Green's Theorem to find Area help

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SUMMARY

The discussion centers on using Green's Theorem to calculate the area of the right leaf of the Lemniscate of Gerono, parametrized by r(t) = for 0 ≤ t ≤ π. The user successfully computed the curl, resulting in (gx - fy) = 1, and applied the theorem to find the area. After performing the necessary integrations, the user concludes that the area is 2/3, suggesting that integrating from t = 0 to t = π/2 and multiplying by 2 is a valid approach to confirm this result.

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  • Knowledge of integration techniques, particularly definite integrals.
  • Basic concepts of area calculation in multivariable calculus.
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Students and educators in calculus, particularly those focusing on vector calculus and area calculations using Green's Theorem.

RJLiberator
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Homework Statement


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.

Homework 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

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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I've been working on this problem this morning.

Is the answer for area simply 0? It seems logical as the positive 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:
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:

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