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Math Help - double integral question

  1. #1
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    double integral question

    Ok. I have a quick question about the limits of integration for a problem:
    *I am going to use @ for theta.*


    Integrate f(x,y) = x + y over the region R bounded by r = cos(@)

    I converted the function to a polar function and got r cos(@) + r sin(@) and a r dr d@ making it:

    r^2[cos(@) + sin(@)]dr d@

    now for the limits of integration I am using 0 to cos(@) and 0 to 2 pi Is this correct or should I use 0 to 1 for the dr since cos can only go from 0 to 1? I am thinking I am correct with my first method but I just wanted to check. Also if I made any other mistakes feel free to point those out as well.
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  2. #2
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    Quote Originally Posted by billbarber View Post
    Ok. I have a quick question about the limits of integration for a problem:
    *I am going to use @ for theta.*


    Integrate f(x,y) = x + y over the region R bounded by r = cos(@)

    I converted the function to a polar function and got r cos(@) + r sin(@) and a r dr d@ making it:

    r^2[cos(@) + sin(@)]dr d@

    now for the limits of integration I am using 0 to cos(@) and 0 to 2 pi Is this correct or should I use 0 to 1 for the dr since cos can only go from 0 to 1? I am thinking I am correct with my first method but I just wanted to check. Also if I made any other mistakes feel free to point those out as well.

    r=\cos \theta\,,\,\,0\leq\theta\leq \pi is the same as the circle \left(x-\frac{1}{2}\right)^2+y^2=\frac{1}{4} ...

    Tonio
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  3. #3
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    so are you saying it would be easier to just convert the bound r = cos(@) to rectangular coordinates?
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  4. #4
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    or are you saying that the theta goes from 0 to pi and the r goes from 0 to 1/2?
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  5. #5
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    Quote Originally Posted by billbarber View Post
    or are you saying that the theta goes from 0 to pi and the r goes from 0 to 1/2?

    No: r goes as cos, as given: r=\cos\theta, and \theta ranges between 0 and \pi

    Tonio
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