# Green's Theorem to find Area help

1. Apr 22, 2015

### RJLiberator

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

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?

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2. Apr 22, 2015

### RJLiberator

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
3. Apr 22, 2015

### RJLiberator

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