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Math Help - Volume of a solid generated by rotating a region about the y-axis

  1. #1
    PTL
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    Volume of a solid generated by rotating a region about the y-axis

    "Sketch the region bounded by the curves y = x^ {\frac{3}{2}}, y = 8, x = 0, and find the volume of the solid generated by revolving the region about the y-axis."

    I've sketched the region, and can visualise what they're looking for with the rotation thing, but I don't know how one actually calculates it...

    I'd suspect integration is needed, but I'm not sure exactly what I'm supposed to be integrating...
    Last edited by PTL; August 1st 2009 at 05:47 PM. Reason: edit: typo
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    Quote Originally Posted by PTL View Post
    "Sketch the region bounded by the curves y = x^ {3}{2}, y = 8, x = 0, and find the volume of the solid generated by revolving the region about the y-axis."

    I've sketched the region, and can visualise what they're looking for with the rotation thing, but I don't know how one actually calculates it...

    I'd suspect integration is needed, but I'm not sure exactly what I'm supposed to be integrating...
    did you mean for your function to be y = x^{\frac{3}{2}} ?
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    PTL
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    Yep, sorry.. Will edit it.
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    Quote Originally Posted by PTL View Post
    "Sketch the region bounded by the curves y = x^ {\frac{3}{2}}, y = 8, x = 0, and find the volume of the solid generated by revolving the region about the y-axis."

    I've sketched the region, and can visualise what they're looking for with the rotation thing, but I don't know how one actually calculates it...

    I'd suspect integration is needed, but I'm not sure exactly what I'm supposed to be integrating...
    two methods.

    cylindrical shells ...

    V = 2\pi \int_0^4 x(8 - x^{\frac{3}{2}}) \, dx<br />

    disks ...

    V = \pi \int_0^8 y^{\frac{4}{3}} \, dy
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    PTL
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    Ok, I think the disks one seems simpler.. I'll go with it..

    I see you're using pi r^2 and stacking the disks on top of one another, from 0 to 8 along the y-axis, that's fine - but where does the y^{4/3} come from? I presume it's something to do with the r^2... but I don't see where it comes from...
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    Quote Originally Posted by PTL View Post
    Ok, I think the disks one seems simpler.. I'll go with it..

    I see you're using pi r^2 and stacking the disks on top of one another, from 0 to 8 along the y-axis, that's fine - but where does the y^{4/3} come from? I presume it's something to do with the r^2... but I don't see where it comes from...
    the radius of each disk is x

    y = x^{\frac{3}{2}}

    x = y^{\frac{2}{3}}

    r^2 = x^2 = \left(y^{\frac{2}{3}}\right)^2 =  y^{\frac{4}{3}}
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  7. #7
    PTL
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    Quote Originally Posted by skeeter View Post
    the radius of each disk is x

    y = x^{\frac{3}{2}}

    x = y^{\frac{2}{3}}

    r^2 = x^2 = \left(y^{\frac{2}{3}}\right)^2 =  y^{\frac{4}{3}}
    Ah, of course - I was thinking {2/3}^2 = 4/9. Gotcha.


    So integrating you get 3/7 * y ^ {7/3} between 0 and 8, gives 54.8 pi

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