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Double Integration in Polar Coordinates

  1. Dec 19, 2016 #1
    1. The problem statement, all variables and given/known data
    Integrate by changing to polar coordinates:

    ## \int_{0}^6 \int_{0}^\sqrt{36-x^2} tan^{-1} \left( \frac y x \right) \, dy \, dx ##

    2. Relevant equations
    ## x = r \cos \left( \theta \right) ##
    ## y = r \sin \left( \theta \right) ##

    3. The attempt at a solution

    So this is a quarter of a circle in the first quadrant of an xy-coordinate system with a radius of 6. So the bounds for the integral in polar coordinates will be r from 0 to 6 and ## \theta ## from 0 to ## \frac{ \pi }{2} ## .

    ## tan^{-1} \left( \frac y x \right) ## is just ## \theta ## so the integral becomes:

    ## \int_{0}^6 \int_{0}^ \frac{ \pi }{2} \theta \, d \theta \, dr ##

    I worked this out and got ## \frac{3 \pi ^2}{4} ## . I checked it on Wolfram Alpha and got the same result. I plugged the original integral into Wolfram Alpha and got ## \frac{9 \pi ^2}{4} ## so somehow I lost a factor of 3 somewhere. Clearly I integrated properly so I must have made a mistake converting to polar coordinates.

    Any help would be appreciated. Thanks.
     
  2. jcsd
  3. Dec 19, 2016 #2

    rock.freak667

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    Homework Helper

    In your conversion to polar the elemental area in Cartesian is dA = dy*dx

    However in polar it should be r*dr*dθ
     
  4. Dec 19, 2016 #3
    You have forgot a factor of ##r##. The area element ##dxdy## should be replaced by ##rdrd\theta##, which is obtained from the Jacobian for the transformation to polar coordinates. I hope this helps.
     
  5. Dec 19, 2016 #4
    Yep, just remembered that right as your replied. Thanks
     
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