# Length of a Cardioid

1. Apr 23, 2007

### 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 $$s = \frac{\pi}{2} - t$$

$$\frac{ds}{dt} = -1$$

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

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

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

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

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

2. Apr 27, 2007

### pizzasky

Note that $$a \int_{\frac{-3\pi}{2}}^{\frac{\pi}{2}} \sqrt{4 \sin^2 (\frac{\phi}{2}) } \ d\phi = 2a \int_{\frac{-3\pi}{2}}^{\frac{\pi}{2}} \mid \sin (\frac{\phi}{2}) \mid d\phi$$

Last edited: Apr 27, 2007
3. Apr 28, 2007

### Bazman

ah got it!

Thanks Pizzasky

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