91Ó°ÊÓ

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.

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

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.

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

Here, R is the radius of the semicircle.

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

$S=4\mathrm{Ï€}{R}^2$

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

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

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