Volumes of Solids of Revolution/Shell Method?

[rgmahoney]

Hey folks, I've been having trouble with this problem and I was wondering if you'd help me out a little.

The problem is:

x = 1 + (y-2)^2 with a boundary at x = 2. Rotated about the x-axis.

I drew the graph but I'm not sure how to draw a cylindrical shell with it.

 

[HallsofIvy]

A cylindrical shell will have radius from the given y value to the x-axis, so r= y. It's "height", as you should be able to see from the left boundary, x= 1+ (y- 2)^2 to the right boundary, x= 2. The length, and so the height of the cylinder, is h= 2- (1- (y-2)^2= (y- 1)^2+ 1. Integrate \(\displaystyle \pi r^2h= \pi(y^2)((y-1)^2+ 1)dy\) from y= 1 to y= 4.

 

[rgmahoney]

I greatly appreciate the answer, but I just have one question.

How exactly is the height in the final integral "( y−1)^2+1". Should it not be "(y-2)^2+1"?