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 Integral

  1. Apr 16, 2015 #1
    1. The problem statement, all variables and given/known data
    ∫∫dydx

    Where the region Ω: 1/2≤x≤1 0 ≤ y ≤ sqrt(1-x^2)

    2. Relevant equations


    3. The attempt at a solution
    The question asked to solve the integral using polar coordinates. The problem I have is getting r in terms of θ. I solved the integral in rectangular ordinates using a trig sub and found the answer to be (π/6) - (√3)/8.


    Thanks for all input.
     
    Last edited: Apr 16, 2015
  2. jcsd
  3. Apr 16, 2015 #2

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Have you drawn a picture? Remember ##r## goes from ##r## on the inner curve to ##r## on the outer curve. What is the equation of ##x = \frac 1 2## in polar coordinates? What is the equation of the circle? And from the picture you should be able to see the ##\theta## limits.
     
  4. Apr 16, 2015 #3

    Mark44

    Staff: Mentor

    You're not solving an equation -- you're setting up and evaluating an integral
    The integrand is pretty straight forward; probably the hardest part is figuring out the limits of integration. A ray extending out from the pole (0, 0) goes through the vertical line, x = 1, to the circle. Convert the equation of the vertical line to polar from. The circle part is simple.
     
  5. Apr 16, 2015 #4
    I have drawn the picture and it is very straightforward. The problem I am having is that the radius is not constant so i need to get r in terms of theta. http://www5b.wolframalpha.com/Calculate/MSP/MSP17811i1c25cda73i33d700001h169b9cb5h0603d?MSPStoreType=image/gif&s=34&w=200.&h=193.&cdf=Coordinates&cdf=Tooltips [Broken]
     
    Last edited by a moderator: May 7, 2017
  6. Apr 16, 2015 #5

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    OK, so you have drawn the picture. Now answer the questions I asked in my reply above.
     
    Last edited by a moderator: May 7, 2017
  7. Apr 16, 2015 #6

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Do you know how to express x in terms of r and θ ?
     
    Last edited by a moderator: May 7, 2017
  8. Apr 16, 2015 #7
    As in x=rcos (θ) and y=rsin (θ) ?
     
  9. Apr 16, 2015 #8

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Good the left hand boundary of the region is x = 2.

    That should give you a relation in terms of r and θ for that boundary.

    The right hand boundary (curved boundary) should be easy in terms of r .
     
  10. Apr 16, 2015 #9

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You mean ##x = \frac 1 2##.
     
  11. Apr 16, 2015 #10

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Right!
     
  12. Apr 17, 2015 #11
    Ok so using the conversion I found the bounds of r from (1/2)sec(θ) to 1

    Thanks for the input
     
  13. Apr 17, 2015 #12

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Yes. You're welcome.
     
  14. Apr 17, 2015 #13

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    And what bounds did you get for ##\theta##?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Polar Integral
  1. Polar Integrals (Replies: 4)

Loading...