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.

  • 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
    J.M. notes the trig identity in a co$m$$m$ent on the question $a$nd in his answer.2013-01-29