5

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.

Krysten
  • 779

2 Answers2

8

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,

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

robjohn
  • 345,667
hawdin
  • 1
  • 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 00:33