What Is the Correct Way to Calculate the Length of a Cardioid?

[Bazman]

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
a \int_0^{2 \pi} \{ 2( 1 - \sin \phi) \}^{\frac{1}{2}} d\phi
to subtitute for the (1 - \sin \phi) without the need to use
s = \frac{\pi}{2} - t.

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

(ii)

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

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

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

using s = \frac{\pi}{2} - t

\frac{ds}{dt} = -1

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

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

a \int_{\frac{-3\pi}{2}}^{\frac{\pi}{2}} \{ 2 \sin (\frac{\phi}{2} \} d\phi

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

= 4a(\frac{1}{2 \sqrt{2}})

 

[pizzasky]

Note that 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

 

[Bazman]

ah got it!

Thanks Pizzasky