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

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.

• You wrote $\int_{\pi/3}^{5\pi/3}$. But this means you go from $\pi/3$ to $5\pi/3$. Really you wanted to go from $-\pi/3$ to $\pi/3$. So this should be your integration limits. – John Doe Feb 13 '18 at 13:25
• Going from $-\pi/3$ to $\pi/3$ would result in including every part of the circumference of the circle besides the parts I want. @JohnDoe – joshuaheckroodt Feb 13 '18 at 13:28
• ... Actually you wanted to go from $-\pi/2$ to $-\pi/3$ and again from $\pi/3$ to $\pi/2$... Those angles correspond to the arc in the question. The angles $\pi/3$ to $5\pi/3$ correspond to the arc starting where it should, going all the way around the circle and ending where it should, i.e. it is by the whole circumference of the circle bigger than desired. That is why you got the result that is bigger than the correct result ($4\pi/3$) by the circumference ($4\pi$). – user491874 Feb 13 '18 at 13:29
• Thank you for the insight, but unfortunately that still does not explain to me why a valid upper bound is $2\pi/3$ :/ @user8734617 – joshuaheckroodt Feb 13 '18 at 13:31
• Because the interval $[-\pi/2, -\pi/3]$ gives you the same result as $[\pi/2, 2\pi/3]$ (the function under the integral is periodic with period $\pi$), so instead of joining intervals $[-\pi/2,-\pi/3]$ and $[\pi/3, \pi/2]$ you can join $[\pi/3, \pi/2]$ and $[\pi/2, 2\pi/3]$ and integrate on $[\pi/3,2\pi/3]$. – user491874 Feb 13 '18 at 13:33