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Polar Coordinate Integration

  1. Nov 8, 2013 #1
    1. The problem statement, all variables and given/known data
    Integrate the double Integral: 6xdydx in polar coordinates

    The y goes from bottom limit of x(3)^(1/2) to the top limit of (1-x^2)^(1/2)
    the x goes from 0 to 1/2

    2. Relevant equations



    3. The attempt at a solution
    So I graphed it, and it looks like a semi circle on the positive y plane with a linear line going through it in the first domain.
    Changing this to polar coordinates I got
    Double Integral: 6r^2cosθdrdθ

    and for dr I evaluated it between 0 and 1
    for dtheta, i'm having trouble figuring it out. It looked to me like it should go from pi/4 to pi/2, because the radius is one and x goes from 0 to 1/2 only. Any advice?
     
  2. jcsd
  3. Nov 8, 2013 #2
    Are you sure you have the lower limit on y written correctly? Did you mean y = x3/2 or y = 3x1/2?

    Either way, neither of those lines are linear. Which means that r isn't going from 0 to 1 but from the lower y limit (converted to polar) to 1.

    To find the upper limit on theta, you have to find where the two equations intersect and the corresponding angle.
     
  4. Nov 8, 2013 #3

    LCKurtz

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    I'm assuming the lower curve is ##y = \sqrt 3 x##. What angle does that line make with the ##x## axis? That should give you a hint about the lower ##\theta## value. It would be good of you to state the exact problem or provide a figure. Does it ask for the region above the line and below the circle? Or above the line and above the ##x## axis and below the circle?? Or something else? Where does the ##x=\frac 1 2## come from?
     
    Last edited: Nov 8, 2013
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