Find the volume of the solid generated by revolving the region enclosed by the curve ${y=4-x^2}$ and the line ${y=2-x}$ about the ${x}$-axis.
I got the answer using the disk/washer method, but I wanted to try solving using the cylindrical method as well. However, I did not manage to obtain the correct volume.
The integral I obtained was this:
$2{\pi}\int_{0}^3{y(\sqrt{\mathstrut4-y}-(2-y))} dy$
Can someone please point out where I went wrong?
If you'd sketch the enclosed, your integral only covers the lower part of the region. What about the part from $y=3$ to $y=4$? In this part, note that the "right function" is the right part of your parabola, and the "left function" is the left part of your parabola.
Hope this helps.