Green's Theorem to find Area help

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The discussion focuses on using Green's Theorem to find the area of the right leaf of the Lemniscate of Gerono, parametrized by r(t) = from t = 0 to pi. The user successfully calculated the curl, determining that (gx - fy) = 1, which is suitable for applying Green's Theorem. They expressed uncertainty about the bounds of integration and whether the area could be zero due to potential cancellation between positive and negative regions. After performing the necessary calculations, they arrived at an area of 2/3 and questioned if they could simplify the process by calculating from t = 0 to pi/2 and then doubling the result. The area calculation appears correct, confirming the area of the right leaf is indeed 2/3.
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:
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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