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Find the area of the region shared by the curves

  1. Feb 24, 2010 #1
    find the area of the region shared by the curves r=6cosx and r=6sinx

    i know that if you graph those two functions you get two circles, both with radius 3, the first one has its center on the cartesian x axis and the other has its center on the y axis.

    i also know that if you draw a line through the center of where they meet the angle of that line is pi/4

    what i dont know is which function to integrate and what would be the lower limit of integration.

    i know that the formula is 1/2 integral from a to b of (g(x)^2 - f(x)^2)

    making 6sinx the second function makes sense since thats the bottom part of the area that i want to find but you cant make the first function 6cosx and integrate from 0 to pi/4 because that would give you more than just where the two functions overlap

    any help or ideas about how to solve this would be appreciated.
     
  2. jcsd
  3. Feb 24, 2010 #2

    Mark44

    Staff: Mentor

    The formula that you "know" is incorrect. A ray going out from the origin should extend to only one or the other of the two circles, so you won't have g(x)^2 - f(x)^2 in your integral. If you have drawn a sketch of the region you will see this.

    An easy way to do this is to exploit the symmetry of the two curves, and integrate along the circle r = 6sint from t = 0 to t = pi/4, then double that value. Note that I changed the x you had to t (for theta), since the curves are (I believe) curves in polar coordinates, in which r and theta appear, but x and y don't.

    Another way is one that doesn't exploit symmetry. To do it this way you need two integrals, one in which you integrate along the sine curve from t = 0 to pi/4, and the other in which you integrate along the cosine curve back to the origin, as t runs from pi/4 to pi/2.

    Both techniques should give you the same area.
     
  4. Feb 24, 2010 #3
    yea i used x instead of theta

    but i completely see that now. i just wasnt thinking straight.

    thanks
     
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