Use Problems 12 and 13 to find the centroids of a semi-circular area and of a semi-circular arc. Hint: Assume the formulas $\mathit{A}\mathbf{=}\mathbf{4}\mathit{\pi }{\mathit{r}}^{\mathbf{2}}$, $\mathit{V}\mathbf{=}\frac{\mathbf{4}}{\mathbf{3}}\mathit{\pi }{\mathit{r}}^{\mathbf{3}}$ for a sphere.

The centroid or geometric center of a flat shape is the arithmetic mean position of all the points in the figure in mathematics and science. Informally, it's the point at which a perfectly balanced cut-out of the shape might be placed on the tip of a pin.

Write the expression for the volume by using the theorem in problem 12.

$V=A\left(2\pi {\overline{y}}_{area}\right)$

Here, A is the semi-circular area and ${\overline{y}}_{area}$is the centroid.

Substitute all the values in the above expression and then rearrange the expression.

$$\begin{gathered} \frac{4}{3}\pi R^3=\frac{1}{2}\pi R^2\left(2\pi {\overline{y}}_{area}\right) \\ {\overline{y}}_{area}=\frac{4R}{3\pi } \end{gathered}$$

Here, R is the radius of the semicircle.

Write the expression for the surface area by using the theorem in problem 13.

$S=4\pi R^2$

Substitute all the values in the above expression and then rearrange the expression.

role="math" localid="1659164929899" $$\begin{gathered} S=L\left(2\pi {\overline{y}}_{arc}\right) \\ =R\pi \left(2\pi {\overline{y}}_{arc}\right) \\ {\overline{y}}_{arc}=\frac{2R}{\pi } \end{gathered}$$

Thus, the centroids of semi-circular area and arc are ${\overline{y}}_{area}=\frac{4R}{3\pi }$and${\overline{y}}_{arc}=\frac{2R}{\pi }$ respectively.