# Find the arc length of the cardioid: r = 3-3cos θ

This is what I have so far:

Using the formula $\mathrm ds = \sqrt{r^2 + \left(\frac{\mathrm dr}{\mathrm dθ}\right)^2}$

$$\frac{\mathrm dr}{\mathrm d\theta} = 3\sin\;\theta$$

$$r^2 = 9 - 18\cos\;\theta + 9\cos^2\theta$$

$$\mathrm ds = \sqrt{9 - 18\cos\;\theta + 9\cos^2 \theta + 9\sin^2 \theta}$$

using $\cos^2 \theta + \sin^2 \theta = 1$,

$$\mathrm ds = \sqrt{18(1-\cos\;\theta)}$$

Then I have

$$\int_0^{2\pi} \sqrt{18(1-\cos\;\theta)}\mathrm d\theta$$

But I'm not sure how to integrate this.

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$1-\cos\;\theta=2\sin^2\frac{\theta}{2}$ is helpful here. –  Ｊ. Ｍ. Apr 29 '11 at 3:11
On another note: it is profitable to exploit any symmetry (usually) present in curves represented in polar coordinates. In the cardioid's case, it's symmetric about the horizontal axis. You can thus just consider the arclength of only the upper portion, and then double that result afterward. –  Ｊ. Ｍ. Apr 29 '11 at 3:23
Does the θ/2 come from the fact that the original trig identity is 1-cos(2x) = 2 sin^2 x and you need the θ/2 to cancel out with the 2x? –  Krysten Apr 29 '11 at 3:31
right on the nose. :) –  Ｊ. Ｍ. Apr 29 '11 at 3:38
alright thanks! –  Krysten Apr 29 '11 at 3:39

So that this does not remain unanswered:

Exploiting the trigonometric identity (which can be obtained from the cosine's double-angle formula):

$$1-\cos\;\theta=2\sin^2\frac{\theta}{2}$$

$$6\int_0^{2\pi} \sin\frac{\theta}{2}\mathrm d\theta$$

which you should be able to handle.

$$2\int_0^{\pi} \sqrt{18(1-\cos\;\theta)}\mathrm d\theta$$

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$$1−\cos(\theta)=2\sin^2\left(\frac{\theta}{2}\right)$$