Parametric equations and polar coordinates

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Homework Help Overview

The discussion revolves around finding the area enclosed by the inner loop of the polar curve defined by the equation r = 1 - 3sin(θ). Participants are exploring the necessary integral and the appropriate limits for integration.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants are attempting to determine the correct interval for integration by finding when r = 0. There is uncertainty regarding the complexity of the resulting angles and whether to use arcsin or another form for the limits.

Discussion Status

Some participants have provided hints and insights into the relationships between sine and cosine, while others are questioning the interpretation of the intervals. The discussion is ongoing with various interpretations being explored.

Contextual Notes

There is mention of potential complications due to the non-special angle involved in the calculations, as well as references to external posts for additional context.

maiad
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Homework Statement


Find the area enclosed by the inner loop of the curve r=1-3sinθ


Homework Equations


A=o.5\int r^2 dθ


The Attempt at a Solution


I found the integral but i don't know how to find the interval at which i will be integrating from. I tried finding when r=0 and it turns out to be sin^-1(1/3) but it's not a special angle so it would be messy if i plugged that interval in cos(θ).Any hints will be appreciated
 
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\sin \theta = \frac{1}{3}
\cos^2 \theta = 1 - \sin^2 \theta = \frac {8}{9}
\cos \theta = \pm \frac {2 \sqrt 2}{3}
So it's not a problem. You have to figure out whether it's + or -, depending on theta.
 
the answer had a sin(1/3)^-1 so i think you have to use that as one of your intervals I am assuming
 
maiad said:
the answer had a sin(1/3)^-1 so i think you have to use that as one of your intervals I am assuming
Do you mean (sin(1/3))^{-1}= 1/sin(1/3)) or sin^{-1}(1/3)= arcsin(1/3)?
 
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