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Polar Coordinants and Double Integrals

  1. Oct 17, 2006 #1
    The problem is to find the area of the piece enclosed by the intersection of the circles r = sin t and r = cos t.

    I tried to set up the integral:

    Integral[0 to Pi/4]Integral[Sin[t] to Cos[t]] r dr d@

    but this doesn't seem work, I get out .25, and just by eyeballing it, I can tell it is less. It seems tricky because the upper bound r = cos [t] is being traced out between the angles Pi/4 and Pi/2, while the lower bound r = sin [t] is being traced out between the angles 0 and Pi/4.
     
    Last edited: Oct 17, 2006
  2. jcsd
  3. Oct 17, 2006 #2

    Hurkyl

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    Don't forget you're working in polar coordinates, not rectangular ones. You shouldn't be thinking of vertical lines; you should be thinking of radial lines.
     
  4. Oct 17, 2006 #3
    I'm struggling to see how I would describe this region using only a single double integral, or would I need two?
     
  5. Oct 17, 2006 #4

    Hurkyl

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    Yah; in polar coordinates, I would expect you either to use two integrals or to take advantage of symmetry.
     
  6. Oct 17, 2006 #5
    Symmetry: Say if I were to describe the region from the center of the cosine circle. The angle would be from Pi/4 to Pi/2. Now, if I flipped the portion of the sin circle corresponding to these angles, it would again look like the region. Would the radius then go from 0 to Cos[t] - Sin[t]? The answer is much more reasonable, but it still seems like I'm doing something that seems wrong.
     
  7. Oct 17, 2006 #6

    Hurkyl

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    Nope. Remember, you're looking at radial lines. What are the endpoints of a radial line lying in your shape?
     
  8. Oct 17, 2006 #7
    I don't get what you're asking. The radial line of the cosine circle would start at [.5, 0] and go to cos[t]?
     
    Last edited: Oct 17, 2006
  9. Oct 17, 2006 #8
    I feel so stupid, It's just the double integral of with angles 0 to pi/4 and radius 0 to sin[t], times 2 (symmetry).
     
  10. Oct 17, 2006 #9

    Hurkyl

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    That sounds right!
     
  11. Oct 17, 2006 #10
    Thanks for the help!
     
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