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Length of a Cardioid

  1. Apr 23, 2007 #1
    Hi,


    The final answer below is wrong. Is should be 8a! Not sure where I made the error below?


    Also is there another substitution that can be used at the earlier line
    [tex] a \int_0^{2 \pi} \{ 2( 1 - \sin \phi) \}^{\frac{1}{2}} d\phi [/tex]
    to subtitute for the [tex] (1 - \sin \phi) [/tex] without the need to use
    [tex] s = \frac{\pi}{2} - t [/tex].

    This would also be very useful for the next part where I have to calculate the area of surface of revolution!!

    (ii)

    [tex] L = \int_0^{2 \pi} \{ r^2 + (\frac{dr}{d \theta})^2 \}^{\frac{1}{2}} d \phi [/tex]

    [tex] \int_0^{2 \pi} \{ a^2(1- \sin \phi)^2 + (-a \cos \phi)^2 \}^{\frac{1}{2}} d\phi [/tex]

    [tex] a \int_0^{2 \pi} \{ 2( 1 - \sin \phi) \}^{\frac{1}{2}} d\phi [/tex]

    using [tex] s = \frac{\pi}{2} - t [/tex]

    [tex] \frac{ds}{dt} = -1 [/tex]

    [tex] -a \int_{\frac{\pi}{2}}^{\frac{-3\pi}{2}} \{ 2( 1 - \cos \phi) \}^{\frac{1}{2}} d\phi [/tex]

    [tex] a \int_{\frac{-3\pi}{2}}^{\frac{\pi}{2}} \{ 4 \sin^2 (\frac{\phi}{2}) \}^{\frac{1}{2}} d\phi [/tex]

    [tex] a \int_{\frac{-3\pi}{2}}^{\frac{\pi}{2}} \{ 2 \sin (\frac{\phi}{2} \} d\phi [/tex]

    [tex] a [ -4 \cos (\frac{\phi}{2} ]_{\frac{-3\pi}{2}}^{\frac{\pi}{2}} [/tex]

    [tex] = 4a(\frac{1}{2 \sqrt{2}}) [/tex]
     
  2. jcsd
  3. Apr 27, 2007 #2
    Note that [tex] a \int_{\frac{-3\pi}{2}}^{\frac{\pi}{2}} \sqrt{4 \sin^2 (\frac{\phi}{2}) } \ d\phi = 2a \int_{\frac{-3\pi}{2}}^{\frac{\pi}{2}} \mid \sin (\frac{\phi}{2}) \mid d\phi [/tex]
     
    Last edited: Apr 27, 2007
  4. Apr 28, 2007 #3
    ah got it!

    Thanks Pizzasky
     
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