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Thread: Finding volume of sphere by integrating Y = rsin ((pi/r)x) from 0 to 2r

  1. #1
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    Finding volume of sphere by integrating Y = rsin ((pi/r)x) from 0 to 2r

    hello,

    I am trying to find the equation of the sphere by integrating the following:

    Y = rsin ((pi/r)x). Integration with limits from 0 to 2r.

    I get pi.r^3, but am missing the 4/3.

    Can someone please give it a go and see if they get the right answer ?

    Thanks a lot!

  2. #2
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    Well, first off, you didn't tell us which variable we're integrating with respect to. From context, I can guess that it's with respect to x, because integrating with respect to r doesn't seem to make much sense in this case. But even given that assumption, I'm still completely uncertain as to how you got your answer, because:

    [tex]\displaystyle \int^{2r}_0 r \cdot \sin \left( \left( \dfrac{\pi}{r} \right) x \right) \: dx = 0[/tex]

    WolframAlpha confirms this result. That said, we cannot troubleshoot work we can't see. You may wish to double check to make sure you copied down the problem correctly from your textbook/problem sheet/whatever, and that you typed it in correctly on here. Also, when you reply back, please share with us any and all work you've done on this problem, even the parts you know for sure are wrong. Thank you.

  3. #3
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    Quote Originally Posted by Ag6012 View Post
    I am trying to find the equation of the sphere by integrating the following:

    Y = rsin ((pi/r)x). Integration with limits from 0 to 2r.

    I get pi.r^3, but am missing the 4/3.
    I don't see the connection between this function and the volume of a sphere. It looks like just the area under a sine curve. At first I thought maybe it was an attempt at using polar coordinates, but that doesn't work out either.

    Please tell us how you obtained this integral, and describe the problem fully. Is the sphere to be obtained by revolving a circle (what circle) about some axis (what axis?)?

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