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Math Help - Volume

  1. #1
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    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?
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  2. #2
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    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!

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  3. #3
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    I worked that integral out and wound up getting zero. That doesn't seem right. Is it?
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  4. #4
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    Quote Originally Posted by Soroban View Post
    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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