1. ## Volume

What is the volume of the solid generated by revolving the region enclosed by (x-2)^2 + y^2 = 1 about the y-axis?

What is the volume of the solid generated by revolving the region
enclosed by $\displaystyle (x-2)^2 + y^2 \:= \:1$ about the y-axis?

Make a sketch. We have a circle with center (2,0) and radius 1
. . revolved about the y-axis.
This results in a torus (donut).

I recommend using Shells: .$\displaystyle V \;=\;2\pi\int^b_a\text{(radius)(height)}\,dx$

The radius is: $\displaystyle x$

The height is: .$\displaystyle 2y \;=\;2\sqrt{1-(x-2)^2}$

We must evaluate: .$\displaystyle V \;=\;4\pi\int^3_1 x\sqrt{1-(x-2)^2}\,dx$

Good luck!

3. I worked that integral out and wound up getting zero. That doesn't seem right. Is it?

4. Originally Posted by Soroban
We must evaluate: .$\displaystyle V \;=\;4\pi\int^3_1 x\sqrt{1-(x-2)^2}\,dx$
Substitute $\displaystyle u=x-2,$

$\displaystyle \int_1^3 {x\sqrt {1 - (x - 2)^2 } \,dx} = \int_{ - 1}^1 {(u + 2)\sqrt {1 - u^2 } \,du} .$

We can kill this quickly: $\displaystyle \int_{ - 1}^1 {u\sqrt {1 - u^2 } \,du} = 0$ by symmetry and $\displaystyle \int_{ - 1}^1 {\sqrt {1 - u^2 } \,du}$ which is a half of circle with radius 1, hence area $\displaystyle \frac\pi2.$ Put these things together yields that $\displaystyle \mathcal V=4\pi^2.$