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Homework Help: Double integrals in polar coordinates

  1. Jul 18, 2009 #1
    1. The problem statement, all variables and given/known data

    [tex]\int{\int_{D}x dA}[/tex]

    where D is the region in Q1 between the circles x2+y2=4 and x2+y2=2x using only polar coordinates.

    3. The attempt at a solution
    Well, the two circles give me r=2 and r=2 cos [tex]\theta[/tex], and the integrand is going to be r2cos [tex]\theta[/tex], but I have no idea how to determine the bounds of integration in this case.
  2. jcsd
  3. Jul 18, 2009 #2


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    Hi compliant! :smile:

    (have a theta: θ and a pi: π :wink:)

    Just integrate θ from 0 to 2π (or -π and π), and integrate r between whatever values it goes between for a fixed value of θ. :wink:
  4. Jul 18, 2009 #3


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    I would recommend first drawing a picture. [itex]x^2+ y^2= 4[/itex] is, of course, a circle with center at (0,0) and radius 2. [itex]x^2+ y^2= 2x= x^2- 2x+ y^2= 0[/itex] or [itex]x^2- 2x+ 1+ y^2= (x- 1)^2+ y^2= 1[/itex] is a circle with center at (1, 0) and radius 1: it is tangent to the y-axis at (0,0) and tangent to the first circle at (2, 0). Now think in terms of polar coordinates. Both equations become very simple in polar coordinates. What drawing the graph tells you is that you will want to handle the integration in three parts: [itex]\theta= 0[/itex] to [itex]\pi/2[/itex], [itex]\theta= \pi/2[/itex] to [itex]3\pi/2[/itex], and [itex]\theta= 3\pi/2[/itex] to [itex]2\pi[/itex].

    suggested, the outside radius (the upper limit of integration) is always 2 and the inner radius (the lower limit of integration, for [itex]\theta= 0[/itex] to [itex]\pi/2[/itex] is
  5. Jul 19, 2009 #4
    tiny-tim, thanks for those. desperately needed.

    hallsofivy, I did draw the diagram, and found that it was rather inconveniently symmetrical, which was why I got stumped. going by your suggestion, from θ = 0 to θ = π/2, I would be integrating along the right side of the curve, where the upper bound is r = 2, and the lower bound is r = 2 cos θ. I would then solve accordingly, with r2 cos θ as the integrand.

    I'm just wondering though, how is the left side of the curve from θ = π/2 to θ = 3π/2 and not θ = π/2 to θ = π ? And as for the third part of the curve that goes from θ = 3π/2 to θ = 2π, that's...a straight line. =/

  6. Jul 21, 2009 #5
    sorry to do this, but bump.
  7. Jul 22, 2009 #6


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    Hi compliant ! :smile:

    I'm confused :confused:

    The area is between two circles, one touching both the edge and the centre of the other.

    So there are two regions:

    the "left" region, which is simply a semicircle, so you know the answer already, and you needn't integrate at all (though if you did, you would integrate a constant, over the whole angle π/2 to 3π/2)

    and the "right" region, which is from -π/2 to π/2, which you seem to be ok with. :smile:
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