How to find the centroid of y=arcsin(x),

Question

I do not know how to continue this problem from what I have, $$y=arcsinx, x=0, y=\frac{\pi}{2}$$ It is asking for the centroid in the region bounded by above. This is what I have so far and after trying to get the answer i keep on getting it wrong. $$Mx=p\frac{1}{2}\int_0^1(arcsinx+\frac{\pi}{2})(\frac{\pi}{2}-arcsinx)$$ $$My=p\int_0^1x(\frac{\pi}{2}-arcsinx)$$ $$M=p\int_0^1(\frac{\pi}{2}-arcsinx)$$ These are the equations I made to determine to get the centroids but it marks my answer to be wrong every single time i solve for them.

Answer 1

Centriod is an intrinsic property. So conveniently interchange axes to get half-sine wave:

$$ y=\sin x, y=(0,1) ; x=(0, \frac{\pi}{2}) $$

The surrounding rectangle has area $ \pi/2 $, area inside sine-wave is $A_1= 1$ and outside sine-wave is $ A_2 = (\pi/2-1)$

Compute $x_1,y_1$ in the usual way for areas inside the sine-curve

$$ \bar y = \frac{\int y dA }{\int dA},\bar x = \frac{\int x dA }{\int dA}. $$

Now find moments about x-axis and y-axis. For surrounding rectangle $( x_R,y_R) $ are known.

$$ x_1 A_1 + x_2 A_2 = x_R ( A_1 + A_2) $$

$$ y_1 A_1 + y_2 A_2 = y_R ( A_1 + A_2). $$