# Help finding the arc length of a polar curve?

Find the arc length of the polar curve $r = 8\sin^3\left(\frac{\theta}3\right)$ from $0\leq\theta\leq \frac{\pi}4$.

I found $\frac{dr}{d\theta}=8\sin^2\left(\frac{\theta}3\right)\cos\left(\frac{\theta}3\right)$ and plugged it into the formula:

$$\int _0^{\frac{\pi }{4}}\sqrt{64\sin^6\left(\frac{\theta }{\:3}\right)+8\sin^2\left(\frac{\theta }{3}\right)\cos\left(\frac{\theta }{3}\right)}\:d\theta$$

(Sorry I'm not sure how to insert this, here is a link)

How would I go about simplifying this integral?

• You should have $$s = \int_0^{\frac \pi 4} \sqrt{r^2 + \left(\frac{\mathrm dr}{\mathrm d\theta}\right)^2} \mathrm d\theta$$, you have seemingly missed the square unless I'm missing something. – George Coote Sep 19 '17 at 9:44

Per my comment, it looks like you've used,

$$s = \int_0^{\frac \pi 4} \sqrt{r^2 + \left(\frac{\mathrm dr}{\mathrm d\theta}\right)} \mathrm d\theta$$

$$s = \int_0^{\frac \pi 4} \sqrt{r^2 + \left(\frac{\mathrm dr}{\mathrm d\theta}\right)^2} \mathrm d\theta$$

Hence you should have,

$$s = \int_0^{\frac \pi 4} \sqrt{64\sin^6\left(\frac \theta 3\right) + 64\sin^4\left(\frac \theta 3\right)\cos^2\left(\frac \theta 3\right)} \mathrm d\theta$$

Using the Pythagorean identity, $$s = \int_0^{\frac \pi 4} \sqrt{64\sin^6\left(\frac \theta 3\right) + 64\sin^4\left(\frac \theta 3\right)\left(1-\sin^2\left(\frac \theta 3\right)\right)}\mathrm d\theta$$

Expanding, $$s = \int_0^{\frac \pi 4} \sqrt{64\sin^6\left(\frac \theta 3\right) + 64\sin^4\left(\frac \theta 3\right) - 64\sin^6\left(\frac \theta 3\right)}\mathrm d\theta$$

Hence, $$s = 8\int_0^{\frac \pi 4} \sqrt{\sin^4\left(\frac \theta 3\right)}\mathrm d\theta$$

Which can be evaluated rather straightforwardly.

• Your $r^2$ should be $1$ – Jan Sep 19 '17 at 11:41
• no, it's correct as-is, note that we're working in polar coords – George Coote Sep 19 '17 at 15:07