Results 1 to 6 of 6

Math Help - Polar Calculus Help Needed

  1. #1
    Newbie
    Joined
    Jan 2007
    Posts
    11

    Polar Calculus Help Needed

    I have a graph of a polar function, r=1-2cos(theta). The graph produces a limacon. I had no problem graphing it, but I'm a bit confused as to how to find the area. The area I need to find is the area within the larger, outer loop and outside the smaller, inner loop. The inner loop isnt really a loop at all, more of an ellipse, so I cant treat it as a circle of known radius and subtract the two functions within the integral. I'm confused because of the loop inside of the loop. Usually, this would happen with two seperate functions and it would just be a matter of subtracting the two areas. Intuition would tell me to integrate from 0 to 2pi for the area, but for some reason I'm not sure if this will work.

    Help
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member ecMathGeek's Avatar
    Joined
    Mar 2007
    Posts
    436
    Someone else correct me if I'm wrong, but if I remember correctly, I believe this is what is happening in your problem:

    As theta goes from 0 to 2pi, the polar graph makes two separate loops. The graph actually completes each loop in pi radians, so integrating from 0 to pi gives you the area of one loop (the outer loop, I believe). Integrating from pi to 2pi gives you the area of the second loop, but notice what is happening between pi and 2pi: the graph seems to loop back on its self (instead of the point at pi radians being located on the negative x axis, it appears on the positive x axis (note that I have not graphed this function so I may be mistaken on its orientation)). This represents a negative radius, which means the integrating from pi to 2pi gives a negative area. If you integrate from 0 to 2pi, you actually are getting the area of the first loop minus the magnitude of the area of the second loop. In other words, to get the area you said you need, just integrate from 0 to 2pi.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Look at picture below.

    Now it seems dangerous to compute the integral from [0,2pi] because we will count the inside loop twice.

    Instead draw the limacon for 0<=theta <= pi
    (Shown below).

    That area is equal to,
    INT[0,pi](1/2)r^2 d(theta)=4.71

    Now draw the limacon for 0<=theta<=pi/3
    (Shown below).

    That area is equal to,
    INT[0,pi/3](1/2)r^2 d(theta)=.2718

    Then if you subtract the area in the first integration from this one you get the area without the half-loop.
    That is, 4.43

    By Symettry we have total area, 8.86

    Note, if you find,
    INT[0,2pi](1/2)r^2 d(theta) = 9.42
    And if you subtract .2718 twice you get,
    8.86 (because we counted the half-loops twice).
    Attached Thumbnails Attached Thumbnails Polar Calculus Help Needed-picture7.jpg  
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member ecMathGeek's Avatar
    Joined
    Mar 2007
    Posts
    436
    Quote Originally Posted by ThePerfectHacker View Post
    Look at picture below.

    Now it seems dangerous to compute the integral from [0,2pi] because we will count the inside loop twice.

    Instead draw the limacon for 0<=theta <= pi
    (Shown below).

    That area is equal to,
    INT[0,pi](1/2)r^2 d(theta)=4.71

    Now draw the limacon for 0<=theta<=pi/3
    (Shown below).

    That area is equal to,
    INT[0,pi/3](1/2)r^2 d(theta)=.2718

    Then if you subtract the area in the first integration from this one you get the area without the half-loop.
    That is, 4.43

    By Symettry we have total area, 8.86

    Note, if you find,
    INT[0,2pi](1/2)r^2 d(theta) = 9.42
    And if you subtract .2718 twice you get,
    8.86 (because we counted the half-loops twice).
    [Edit] I just noticed that you integrated 1/2r^2d(theta)... why?

    The function is r=1-2cos(theta). Should we not be integrating 1-2cos(theta)?

    Suppose we integrate 1-2cos(theta) from 0 to 2pi:

    INT[0,2pi](1-2cos(theta))d(theta) = (theta)-2sin(theta)[0,2pi] = (2pi-0)-(0) = 2pi = 6.2831

    If, instead, we integrate from pi/3 to pi, subtract the absolute value of the integration from 0 to pi/3, and multiply the result by 2, we get:

    2{ INT[pi/3,pi](1-2cos(theta))d(theta) - | INT[0,pi/3](1-2cos(theta))d(theta) | }
    2{ (theta-2sin(theta))[pi/3,pi] - | (theta-2sin(theta))[0,pi/3] | }
    2{ [(pi-0)-(pi/3-sqrt(3))] - | [(pi/3-sqrt(3))-(0-0)] | } = 2{ (3.826446) - |-.68485|} = 6.2831

    The two integratations are equal because the integration between 0 and pi/3 is negative, thus it subtracts from the total integration

    [Edit 2] Given what I did just above, I noticed that I made two mistakes in my first post: 1) the integrations from 0 to pi gives the area of the inner loop, and 2) the inner loop has the negative radius and thus gives a negative area.
    Last edited by ecMathGeek; March 20th 2007 at 06:09 PM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member ecMathGeek's Avatar
    Joined
    Mar 2007
    Posts
    436
    Quote Originally Posted by ThePerfectHacker View Post
    Thank you
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: December 13th 2011, 09:11 PM
  2. More polar help needed
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 28th 2009, 03:08 PM
  3. Calculus help needed
    Posted in the Calculus Forum
    Replies: 2
    Last Post: November 18th 2007, 03:44 PM
  4. Calculus Help Needed!
    Posted in the Calculus Forum
    Replies: 2
    Last Post: January 6th 2007, 03:27 PM
  5. Help In Calculus Needed
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 18th 2006, 10:39 PM

Search Tags


/mathhelpforum @mathhelpforum