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Area inside r=6sin(theta) but outside r=3

  1. Jul 6, 2010 #1
    1. The problem statement, all variables and given/known data

    The title really says it all. Find the area of the region inside r=6sin(theta) but outside r=3.

    2. Relevant equations



    3. The attempt at a solution

    I first found the x values where the two curves intersect and came up with .52359877 and 2.61799388. I then integrated 6sin(theta) (or, -6cos(theta)) from .52359877 to 2.61799388. The answer I got was 10.3923, but that's incorrect.
     
  2. jcsd
  3. Jul 6, 2010 #2
    Anyone?
     
  4. Jul 6, 2010 #3

    LCKurtz

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    Science Advisor
    Homework Helper
    Gold Member

    What's with all the decimals? Leave things in terms of pi. I don't believe those are x values. This is a polar coordinate problem, in terms of r and theta. You might start by stating the correct formula for the area between two polar curves expressed in terms of polar coordinates.
     
  5. Jul 7, 2010 #4
    Okay, I found the formular for the area between two polar curves, which is:

    1/2 Integral from B to A of (f2(theta)^2 - f1(theta)^2)

    But what do I do about the r=3? And how do I find out what B and A are?
     
  6. Jul 7, 2010 #5

    Mark44

    Staff: Mentor

    The two polar curves are symmetric about the y-axis, so you can find the area in the first quadrant and double it to get the entire area.

    For b, you can use pi/2. For a, find the point of intersection in quadrant 1 of the two curves - use the exact value, not a decimal approximation.

    For the outer curve, r = 6 sin(theta). For the inner curve, r = 3. These are the functions you're calling f1 and f2 (not necessarily in order).

    This problem is very similar to the other problem you posted.
     
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