Average distance of two points on a circle I stumbled upon the question of the average distance of two points on a circle, I learned how to calculate this with polar cordinates (find the distance as a function of $\theta$, itegrate the general distance for $\theta$  in [0, $\pi$] and divide by $\pi$), I wondered though if the same calculations be done in cartesian coordinates?
I tried but instead of $\frac{4R}{\pi}$ I got $\frac{4R}{3}$, can it be done or is there so fundamental flaw in using cartesian?
Here is how I solved it.
I used the same symmetry argument of equal avarage distances for all the points to reduce the problem for integrating the average distance from the point (R;0).
Than I used the of symmetry of equal distances from the point around the x-axis to reduce the problem to itegrating only along the upper semi circle.
The general point on the circle is ($x$;$\sqrt{R^2 -x^2}$) where $x$ can be any value between -R and R, so the average distance integral is.
    $\int_{-R}^R \frac{(\sqrt{(R - x)^2 + R^2 - x^2})}{2R} \, dx$ = $\int_{-R}^R \frac{\sqrt{2 * R^2 - 2x*R}}{2R} \, dx$ (the 2R in the denominator comes from the fact that for average integral we need to divide by the size of the domain of the variable we itegrate over, which is  R - -R = 2R).
which pans out to be $\frac{4R}{3}$.
 A: What you did differently in Cartesian coordinates was you weighted the average in favor of the parts of the semicircle that are parallel to the  $x$ axis, about halfway between the ends. 
That is, you are (in effect) projecting all the points into the $x$ axis before averaging them by the length of each projected piece. If you take an arc of the semicircle of length $0.01$ near $(0,R)$, it projects onto a segment almost that long on the $x$ axis. But compare this to arcs of the same size near $(R,0)$ or $(-R,0).$ They project onto much smaller segments. Since you’re just simply integrating over $x$, the segment near $(0,R)$ gets a lot more “votes” toward the average than the other two segments do. 
There is more than one way to fix this. 
One way is, Instead of just $dx$, you can write $\frac1{\sqrt{1-x^2}}dx.$ This equalized the weights given to each part of the semicircle.
Another way is to integrate parametrically along the semicircle. For example, let the Cartesian coordinates of a point on the semicircle be $(R\cos(\pi t),R\sin(\pi t))$, and integrate over $t$ from $t=0$ to $t=1$.

It's not really the coordinates, it's how you use them that makes the "average distance" calculation valid or flawed.
The thing that makes an average distance an "average" is that every part of the circle gets counted equally. This is easy to do in polar coordinates for a circle around the origin because the size of an arc of that circle is proportional to $\Delta \theta$. 
(If you put the circle somewhere else, this might not be true.)
But if you're going to use Cartesian coordinates, isn't $dy$ just a valid as $dx$? Let's set this up as an integral over $y$.
By symmetry, again, we can again look just at the semicircle where $y$ is positive. But now we have to consider two formulas for a point on the semicircle:
$(-\sqrt{R^2 - y^2}, y)$ or $(\sqrt{R^2 - y^2}, y)$.
The first formula gives the left quarter circle, the second formula gives the right quarter circle. 
The distance formula on the left is
$$ \sqrt{\left(-\sqrt{R^2 - y^2} - R\right)^2 + \left(y - 0\right)^2}
= \sqrt{2R\left(R + \sqrt{R^2 - y^2}\right)}
$$
and on the right it is
$$ \sqrt{\left(\sqrt{R^2 - y^2} - R\right)^2 + \left(y - 0\right)^2}
= \sqrt{2R\left(R - \sqrt{R^2 - y^2}\right)}.
$$
So although we only have to integrate over positive $y$, we're doing it twice, so the denominator for the average is $2R$. This gives us
\begin{multline}
 \frac1{2R}\left(\int_0^R \sqrt{2R\left(R + \sqrt{R^2 - y^2}\right)}\,dy
+ \int_0^R \sqrt{2R\left(R = \sqrt{R^2 - y^2}\right)}\,dy \right)\\
= \frac1{2R}\left(\frac{4\sqrt2}3 R^2 +
       \frac43\left(\sqrt2 - 1\right)R^2\right)
= \frac43\left(\sqrt2 - \frac12\right) R
\end{multline}
which is less than $\frac43 R.$ Even though we used the same coordinate system we got a different answer.
You could also put the reference point at $(0,R)$ and integrate over $x$ (one integral for the part below the $x$ axis, another above) and this also would produce $\frac43\left(\sqrt2 - \frac12\right) R.$

But let's try fixing the Cartesian calculation as I suggested. Errors are introduced because one part of the curve gets "counted" more than another part. So let's make sure we count all parts equally.
The length of a circular arc between the lines $x= x_0$ and $x=x_0+\Delta x$
is approximately $\frac1{\sqrt{1-x^2}}\Delta x$ when $\Delta x$ is small;
so to count each piece in equal proportion to its length,
the correct length of the individual "infinitesimal arc" is not $dx,$
it is $\frac1{\sqrt{1-x^2}}dx$.
So to get the total length of the semicircle we're averaging over
(the denominator of the average), we should integrate
$$\int_{-R}^R  \frac1{\sqrt{1-x^2}}\, dx = \pi$$
and to get the numerator we want
$$\int_{-R}^R \frac{\sqrt{2R^2 - 2Rx}}{2R}
 \cdot \frac1{\sqrt{1-x^2}}\, dx = 4.$$
And what do you know, this leads the same answer as the polar coordinates gave, $\frac4\pi.$
So what it comes down to is that Cartesian coordinates make it easy to average things along a curve provided that the "curve" is actually just a straight line. For anything else you have to take care to make sure that you put the right amount of "weight" on each part of the curve.
Polar coordinates make it easy to average over a line provided that the line (extended if necessary) goes right through the origin, along a circular arc that has its center at the origin, and in a few other special cases. Anything else requires special care again.
