Originally Posted by

**earboth** Hello,

I don't know how to get the formula but I can give you possible way to get a valid solution:

The "centre of mass" of a plane area of homogenous material which is bounded by the x-axis, the curve of y = f(x), the straight lines $\displaystyle x = x_1$ and $\displaystyle x = x_2$ is the centroid of this area. The coordinates of the centroid are calculated by:

$\displaystyle x_c=\frac{\int_{x_1}^{x_2}x \cdot y dx}{\int_{x_1}^{x_2}y dx}$ and $\displaystyle y_c=\frac{\int_{x_1}^{x_2}y^2 dx}{2 \int_{x_1}^{x_2}ydx}$

With your problem:

$\displaystyle y = \sqrt{r^2-x^2}$ , that menas the center of the semi-circle is (0, 0), $\displaystyle x_1 = -r$, $\displaystyle x_2 = r$

Obviously $\displaystyle \int_{-r}^{r}ydx = \frac12 \cdot \pi \cdot r^2$

Plug in all values you know:

$\displaystyle x_c=\frac{\int_{-r}^{r}x \cdot \sqrt{r^2-x^2} dx}{\frac12 \cdot \pi \cdot r^2} = 0$ Use integration by substitution

$\displaystyle y_c=\frac{\int_{-r}^{r}(r^2-x^2)dx}{\pi \cdot r^2}=\frac{\frac43 r^3}{\pi \cdot r^2}=\frac{4}{3\pi}r$

Therefore the centre of mass has the coordinates $\displaystyle \left(0, \frac{4}{3\pi}r \right)$