# Why does the integral for the arc length of a polar curve have the boundaries it has?

by joshuaheckroodt   Last Updated February 13, 2018 13:20 PM

So I've been working on the question below for a bit, and after arriving at what I thought was the answer, I found out that I was wrong, because the bounds I used for integration was incorrect.

Here is the problem statement.

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So, the first thing I did was solve for the values of $\theta$ where an intersection occurs, and obviously these values are $\theta_1=\frac{\pi}{3}$ and $\theta_2=\frac{5}{3}\pi$. From here, I found the value of $(4\cos\theta)^2$ as well as the value of $\left(\frac{d}{dx}(4\cos\theta)\right)^2$, and then substituted these values into the formula for the arc length of a polar curve:

$$\implies \int_\alpha^\beta\sqrt{f(\theta)^2+f'(\theta)^2}~d\theta=\int_{\frac{1}{3}\pi}^{\frac{5}{3}\pi}\sqrt{16cos^2\theta+16\sin^2\theta}~d\theta$$

$$=4\int_{\frac{1}{3}\pi}^{\frac{5}{3}\pi}~d\theta$$

$$=4\left(\frac{5}{3}\pi-\frac{1}{3}\pi\right)$$

$$=\frac{16}{3}\pi$$

Although this answer was wrong, because according to the marking scheme, the correct bounds did not include $\frac{5}{3}\pi$, but rather included $\frac{2}{3}\pi$. Can someone perhaps explain to me why this is?

Any help is appreciated, thank you.

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