5
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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.

  • 3
    $1-\cos\;\theta=2\sin^2\frac{\theta}{2}$ is helpful here.2011-04-29
  • 1
    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.2011-04-29
  • 0
    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?2011-04-29
  • 0
    right on the nose. :)2011-04-29

2 Answers 2

7

So that this does not remain unanswered:

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

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

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

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

-1

This answer is correct and equal to 24.

Hint: Also note,

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

  • 0
    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. Regards2013-01-28
  • 0
    J.M. notes the trig identity in a comment on the question and in his answer.2013-01-29