanks!~

2. Originally Posted by shannon1111
The region R enclosed by the curves y=5x−6 and y=−x2+12x−16 is rotated about the x-axis. Find the volume of the resulting solid.

Thanks!~
Those graphs cross at (2, 4) and (5, 19). Each cross section of the figure is a "washer" whose outer diameter is $\displaystyle -x^2+ 12x- 16$ and the inner diameter is $\displaystyle 5x- 6$.

The area of each such "washer" is the area of the outer circle minus the area of the inner circle, and each of those is $\displaystyle \pi r^2$: $\displaystyle \pi((-x^2+ 12x- 16)^2- (5x-6)^2)$. Taking the thickness of each cross section to be "dx", each cross section has volume $\displaystyle \pi((-x^2+ 12x- 16)^2- (5x-6)^2)dx$ and "adding" (integrating) those cross sections gives the entire volume:
$\displaystyle \pi\int_2^5 ((-x^2+ 12x- 16)^2- (5x-6)^2)dx$
The simplest way to integrate that is to multiply it all out before integrating.