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Area of a Polar Graph

  1. May 13, 2009 #1
    This is an extra credit problem for a take home test, so i will understand if no one feels comfortable helping me out, but any advice is greatly appreciated :biggrin:



    1. The problem statement, all variables and given/known data

    Compute the area enclosed by one loop of the graph given by r = sqrt(sin(3{theta}))


    2. Relevant equations

    see above

    3. The attempt at a solution

    The graph makes 3 loops, so i tried finding the area from (0, 2{pi}) but all come up with is 0


    Thanks again!
     
  2. jcsd
  3. May 13, 2009 #2

    Mark44

    Staff: Mentor

    You get one loop as theta ranges from 0 to pi/3.
     
  4. May 13, 2009 #3
    Thank you Mark. Is finding the finding the area of the entire graph, then dividing that answer by the number of loops a viable method for this type of problem?
     
  5. May 13, 2009 #4

    Mark44

    Staff: Mentor

    You could do it that way, I suppose, but it makes more sense to me to get the area within one loop and multiply it by the number of loops. Keep in mind that the integrand is undefined for theta in [pi/3, 2pi/3], because of the square root.
     
  6. May 13, 2009 #5
    The formula for the area [ A = (1/2)*r2 ] effectively eliminates the square root, so i dont quite understand how its undefined from [pi/3, 2pi/3] because of the square root.

    However after proving to myself that r = 0 at 0 and pi/3, I have successfully found the area of one loop. Thanks again for all the help Mark :smile:
     
  7. May 13, 2009 #6

    Mark44

    Staff: Mentor

    I was just looking at r in your first post, and wasn't thinking about the integral.
     
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