1. Not finding help here? Sign up for a free 30min 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 - Areas

  1. May 7, 2012 #1
    I am asked to consider the following graph:

    r2=a+sin(θ), where a=2

    I have a picture of this plot, which I have attached,

    We are asked to find the area of the upper 'cresent' of the curve, contained at the top

    How would I go about calculating that?

    I've found that if I plot r=√(2+sinθ) and r=-√(2+sinθ) that this gives me the separate graphs individually, and that if I integrate from 0-π on the first one, then integrate from π-2π on the second one, if i subtract, i get the correct answer geometrically, using my graphing programme, but i dont know how to do this analytically? (i have attached another image to show the to sections - its the blue section i need)

    Any help would be vastly appreciated, thank you
     

    Attached Files:

  2. jcsd
  3. May 7, 2012 #2
    From my graphing programme, ive found the area should be 4.06-2.139 = 1.921 approximately, if this helps anyone?
     
  4. May 7, 2012 #3
    I think your going to have to use cylindrical coordinates. From the graph it's evident that the limits of theta would be 0 to pi. The limits of r would be sqrt(2) to sqrt(3) (if your confused about how did I get these limits try finding the max and min values of r by playing around with the theta value).
     
  5. May 7, 2012 #4
    I get where you have your limits by maximising sin in the range of 0-pi, giving r^2= 2 or 3... Ive never done cylindrical coordinates before? We've never encountered them in lessons yet. Worrying
     
  6. May 7, 2012 #5
    Never mind then lol. I know its definitely not cylindrical coordinates. Though.
     
  7. May 10, 2012 #6
    Is there any reason you are plotting [itex]r = -\sqrt{2+sinθ}[/itex]? It would be easier if you use [itex]r = \sqrt{2-sinθ}[/itex]. It plots the same circle but the intersecting points are coincident. The you could simply integrate

    [itex]∫\frac{1}{2}((2+sinθ)-(2-sinθ))[/itex]dθ

    Since these two curves have the same period and starting point you can integrate from 0 to ∏
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Polar Coordinates - Areas
  1. Polar Coordinates (Replies: 1)

  2. Polar Coordinates (Replies: 7)

  3. Polar coordinates (Replies: 1)

  4. Polar Coordinates (Replies: 1)

  5. Polar Coordinates (Replies: 9)

Loading...