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Area Integral for unknown portion of circle.

  1. Apr 30, 2008 #1
    My problem requires me to write an iterated integral for the area inside the circle r=2cos(theta) for -pi/2 <theta< pi/2 (circle of radius 1 centered at (1,0)) and outside the circle r=1 centered at the origin.

    So I can write an iterated integral for the circle of radius 1 centered at (1,0) and then subtract the area portion of the circle at the origin. But what is this portion of area? Can I write another iterated integral for this area portion and then subtract it from the iterated integral? How would I approach this?

    Any help would be greatly appreciated! Thanks!
  2. jcsd
  3. Apr 30, 2008 #2
    So i was able to write an integral for the area of the circle, and tried integrating for the area between the curves of each circle. I am getting a value for the requested area (as described above) that seems unreasonable and do not know where I am making the mistake, or if I am evening doing it correctly. Any suggestions?
  4. Apr 30, 2008 #3
    The way I would do this would just be to find the area of the part of the circle centered at the origin which overlaps the other circle and then subtract that from the area of the circle centered at (1,0). To get the area in the overlap region you need to see where the two circles intersect, i.e. solve 2cos(theta) = 1. This should give you two values and then you can either integrate dA = r*dr*dtheta or you can just get the area geometrically since it is just a wedge of the circle. I think the answer should be 2pi/3.
  5. Apr 30, 2008 #4
    I think 2pi/3 would be the answer if the left side of the shape were bounded by straight lines as opposed to curved ones. Since this is an approximation to the area, I'll probably use it if I cannot figure something else out. Thanks so much for your help!
  6. Apr 30, 2008 #5
    I think this integral may work for the Area outside the circle, r=1 and inside the circle, r=2cos(theta)

    polar coordinates:

    The double integral (where 1 <= r <= 2cos(theta) and -pi/3 <= theta <= pi/3 ) of differential Area dA=rdrdtheta

    This seems to make sense when visualizing the double integral.

    Solving, I got Area = 1.91 which seems reasonable since it is about sixty percent of a circle of radius 1 and area pi.

    Does this seem like a valid method?
  7. Apr 30, 2008 #6
    Ahh, you are right. Let me try again: I think you want to integrate dA = r*dr*dtheta over the region -pi/3 < theta < pi/3 and 1 < r < 2*cos(theta). This is becasue +-pi/3 are where the two circles intersect and for any fixed theta, the value of r in the region under consideration goes from the inner circle to the outer circle. When I evaluted this integral I got something which was fairly close to 2pi/3, as it should be.
  8. Apr 30, 2008 #7
    haha, you beat me to it by a minute!!! good work.
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