Volume

**chaddy** · February 12th 2008, 12:59 PM

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

**Soroban** · February 12th 2008, 01:14 PM

Hello, chaddy!

What is the volume of the solid generated by revolving the region
enclosed by $(x-2)^2 + y^2 \:= \:1$ 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: . $V \;=\;2\pi\int^b_a\text{(radius)(height)}\,dx$

The radius is: $x$

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

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

Good luck!

**chaddy** · February 14th 2008, 06:56 PM

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

**Krizalid** · February 14th 2008, 07:04 PM

Originally Posted by Soroban

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

Substitute $u=x-2,$

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

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

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