Mean value of $r$ with respect to $\theta$ between two limits of a polar curve

Question

Given a polar equation, what is the equation (and how does one derive it) for the mean value of $r$ with respect to $\theta$ between two limits, say $a$ and $b$?

Answer 1

If you've studied calculus, you'd know about the mean value of a function. In rectangular coordinates, it's quite simple: Area/Length = Average Height. ($\frac{\int_a^b f(x) dx}{(b-a)}$)

For example: What's the average of the function $\sin(x)$ between $0$ and $\pi$?

Our first step should be to integrate: $\int_0^{\pi} \sin(x) dx = -\cos(x) |_0^{\pi} = -\cos(\pi) + \cos(0) = 2$. Then we have to divide by the length of the function we are considering to get the average height.
Final answer: $\frac2\pi$.

It's quite the same with polar coordinates. Integrate between your two angles and divide by their difference. Let's try it again with $r=\sin(\theta)$ now.

We have to integrate from $0$ to $\pi$. The formula for integrating in polar coordinates is $\int \frac12r^2 d\theta$.

So first step: $\int_0^{\pi} \frac12\sin^2(\theta)\ d\theta$. Use a trig identity to expand $\sin^2(\theta)$.
$\int_0^{\pi}\frac12\left(\frac{1-\cos(2\theta)}2\right)\ d\theta$
$=\frac14\int_0^{\pi}1-\cos(2\theta)\ d\theta$
$=\frac14[\theta-\frac12\sin(2\theta)]_0^{\pi} = \pi/4$

Bonus knowledge: How do you get the formula for integrating polar coordinates? It looks similar to the formula for area of a triangle! In fact, that's it. The triangle has a height of $r$, and its base has a length of $r d\theta$, and we sum up all these $d\theta$'s, and that's how we get $\int \frac12 \cdot r \cdot rd\theta = \int \frac12r^2\ d\theta$