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Math Help - Integral

  1. #1
    Senior Member vincisonfire's Avatar
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    Integral

    I want to evaluate the volume inside r=cos(\theta) .
    Thanks.
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  2. #2
    o_O
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    \begin{aligned}r & = \cos \theta \\ r^2 & = r\cos \theta \\ x^2 + y^2 & = x \\ (x-\tfrac{1}{2})^2 + y^2 & = \tfrac{1}{4} \end{aligned}

    Revolving it around the x-axis produces a sphere with radius \tfrac{1}{2} .
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  3. #3
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    That is just a circle of radius 1 centered at (1/2,0)

    A sphere if you revolve

    If you must do this with integration, you can just move it over the the origin.

    Now, revolve a circle of radius 1/2.

    2{\pi}\int_{0}^{\frac{1}{2}}(\frac{1}{4}-x^{2})dx

    or leave it where it is and:

    {\pi}\int_{0}^{1}\left(\frac{1}{4}-(x-\frac{1}{2})^{2}\right)dx
    Last edited by galactus; December 17th 2008 at 02:45 PM.
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    Senior Member vincisonfire's Avatar
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    I want the volume within  \sqrt{x^2+y^2+z^2} = \frac{x}{\sqrt{x^2+y^2}} *
    Here is a picture of the solid.
    Attached Thumbnails Attached Thumbnails Integral-rcoso.jpg   Integral-rcoso-2-.jpg  
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  5. #5
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    I want the volume within *
    I suggest you check that you have the right definition of \theta because my book used it for the other angle. Anyway, using your notation:
    \int\int\int dV=\int_0^{\pi}\int_0^{2\pi}\int_0^{cos(\theta)}r^  2\sin(\phi)dr d\theta d\phi
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  6. #6
    Senior Member vincisonfire's Avatar
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    Physicists and mathematicians use the opposite definition right.
    If you use the physicists' you get a sphere.
    If you the mathematicians' you get this weird figure I've showed above.
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