1. ## volume of revolution

Find the volume of revolution when the region bounded by the curve y=x^2+2 and y=6 is rotated about the y-axis.

I am confused when the region rotated is at both sides of the axis in which the region is rotated upon. If rotate one part of the region, the volume is
pi* integrate from 2 to 6 (y-2) dy.

Is it the same in this case, when the rotated region is at both sides? It seems when the region is rotated 2pi about the y-axis, it covers double the volume of rotating only one part of the region but if overlapping is to be considered, shouldn't both cases generate the same result?

2. Yes you only need to worry about one half of the region being rotated.

By the way, this is clearly a Calculus question - why have you posted in Pre-Calculus?

3. Originally Posted by Prove It
Yes you only need to worry about one half of the region being rotated.
Thanks, so does it mean rotating the shaded region 360 degrees in both the attachments about the y-axis generate the same volume?

4. Yes, but the second would cover the region twice- and you don't want that.

5. Originally Posted by HallsofIvy
Yes, but the second would cover the region twice- and you don't want that.
Thanks, so volume of revolution simply means the effective volume of the region covered, in which the overlapped volume is not taken into consideration?

Would the volume be: $\displaystyle \int^{6}_{2}(y-2) dy = 8$ ?

6. Volume is volume is volume. There is no "overlapping volume".

You forgot "$\displaystyle \pi$".