Finding the Area Enclosed by a Polar Curve

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    Area Curve Polar
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Homework Help Overview

The discussion revolves around finding the area enclosed by a polar curve, specifically involving the integration of a function defined in polar coordinates. Participants are exploring the correct limits of integration and the setup for the problem.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the integration limits and question the choice of limits from -π/2 to 0. There is an exploration of whether a sketch has been made and what values of θ correspond to negative r values for the inner loop. Some participants suggest alternative limits involving arcsin, while others confirm the correctness of these limits.

Discussion Status

The discussion is active, with participants providing guidance on the limits of integration and confirming the correctness of proposed values. There is a mix of interpretations regarding the setup, but productive dialogue is ongoing.

Contextual Notes

Participants mention potential mistakes in their initial approaches and express uncertainty about the limits of integration, indicating a need for clarification on the polar curve's behavior.

amninder15
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Can someone please help me on this question. I tried to solve it by integrating 0.5*(1-3sin(θ)^2 from -Pi/2 to 0 but I didnt get the answer.
 

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amninder15 said:
Can someone please help me on this question. I tried to solve it by integrating 0.5*(1-3sin(θ)^2 from -Pi/2 to 0 but I didnt get the answer.

Why did you choose ##-\frac \pi 2## to ##0##? Just guessing? Have you drawn a sketch? Do you know what ##\theta## give negative ##r## values for the inner loop?
 
Yea I made some silly mistake I think It goes from sin^-1(1/3) to Pi-sin^-1(1/3). But that looks ugly. Am I on right path.
 

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Your graph is correct for ##\theta## varies between 0 and 2pi. Now, generally how do you find the area of a polar region?
 
amninder15 said:
Yea I made some silly mistake I think It goes from sin^-1(1/3) to Pi-sin^-1(1/3). But that looks ugly. Am I on right path.

Yes. Those are the correct limits. Ugly or not, just plow ahead and you will get one of the answers listed.
 

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