1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Polar Coordinates Area

  1. Mar 12, 2009 #1
    1. The problem statement, all variables and given/known data
    Find the area inside the lemniscate r = 2sqrt(sin(2theta))

    2. Relevant equations
    Integral from a to b of (1/2)[f(theta)]^2 d(theta)

    3. The attempt at a solution
    I tried integrating from 0 to 2pi and got an area of 0. Then I tried integrating from 0 to pi and still got an area of 0. I looked at the answer and they integrated from 0 to pi/2 and then multiplied the integral by 2. I don't understand why they chose to integrate from 0 to pi/2 or why they multiplied the integral by 2? Any help would be greatly appreciated.
  2. jcsd
  3. Mar 12, 2009 #2


    Staff: Mentor

    y = sin(2x) has a complete cycle between 0 and pi. If you wanted the area between this curve and the x-axis, it wouldn't do you any good to integrate between 0 and pi -- you would get 0. So instead you would integrate between 0 and pi/2 to get the area under one arch, and then double it, to get the area of both regions.

    A similar thing is happening with your polar curve.
  4. Mar 12, 2009 #3
    Ahh that makes sense, thank you!
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook