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Polar Double integrals

  1. Jun 4, 2009 #1
    1. The problem statement, all variables and given/known data
    Find area bounded by x^2 + y^2 = 1 and x^2 + y^2 = x + y


    2. Relevant equations



    3. The attempt at a solution

    from the second circle, we can see r^2 >= r cos t + r sin t
    so r >= cos t + sin t

    Limits are:
    cos t + sin t <= r <= 1
    -pi/4 <= t <= 3pi/4

    Doing the integrals however, I always seem to get zero:

    [tex]\int^{3pi/4}_{-pi/4}\int^{1}_{cos t+ sin t} r dr dt [/tex]
    this gives me
    [tex]\int^{3pi/4}_{-pi/4}-sin(2t)/2 dt [/tex]
    which is zero. What am I doing wrong?
    Thanks
     
  2. jcsd
  3. Jun 4, 2009 #2

    Mark44

    Staff: Mentor

    If you integrate a periodic function over one complete period, you don't get the area between the graph and the horizontal axis - you get zero. You are integrating a multiple of sin(2t) over an interval of length pi, so naturally you'll get zero. If you want the area, you have to break up the integral into two integrals - one for the part where sin(2t) is above the horizontal axis, and the other where it is below. Then you'll get the area between the curve and the horizontal axis.
     
  4. Jun 4, 2009 #3
    Thanks, I just found that if I split the integral up into the first circle and two segments of the other circle, it ends up giving me a non-zero answer (not sure if right thought)
    Area within first quadrant = pi/4

    Area in 2nd and 4th quadrant =
    [tex]\int^{0}_{-pi/4}\int^{cos t + sin t}_{0} r dr dt[/tex]

    which gives me
    [tex]\int^{0}_{-pi/4} (1+sin2t)/2 dt [/tex]
    which is pi/8 + 1/4

    so in total i have pi/4 + 2(pi/8 + 1/4)
    = (pi + 1)/2

    But yeah, what you said makes sense. Thanks
     
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