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Math Help - Find circumference of cardioid

  1. #1
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    Find circumference of cardioid

    The cardioid r=a(1-\sin\theta); find the length of its circumference.

    The formula for polar equations:
    \displaystyle s=\int \left( r^2+\left[\frac{dr}{d\theta}\right]^2\right)^{\frac{1}{2}} d\theta

    \frac{dr}{d\theta}=-a\cos\theta
    \displaystyle s=2\int^{\pi}_0\left(a^2(1-\sin\theta)^2+a^2\cos^2\theta\right)^{\frac{1}{2}}  d\theta
    \displaystyle s= 2a\int^{\pi}_0 \left(1-2\sin\theta+\sin^2\theta+\cos^2\theta\right)^{\fra  c{1}{2}}
    \displaystyle s=2a\int^{\pi}_0\left(2(1-\sin\theta)\right)^{\frac{1}{2}}d\theta

    (1-\sin\theta)^{1/2} is what stumps me, how do i convert that to something that can be integrated?
    Thanks
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  2. #2
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    Quote Originally Posted by arze View Post
    The cardioid r=a(1-\sin\theta); find the length of its circumference.

    The formula for polar equations:
    \displaystyle s=\int \left( r^2+\left[\frac{dr}{d\theta}\right]^2\right)^{\frac{1}{2}} d\theta

    \frac{dr}{d\theta}=-a\cos\theta
    \displaystyle s=2\int^{\pi}_0\left(a^2(1-\sin\theta)^2+a^2\cos^2\theta\right)^{\frac{1}{2}}  d\theta
    \displaystyle s= 2a\int^{\pi}_0 \left(1-2\sin\theta+\sin^2\theta+\cos^2\theta\right)^{\fra  c{1}{2}}
    \displaystyle s=2a\int^{\pi}_0\left(2(1-\sin\theta)\right)^{\frac{1}{2}}d\theta

    (1-\sin\theta)^{1/2} is what stumps me, how do i convert that to something that can be integrated?
    Thanks
    integrate Sqrt[1 - Sin[x]] - Wolfram|Alpha

    Click on Show steps.
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    Ok, so i did that, and got:
    2a\sqrt{2} [2\sqrt{\sin\theta+1}]^{\pi}_0
    which gives me 0.
    I tried adding the integrals between \frac{\pi}{2} and 0, and \pi and \frac{\pi}{2} they still add to 0.
    Thanks!
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  4. #4
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    Hello, arze!

    The cardioid r=a(1-\sin\theta); find the length of its circumference.

    \displaystyle s \;=\;2\sqrt{2}\,a\int^{\pi}_0\left(1-\sin\theta\right)^{\frac{1}{2}}d\theta

    We have: . \sqrt{1-\sin\theta}

    Multiply by \frac{\sqrt{1+\sin\theta}}{\sqrt{1+\sin\theta}}:\;  \;\dfrac{\sqrt{1-\sin\theta}}{1}\cdot\dfrac{\sqrt{1+\sin\theta}}{\s  qrt{1+\sin\theta}} \;=\;\dfrac{\sqrt{1-\sin^2\theta}}{\sqrt{1+\sin\theta}}

    . . . . . . . . . . . . . =\;\dfrac{\sqrt{\cos^2\theta}}{\sqrt{1+\sin\theta}  } \;=\;\dfrac{\cos\theta}{\sqrt{1+\sin\theta}}


    \text{The integral becomes: }\;\displaystyle 2\sqrt{2}\,a\int (1 + \sin\theta)^{-\frac{1}{2}}\,\cos\theta\,d\theta
    \text}Let }\,u \:=\:1 + \sin\theta

    Got it?
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  5. #5
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    yes, got it!
    Thanks!
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  6. #6
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    ok, now i have another problem. between pi and 0, u has the same values, what values should i use? for the angles?
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