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Math Help - Integrating using Polar Coordinate

  1. #1
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    Integrating using Polar Coordinate

    Using polar coordinates, evaluate the double integral over r sin(x^2+y^2)dA where R is the region 4<x^2+y^2<36

    Need to know where to start, or an example? I keep getting 8.something and it's incorrect.
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  2. #2
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    Re: Integrating using Polar Coordinate

    Quote Originally Posted by Bracketology View Post
    Using polar coordinates, evaluate the double integral over r sin(x^2+y^2)dA where R is the region 4<x^2+y^2<36
    The region you are integrating over is the region enclosed by (and not including) the circles centred at the origin of radii 2 and 6. So clearly your r bounds are 2 and 6, and since you are sweeping over the whole circle, your theta bounds are 0 and 2pi.
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    Re: Integrating using Polar Coordinate

    I get that. But when I integrate and plug in, I get the wrong answer. So I integrate sin(r) from r=2 to 6? that give me -cos(r) from r=2 to 6. Then, integrate -cos(2)+cos(6) from theta=0 to 2pi?

    This gives me 8.647 which is incorrect.
    Last edited by Bracketology; December 5th 2011 at 05:50 AM.
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    Re: Integrating using Polar Coordinate

    I think that this is the expression that you should integrate:

    r\sin\left(x^2+y^2\right)\mathrm{d}A=r\sin\left(r^  2)r\mathrm{d}r\mathrm{d}\theta
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    Re: Integrating using Polar Coordinate

    Quote Originally Posted by corsica View Post
    I think that this is the expression that you should integrate:

    r\sin\left(x^2+y^2\right)\mathrm{d}A=r\sin\left(r^  2)r\mathrm{d}r\mathrm{d}\theta
    Thanks for the imput, but how do I integrate rsin(r^2)*r? Or is the integral suppose to be rsin(r^2)?

    I think you meant rsin(r^2), because that would make sense.
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    Re: Integrating using Polar Coordinate

    The width of an infinitessimal element expressed in polar parameters is r\mathrm{d}\theta, and its length is \mathrm{d}r. So the area of the infinitessimal element is:
    \mathrm{d}A=r\mathrm{d}r\mathrm{d}\theta
    That's where the 'extra' factor r comes from.

    As far integrating it, I don't have time to check it myself right now, but integration by parts might be the way to go.
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  7. #7
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    Re: Integrating using Polar Coordinate

    I have a feeling your original integrand must have been \displaystyle \begin{align*} \sin{\left(x^2 + y^2\right)} \end{align*} , not \displaystyle \begin{align*} r\sin{\left(x^2 + y^2\right)} \end{align*} seeing as it doesn't make sense to have mixed polars with cartesians in the first place. Also, \displaystyle \begin{align*} \int{r^2\sin{\left(r^2\right)}\,dr} \end{align*} doesn't have a closed form answer, whereas the first will.
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