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Math Help - evaluating a double integral using polar coordinates

  1. #1
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    evaluating a double integral using polar coordinates

    Evaluate the integral given that ∫ ∫ x dA, where D is the region in the first quadrant that lies between the circles x^2 + y^2 = 4 and x^2 + y^2 = 2x (D is under the double integral)

    so I decided it would be easier to use polar coordinates for this problem. My limits for theta are 0 to Pi/2 and I used pythag to find the first limit of r which is 2, My professor gave us a hint and told us for x^2 + y^2 = 2x , r = 2cos(theta). Is there a way to actually find the other limit or is this just some kind of rule?

    I also set up the double integral

    Pi/2 2
    ∫ . . ∫ rcos(theta) *r drd(theta)
    0 . . 2cos(theta)

    I integrated the r but I think I integrated incorrectly, I got this

    Pi/2 . . . . . . . . . . . . . . . . . . 2
    ∫ (r^2/2)cos(theta) *(r^2/2)| d(theta)
    0 . . . . . . . . . . . . . . . . . . . . 2cos(theta)

    I'm not sure how I should go about integrating the first r in front of the cosine.
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  2. #2
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    Re: evaluating a double integral using polar coordinates

    Hey mrb165.

    I think that if you are integrating between x^2 + y^2 = 4 and x^2 + y^2 = 2x then shouldn't the limit for r be 2cos(theta) to 2?

    Doing 0 to 2cos(theta) should (I think) find the area between the first curve and the x/y axes.
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  3. #3
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    Re: evaluating a double integral using polar coordinates

    I find it very difficult to read your math "expressions". You say you want to "Evaluate the integral given that ∫ ∫ x dA, where D is the region in the first quadrant that lies between the circles x^2 + y^2 = 4 and x^2 + y^2 = 2x".

    Okay, in the first quadrant \theta goes from 0 to \pi/2. Converting the first circle to polar coordinates again is easy- r^2= 4 means that r= 2 (r is never negative).
    The other circle is not much harder: x^2+ y^2 is again r^2 and 2x= 2r cos(\theta) so the equation is r^2= 2rcos(\theta) or ( if r is not 0, r= 2cos(\theta). So the double integral is \int_{\theta= 0}^{\pi/2}\int_{r= 2}^{2cos(\theta)} (r cos(\theta)) r dr d\theta
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    Re: evaluating a double integral using polar coordinates

    Quote Originally Posted by HallsofIvy View Post
    I find it very difficult to read your math "expressions". You say you want to "Evaluate the integral given that ∫ ∫ x dA, where D is the region in the first quadrant that lies between the circles x^2 + y^2 = 4 and x^2 + y^2 = 2x".

    Okay, in the first quadrant \theta goes from 0 to \pi/2. Converting the first circle to polar coordinates again is easy- r^2= 4 means that r= 2 (r is never negative).
    The other circle is not much harder: x^2+ y^2 is again r^2 and 2x= 2r cos(\theta) so the equation is r^2= 2rcos(\theta) or ( if r is not 0, r= 2cos(\theta). So the double integral is \int_{\theta= 0}^{\pi/2}\int_{r= 2}^{2cos(\theta)} (r cos(\theta)) r dr d\theta
    I think your r limits should be around the other way...
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