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Green's Theorem to find Area help

  1. Apr 22, 2015 #1

    RJLiberator

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    Gold Member

    1. The problem statement, all variables and given/known data
    Find the area of the right leaf of the Lemniscate of Gerono (the ∞ sign, see figure below) parametrized by
    r(t)= <sin(t), sin(t)cos(t)>
    from 0=<t=<pi

    Picture is uploaded.

    2. Relevant equations
    Green's theorem: integral of fdx+gdy = double integral (over the region) of (gx-fy) dA
    Green's theorem used to compute the area of R = double integral (over the region R) of 1dA

    3. The attempt at a solution
    1. First, I found the curl, I let g=x and f=0 to make the curl (gx-fy) =1.
    So now this is in Green's Theorem area form.
    2. Now I need to find the bounds of integration.
    I am stuck here and not entirely sure what to do. Any guidance on how to proceed?
     

    Attached Files:

  2. jcsd
  3. Apr 22, 2015 #2

    RJLiberator

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    I've been working on this problem this morning.

    Is the answer for area simply 0? It seems logical as the postive y area and the negative y area would cancel each other out.

    What I did was use:
    F = <0, sin(t)>
    r'(t) = <cos(t), -sin^2(t)+cos^2(t)>

    I then performed a dot product between them and took the integral with t going from 0 to pi as the problem stated.

    I ended up after integration with
    -(2/3)cos^3(t)+cos(t) from t=0 to t=pi to equal -2/3
    The answer then would just be 2/3 = area?

    Correct?
     
    Last edited: Apr 22, 2015
  4. Apr 22, 2015 #3

    RJLiberator

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    Gold Member

    Could I just go from t =0 to t=pi/2
    then multiply the answer by 2, to get 2/3 as the area?
     
    Last edited: Apr 22, 2015
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