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Problem with limits of integration - converting double integral to polar form

  1. Dec 5, 2012 #1
    1. The problem statement, all variables and given/known data
    [itex]\int_0^2 \int_0^\sqrt{2x-x^2} xy,dy,dx[/itex]

    I know the answer, but how does the 2 in the outer integral become pi/2?? I'm fine with everything else, I just can't get this...
     
  2. jcsd
  3. Dec 5, 2012 #2
    Do you have a particular reason for calculating this integral in polar form? It's much easier with Carteesian coordinates. Also I don't understand your question. What is "outer integral"?
     
  4. Dec 5, 2012 #3
    The outer integral is the dx integral, and it's a practice problem out of the book. I know the upper limit for dy in polar is r=2cos(theta) which I got by taking y=root(2x-x2) and converting to polar and solving for r. Why won't this work for dx?
     
  5. Dec 5, 2012 #4
    But what good is that? You need the limits for r and theta, not for x and y. If you want to change to polar coordinates, start by figuring out how x and y are related to r and θ (hint, x≠rcosθ)
     
  6. Dec 5, 2012 #5
    I don't think you read very carefully.
     
  7. Dec 5, 2012 #6

    haruspex

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    You mean, the upper limit for r is 2cos(θ).
    To understand the range for θ, it's easiest to draw a picture. The region is bounded above by an inverted parabola passing through (0,0) and (2,0). Since it is wholly within the first quadrant, you have 0 ≤ θ ≤ π/2. For any given θ in that range, r can be anything from 0 to the point on the parabola at that θ. Allowing θ to go from 0 to π/2 therefore captures the whole region and nothing but.
     
  8. Dec 6, 2012 #7

    HallsofIvy

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    Well said, haruspex, but the area is not an "inverted parabola". It is the upper semi-circle of the circle with center at (1, 0) and radius 1.

    [itex]y= \sqrt{2x- x^2}[/itex] and, squaring, [itex]y^2= 2x- x^2[/itex].
    [itex]x^2- 2x+ y^2= 0[/itex], [itex]x^2- 2x+ 1+ y^2= 1[/itex],
    [itex](x- 1)^2+ y^2= 1[/itex].
     
  9. Dec 6, 2012 #8

    haruspex

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    Good catch. I must have dropped the square root.
     
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