Prior to this question, I had asked another question about how to take average of the "x, y" co-ordinates of a continuous function which was difficult compared to a system of discrete points. And I was met with this answer :-
$$mean = \frac{1}{b-a} \int_a^b f(x)dx$$
Say we have a semi circle $$y = \sqrt{1-x^2} $$ where the radius of the semi circle is 1.
And a set, S' = {(x, y): where (x, y) are all the co-ordinates contained within and on the semi-circle}
How can I find the average all the y co-ordinates in set S'.
I already know that the average of X co-ordinates on a semi circle is "0" and for Y it is $$Y = \frac{2}{π} $$ (by using the mean formula given above). However, for the Avg. of X and Y in the set S' I did not know how to apply the given formula. So I thought that all the Coordinates in the set S' can be represented by equation for semi-circle where the radius changes from 0 to 1.
so, for Avg of all X in set S' was as such :- $$X.Avg = \frac{{ \frac{1}{ π } \int_0^π (1)cos(x)dx + \frac{1}{ π } \int_0^π (0.99)cos(x)dx + .. + \frac{1}{ π } \int_0^π (0)cos(x)dx}}{ π(0.5)}$$ I Hope I properly conveyed what I was thinking. Anyways, each integral would be a zero so, $$X.Avg = 0$$
I am a bit unsure whether the denominator would be the area of the semi-circle.
for Y, $$Y.avg = \frac{{ \frac{1}{ π } \int_0^π (1)sin(x)dx + \frac{1}{ π } \int_0^π (0.99)sin(x)dx + .. + \frac{1}{ π } \int_0^π (0)sin(x)dx}}{ π(0.5)}$$
this comes out to be $$ Y.Avg = \frac{2}{π^2} $$
which I know is wrong, because my physics teacher had already derived it
$$ Y.Avg = \frac{4}{3π} $$
Help.
Divide your half-disk into $n$ equal sectors of amplitude $\Delta\theta=\pi/n$. For large values of $n$ each sector can be approximated by an isosceles triangle, whose centre of mass lies on the altitude, at a distance $2/3$ from the centre of the half-disk. The centers of mass of those sectors, in the limit $n\to\infty$, form then a half-circle of radius $2/3$.
But the centre of mass of $n$ bodies can be computed as if all their mass is concentrated at their centre of mass. Hence the center of mass of a half disk of radius $1$ is the same as the center of mass of a half-circle (perimeter only) of radius $2/3$, which you have already computed. In particular: $$ Y={2\over3}\cdot{2\over\pi}={4\over3\pi}. $$
Otherwise, if you want to use an integral but want to avoid double integrals, you can slice the half-disk into $n$ small rectangles of width $\Delta x$ and height $y=\sqrt{1-x^2}$, with $x\in[-1,1]$. The center of mass of each rectangle lies at $y_{CM}={1\over2}\sqrt{1-x^2}$.
The center of mass of the disk is the weighted average of the centers of mass, where the weight is the area of each rectangle: $$ Y={\displaystyle\sum_{k=1}^n (y_{CM})_k\cdot y_k \Delta x \over\displaystyle\sum_{k=1}^n y_k \Delta x} $$ which in the limit $n\to\infty$ becomes $$ Y={\displaystyle\int_{-1}^1 {1\over2}\sqrt{1-x^2} \cdot \sqrt{1-x^2} dx \over \pi/2} ={4\over3\pi}. $$