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Math Help - double integrals

  1. #1
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    double integrals

    hello everyone
    can somebody help me with this question?
    Attached Thumbnails Attached Thumbnails double integrals-untitled-1.gif  
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  2. #2
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    Quote Originally Posted by sbsite View Post
    hello everyone
    can somebody help me with this question?
    Introduce polar variables (r,\theta) such that x=r \cos(\theta),~y=r \sin(\theta), then the integaral becomes:

    <br />
\int_{\theta=0}^{2\pi} \int_{r=0}^R r^4 \cos^2(\theta) \sin^2(\theta) ~r ~d\theta dr<br />

    RonL
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  3. #3
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    Smile Thank you



    in the answer to this question they got "pi/6"
    where the R disappear ??
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by sbsite View Post


    in the answer to this question they got "pi/6"
    where the R disappear ??
    I get \frac{\pi R^6}{24}

    -Dan
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  5. #5
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    Thank you

    There is a problem with the solutions
    how do you integrate the INTEGRAL(cos^2(theta)*sin^2(theta))
    Last edited by sbsite; August 10th 2007 at 03:48 PM.
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  6. #6
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    Rewrite the it as cos^{2}{\theta}sin^{2}{\theta}=\frac{1}{8}(1-cos{4\theta})
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  7. #7
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by sbsite View Post
    There is a problem with the solutions
    how do you integrate the INTEGRAL(cos^2(theta)*sin^2(theta))
    Or, alternately and a bit more complicated, you could do
    (sin(\theta)cos(\theta))^2 = \left ( \frac{1}{2} sin(2\theta) \right ) ^2

    -Dan
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