Am I Doing This Area Problem Correctly?

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Homework Help Overview

The discussion revolves around finding the area enclosed by the polar curve r=1-sin(θ) while also being outside the circle r=1. The original poster is attempting to set up the problem correctly and has provided an integral for evaluation.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster describes their approach of integrating over half of the area due to symmetry and presents their integral setup. Some participants question the completeness of the problem setup and whether a diagram is necessary. Others offer alternative results and suggest checking the calculations.

Discussion Status

The discussion is ongoing, with various participants providing feedback on the setup and results. While some guidance has been offered regarding the correctness of the setup, there is no explicit consensus on the final answer, and multiple interpretations of the results are being explored.

Contextual Notes

Participants note the absence of a diagram or complete problem statement, which may affect the clarity of the discussion. The original poster mentions using a sketch and a tool for verification.

Lancelot59
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Hi there. The book doesn't have an answer in the back for this problem, so I wanted to know if I was setting everything up correctly.

I need to find the area enclosed by r=1-sin(theta) but outside r=1. This is polar by the way. So a nice simple cardioid and circle. I decided to only do half and then double it, because of the symmetry.

Here is the integral I got.
[tex] 2*\frac{1}{2}[\int_{\frac{-\pi}{2}}^{0}{(1-sin(\theta))^{2} d\theta} - \int_{\frac{-\pi}{2}}^{0}{(1)} d\theta}][/tex]

My answer was [tex]\frac{5\pi}{2}+\frac{9}{2}[/tex]. I'm fine with integrating, I'm just wondering if I set it up correctly.
 
Last edited:
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I am not that good at calculus,but perhaps if you could send me the complete question i might be able to help. Ezim.
 
That is the complete question. Find the area enclosed by r=1-sin(theta) but outside r=1.
 
you mean there is no diagram or figure?
 
EzimOsai said:
you mean there is no diagram or figure?
Nothing is given. I just drew a quick sketch and checked it on wolfram alpha.
 
You set-up looks fine, but your final result isn't correct. Check your work.
 
Oh...5/2 + pi/4 I guess.
 
I get 2+pi/4.
 
...This is getting really silly really fast. Well we established that it's just an algebra issue. I'll work it out later. Thanks.
 
Last edited:
  • #10
yes. I solved it with double integral and I got 2+pi/4.
 

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