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October 25th, 2014, 09:15 AM   #1
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Finding the area inside a cardiod

I messed up towards the end but can't find my error.

$\displaystyle 2[1/2∫ (π/2) to (-π/2) , (1 -2sin(θ) + sin^2(θ))$d(θ)

I made two integrals

$\displaystyle 2[1/2∫(π/2) to (-π/2), (1-sin(θ))] dθ+ 2[1/2∫(π/2) to (-π/2), (sin^2(θ))] dθ $

Then I derived it.

2[1/2 (((θ)+2cos(θ) + 1-cos(2θ)))] π/2 to - π/2




When evaluate each side I keep getting the wrong answer.

I keep getting 0 or π, I have been working on it for 3 days already.
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Last edited by j423; October 25th, 2014 at 09:17 AM.
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October 25th, 2014, 09:45 AM   #2
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Quote:
Originally Posted by j423 View Post
I messed up towards the end but can't find my error.

$\displaystyle 2[1/2∫ (π/2) to (-π/2) , (1 -2sin(θ) + sin^2(θ))$d(θ)

I made two integrals

$\displaystyle 2[1/2∫(π/2) to (-π/2), (1-sin(θ))] dθ+ 2[1/2∫(π/2) to (-π/2), (sin^2(θ))] dθ $

Then I derived it.

2[1/2 (((θ)+2cos(θ) + 1-cos(2θ)))] π/2 to - π/2




When evaluate each side I keep getting the wrong answer.

I keep getting 0 or π, I have been working on it for 3 days already.
how did you get 1-cos(2θ) as an antiderivative?

For your last term in your original integrand ...

$\displaystyle \sin^2{\theta} = \frac{1-\cos(2\theta)}{2}$

antiderivative is $\displaystyle \frac{1}{2}\left(\theta - \frac{\sin(2\theta)}{2}\right)$
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October 25th, 2014, 10:11 AM   #3
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I can't see where the 2 in the $2\cos \theta$ term came from either.
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October 25th, 2014, 01:08 PM   #4
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Yeah, I see that now. I did something weird on my paper too! thanks
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