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Thread: Solid of Revolution about x = 4

  1. #1
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    Solid of Revolution about x = 4

    I have the following question:
    Determine the volume of the solid when the region bounded by $\displaystyle y = \sqrt{x}$ and the lines $\displaystyle y = 2$ and $\displaystyle x = 0$ is rotated about the line $\displaystyle x = 4$

    I went about it as follows:

    $\displaystyle x = y^2$ therefore...
    $\displaystyle \int_0^2 \pi R(y)^2dy - \int_0^2 \pi r(y)^2dy$

    $\displaystyle \int_0^2 \pi4^2 dy - \int_0^2 \pi(4 - y^2)^2 dy$

    $\displaystyle \pi[16y]_0^2 - \pi[16y - \frac{8}{3}y^3 + \frac{1}{5}y^5]_0^2$

    giving me a final answer of $\displaystyle \frac{224\pi}{15}$

    Have I done anything wrong?
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  2. #2
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    Quote Originally Posted by blackhug View Post
    I have the following question:
    Determine the volume of the solid when the region bounded by $\displaystyle y = \sqrt{x}$ and the lines $\displaystyle y = 2$ and $\displaystyle x = 0$ is rotated about the line $\displaystyle x = 4$

    I went about it as follows:

    $\displaystyle x = y^2$ therefore...

    $\displaystyle \int_0^2 \pi R(y)^2dy - \int_0^2 \pi r(y)^2dy$

    $\displaystyle \int_0^2 \pi4^2 dy - \int_0^2 \pi(4 - y^2)^2 dy$

    $\displaystyle \pi[16y]_0^2 - \pi[16y - \frac{8}{3}y^3 + \frac{1}{5}y^5]_0^2$

    giving me a final answer of $\displaystyle \frac{224\pi}{15}$

    Have I done anything wrong?
    You are correctly finding the surface area of washers and integrating along the y-axis.
    Your first line doesn't really require the y in brackets part.
    You just need to know the two radii, which you show you do.
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  3. #3
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    Seems good to me.
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  4. #4
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    Awesome thanks very much for the quick responses. Wasn't sure if I had gone about this the right way.
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