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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?

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  • $\begingroup$ 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. $\endgroup$ – George Coote Sep 19 '17 at 9:44
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Per my comment, it looks like you've used,

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

Instead of

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

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.

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

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