Am I Doing This Area Problem Correctly?

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SUMMARY

The discussion centers on calculating the area enclosed by the polar curve r=1-sin(θ) while excluding the area of the circle defined by r=1. The user initially set up the integral correctly but arrived at an incorrect final answer of 5π/2 + 9/2. Other participants confirmed the setup and provided the correct result of 2 + π/4, indicating that the discrepancy was due to an algebraic error. The use of symmetry to simplify the calculation by integrating half of the area is noted as a valid approach.

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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.
<br /> 2*\frac{1}{2}[\int_{\frac{-\pi}{2}}^{0}{(1-sin(\theta))^{2} d\theta} - \int_{\frac{-\pi}{2}}^{0}{(1)} d\theta}]<br />

My answer was \frac{5\pi}{2}+\frac{9}{2}. 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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