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Math Help - Volume of solids

  1. #1
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    Volume of solids

    Consider region enclosed by the curves y=sinx and and y=0 for x between 0 and \pi.Find the volume of the solid obtained when this region is revolved about the y-axis.


    Attempt to solution:

    Use shells method

    r=x-0

    v=2\pi\int_0^{\pi} xsinx dx

    Use integration by parts

    \int udv=-xcosx+\int cosx dx

    \int udv=2\pi(-xcosx+sinx)

    How come the answer book is saying l am wrong ? What did l do wrong ?
    Last edited by nyasha; March 22nd 2009 at 05:56 PM.
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  2. #2
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    Quote Originally Posted by nyasha View Post
    Consider region enclosed by the curves y=sinx and and y=0 for x between 0 and \pi.Find the volume of the solid obtained when this region is revolved about the y-axis.


    Attempt to solution:

    Use shells method

    r=x-0

    v=2\pi\int_0^{\pi} xsinx dx

    Use integration by parts

    \int udv=-xcosx+\int cosx dx

    \int udv=-xcosx+sinx

    How come the answer book is saying l am wrong ? What did l do wrong ?
    I tried really hard, but i just couldn't see your answer book on my PC screen.

    so, did you evaluate the definite integral from 0 to \pi ?
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  3. #3
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    Quote Originally Posted by nyasha View Post
    Consider region enclosed by the curves y=sinx and and y=0 for x between 0 and \pi.Find the volume of the solid obtained when this region is revolved about the y-axis.


    Attempt to solution:

    Use shells method

    r=x-0

    v=2\pi\int_0^{\pi} xsinx dx

    Use integration by parts

    \int udv=-xcosx+\int cosx dx

    \int udv=-xcosx+sinx

    How come the answer book is saying l am wrong ? What did l do wrong ?


    when you reflect it you get a cylinder, unroll it to make a rectangle and the dimensions are 2piX by sinx

    multiply them and take out the 2pi

    2pi (XsinX)

    then integrate on your calculator or by hand

    either way the answer should be

    19.739 or 6.28 pi

    is that right?
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  4. #4
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    Quote Originally Posted by skeeter View Post
    I tried really hard, but i just couldn't see your answer book on my PC screen.

    so, did you evaluate the definite integral from 0 to \pi ?

    I had evaluated from 0 to \pi but l forgot to multiply by 2\pi
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