Find the centroid of the first quadrant part of the arc ${\mathbf{x}}^{\raisebox{1ex}{$\mathbf{2}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{3}$}\right.}\mathbf{+}{\mathbf{y}}^{\raisebox{1ex}{$\mathbf{2}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{3}$}\right.}\mathbf{=}{\mathbf{a}}^{\raisebox{1ex}{$\mathbf{2}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{3}$}\right.}$ . Hint: Let$\mathbf{x}\mathbf{=}\mathbf{a}\mathbf{}\mathbf{cos}\mathbf{3}\mathbf{}\mathbf{\theta }\mathbf{,}\mathbf{y}\mathbf{=}\mathbf{a}\mathbf{}\mathbf{sin}\mathbf{3}\mathbf{}\mathbf{\theta }$ .

The arc is ${\mathrm{x}}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}+{\mathrm{y}}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}={\mathrm{a}}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$

If a region lies between two functions or curves defined by $\mathbf{y}\mathbf{=}\mathbf{f}\left(x\right)$ and$\mathbf{y}\mathbf{=}\mathbf{g}\left(x\right)$ and$\mathbf{f}\left(x\right)\mathbf{\ge }\mathbf{g}\left(x\right)$ then the centroid of the region is given by the coordinates$\left(\overline{x},\overline{y}\right)$ where

role="math" localid="1659155292806" $\overline{\mathbf{x}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{A}}{\mathbf{\int }}_{\mathbf{a}}^{\mathbf{b}}\mathbf{x}\mathbf{[}\mathbf{f}\left(x\right)\mathbf{-}\mathbf{g}\left(x\right)\mathbf{]}\mathbf{dx}\mathbf{,}$

$\overline{\mathbf{y}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}\mathbf{A}}{\mathbf{\int }}_{\mathbf{a}}^{\mathbf{b}}\left({\left[f\left(x\right)\right]}^2-{\left[g\left(x\right)\right]}^2\right)\mathbf{dx}$

A is the area of the region bounded by the two curves$\mathbf{f}\left(x\right)$ and$\mathbf{g}\left(x\right)$ . Its value is given by the formula

$\mathit{A}\mathbf{=}{\mathbf{\int }}_{\mathbf{a}}^{\mathbf{b}}\mathbf{\left[}\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{\right]}\mathbf{-}\mathbf{\left[}\mathbf{g}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{\right]}\mathit{d}\mathit{x}$

The length of the arc is given by$\mathrm{L}={\int }_0^{\mathrm{a}}\sqrt{1+{\left(\mathrm{y}\text{'}\left(\mathrm{x}\right)\right)}^2\mathrm{dx}}$

The derivative is as follows

$$\begin{gathered} y\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.=a\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.-x\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right. \\ y={\left(a\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.-x\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \\ y\text{'}=\frac{3}{2}{\left(a\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.-x\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{2}{3}{x}^{-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.} \\ y\text{'}={\left(a\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.-x\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{x}^{-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.} \end{gathered}$$

Thus the value is as follows

$$\begin{gathered} \mathrm{L}={\int }_0^{\mathrm{a}}\sqrt{1+{\mathrm{x}}^{-\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\left({\mathrm{a}}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}-{\mathrm{x}}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\right)}\mathrm{dx} \\ ={\int }_0^{\mathrm{a}}{\left(\frac{\mathrm{a}}{\mathrm{x}}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\mathrm{dx} \\ ={\mathrm{a}}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\frac{3}{2}{\mathrm{x}}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\overline{){}_0^{\mathrm{a}}} \\ =\frac{3}{2}\mathrm{a} \end{gathered}$$

The x- coordinate of the centroid is found by the following formula.

$$\begin{gathered} \overline{x}=\frac{1}{L}{\int }_0^a\sqrt{1+{\left(y\text{'}\left(x\right)\right)}^2}xdx \\ =\frac{a^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}{L}{\int }_0^a{x}^{-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}xdx \\ =\frac{a^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}{L}\frac{3}{5}{x}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\overline{){}_0^a} \\ =\frac{2}{3a}{a}^2\frac{3}{5} \\ =\frac{2a}{5} \end{gathered}$$

Hence, the centroid of the first quadrant part of the arc is $\overline{\mathrm{x}}=\overline{\mathrm{y}}=\frac{2\mathrm{a}}{5}$