The Problem
There's an exercise in the MIT OCW 18.01SC course:
What is the average distance from the $x$-axis of a point chosen at random on the cardioid $r = a (1 - \cos (\theta))$, if the point is chosen b) by letting a point $P$ travel around the cardioid at uniform velocity, stopping at random;
I managed to find the answer correctly, my solution is long, but straightforward. The solution given in the answers is much shorter, but I don't understand it.
MIT Solution
In the proposed solution they are averaging the expression
$$ \frac 1 {8a} \int_{-\pi}^{\pi} {| r \sin \theta | a \sqrt {2 - 2 \cos \theta} d\theta} $$
which is actually
$$ \frac 1 {8a} \int_{-\pi}^{\pi} { d(\theta) \frac {dw} {d\theta} d\theta }$$
as I understand because it stems from finding the arc length element $dw$ as:
$$ dw = {\sqrt {\left(\frac{dx}{d\theta}\right)^2 + \left(\frac{dy}{d\theta}\right)^2 } } d\theta \\ = \sqrt { \left(\frac {dr} {d\theta}\right)^2 + r^2 } d\theta \\ = a \sqrt {2 - 2 \cos \theta} d\theta$$
I don't quite understand their idea of finding the average.
My Solution
I considered just the upper part of the cardioid because of the symmetry: $\theta: 0 \dots \pi$.
I aimed to find the distance as a function of the arclength: $d(w)$ and calculate the average as $$ \frac {\int_{0}^{w(\pi)}{d(w) dw}} {\int_{0}^{w(\pi)}{dw}} $$
As I know the distance as a function of $\theta$: $d(\theta) = r(\theta) sin(\theta)$, I wanted to find $\theta$ as the function of the arclength: $\theta(w)$.
I found at first $$w(\theta) = \int_0^{\theta} {\sqrt {\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 } dt \\ = 4a (1 - \cos \frac \theta 2) } $$
Hence
$$ \theta = 2 \cos ^ {-1} (1 - \frac w {4a}) $$
Plugging it into $$ {\int_{0}^{w(\pi)}{d(w) dw}} = \\ {\int_{0}^{w(\pi)}{r(\theta) * \sin(\theta) dw}} $$
I integrated it using some trigonometry into the correct answer $\frac {4a} 5$
The MIT course result looks ok to me. A differential approach is shorter:
$$ r= a- a \cos \theta \tag1$$
Differentiate with respect to arc ($ \dfrac {dr}{ds}=\cos \psi )$ which is angle between tangent and and radius vector.
$$\cot \psi= a \sin \theta /r \tag2$$
where $\sin \theta $ is understood to be taken absolute value.
Average y-value numerator denominator is (total arc of cardioid is $8a$ ):
$$\int r \sin \theta \, ds =\int r \sin \theta \, \frac{ds}{d\theta} d\theta =\int r \sin \theta \, \frac {r}{\sin \psi} d\theta \tag3 $$
Plug in from (2) numerator $$ \int r^2 \frac{\sin \theta}{\sin \psi} d\theta=\int r^2 \sqrt{1+\cot{^2}\psi} \sin \theta d\theta = r \sqrt{r^2+ a^2 \sin^2 \theta}\, \sin \theta \,d \theta \tag4 $$
Plug in from (1) and simplify $$ \bar y = \int a^2 \sqrt2 \, (1-\cos \theta)^{3/2}\,|\sin\theta| \,d \theta / (8a)\tag5 $$ and further.