I found a better way (imo) to calculate the integral without relying on the symmetry argument at 3:50, or at least, I found a more formal way to justify it. I INTENTIONALLY LEFT THE 2/πR^2 CONSTANT OUT OF THE CALCULATIONS BECAUSE IT'S TOO TIRESOME TO WRITE, BUT THE INTEGRAL CALCULATIONS ARE ALL THE SAME REGARDLESS. Notice that if dA is the area of an infinitesimal square, then dA = dydx. We'll write the half disk domain as a double integral, int{x=-R, x=R}( int{y=0, y=sqrt(R^2 -x^2)}( xX + yY)dy )dx, where X is x-hat and Y is y-hat. Calculating the inner Integral gives int{-R, R}( x*sqrt(R^2 - x^2)X +((R^2 - x^2)/2)Y )dx. Now, since x*sqrt(R^2 - x^2) is an odd function, and integral over a symmetric domain [-R, R] would he equal to 0, and (R^2 - x^2)/2 is even, so it equals to 2/2*int{0, R}(R^2 - x^2)dx*Y = 2/3*R^3*Y. After multiplying it by the constant, we get the correct answer of (4R/3π)Y.
Yes! This is also a totally valid way of computing the integral out, and I like how you noticed the even/odd functions which popped out in Cartesian coordinates to make your integrals fast and easy. Exploiting odd and even functions in symmetric domains is absolutely another key skill in physics. If this problem were on an exam, for example, I think the "better" method is whatever you can do fastest haha (and with proper justification, of course)! Just like using properties of even/odd functions, In physics, being able to point out physical symmetries in advance is absolutely critical and valid as well. I do like your method though, thanks for sharing! :)
I solved it differently. Assuming uniform density, and understanding that the x component of the center of mass is zero, we can apply the coordinates to the plane. Because of symmetry, we can knock of the negative x-axis and solve the following equation; The integral of the function (✓r^2 -y^2) from the bounds; 0 to a variable p with respect to y = integral of the same function from p to r with respect to y. That way, we get the y-coordinate. It's long winded and unnecessary but that's how I knew to solve it.
I suggest a different approach :take the positive quadrant of the circle : y=√(R^2 - x^2). Then the mass-elements with same distance from the y - axis have the mass y dx ,hence we need ∫√(R^2 -x^2)*x dx (x from 0 to R. Make the substitution x = R* cos t , then dx = -R sin t dt , y= R*sin t ,and we are left with ∫ sin t^2 *cos t dt . Now set sint = s and you' re left with ∫ s^2 ds ( 0 to 1), which is 1/3. Finally you have to take the corresponding mass M =1/4 *π *R^2 to find xs = (R^3/3 ) /M = 4/(3*π) *R ,or for the original half-circle : xs = 0 ,ys = 4 R/ (3* π).
It's an area element in polar coordinates: see this video for a quick proof on how it's derived. ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-fwHQoQTECpU.html
