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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
up vote 7 down vote accepted

So that this does not remain unanswered:

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


your integral turns into

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

which you should be able to handle.

Alternatively, since the cardioid is symmetric about the horizontal axis, you can instead start with the integral

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

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This answer is correct and equal to 24.

Hint: Also note,


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Welcome to MSE! Your answer is correct, but you may want to format it in a clearer way for the reader since you are making two individual statements. Regards – Amzoti Jan 28 '13 at 22:32
J.M. notes the trig identity in a comment on the question and in his answer. – robjohn Jan 29 '13 at 0:33

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