The region enclosed between the curves $y=x^2−2x+3$ and $y=x+1$ is now rotated about the $x$-axis to form a solid of revolution. Find its volume.
I have tried integration but i am unsure what area the volume would be bounded by, is it $a=1$ and $b=2$ for the integral or do i need to do two separate integrals for the different graphs. My integral so far is
$\displaystyle V= \int_a^b \pi(x+1)^2 - \pi(x^2-2x+3)^2 \, dx$ - Is this correct for the integral?