Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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
  4. Jul 6, 2010 #3


    User Avatar
    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


    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook