In Problems 17 to 30, for the curve $\mathit{y}\mathbf{=}\sqrt{\mathbf{x}}$, between$\mathit{x}\mathbf{=}\mathbf{0}$and $\mathit{x}\mathbf{=}\mathbf{2}$,

find:

The centroid of the arc.

The area, inside a closed curve in the $\left(x,y\right)$ plane, $\mathit{y}\mathbf{\ge }\mathbf{0}$, is revolved around the x -axis. The volume of the solid generated is equal to times the circumference of the circle traced by the centroid of A .

From the problem 3.18 the arc length of the curve is given as follows,

$L\approx 2.562$

From the problem 3.20, the surface of the curve rotated about the -axis is given as follows,

$S=\frac{13\mathrm{\pi }}{3}$

Apply the theorem.

$$\begin{gathered} S=2\mathrm{\pi }\overline{\mathrm{y}}\mathrm{L} \\ \overline{\mathrm{y}}\mathrm{L}=\frac{\mathrm{S}}{2\mathrm{\pi }} \\ =\frac{13}{6} \\ \overline{\mathrm{y}}\approx 0.846 \end{gathered}$$

Draw the bounded region for the curve $y=\sqrt{x}$, between $x=0$and $x=2$.

Calculate the value of surface.

$$\begin{gathered} S={\int }_0^{\sqrt{2}}x\left(y\right)2\mathrm{\pi dsy} \\ =2\mathrm{\pi }{\int }_0^{\sqrt{2}}{\mathrm{y}}^2\sqrt{1+{\left({\mathrm{x}}^{\mathrm{\varphi }}\left(\mathrm{y}\right)\right)}^2}\mathrm{dy} \\ =2\mathrm{\pi }{\int }_0^{\sqrt{2}}{\mathrm{y}}^2\sqrt{1+4{\mathrm{y}}^2}\mathrm{dy} \end{gathered}$$

Substitute $\frac{1}{2}$for y, and $\frac{1}{2}$for in the above expression.

$$\begin{gathered} S=2\mathrm{\pi }{\int }_0^{{\sinh }^{-1}\left(2\sqrt{2}\right)}\frac{1}{8}{\sinh }^2{\mathrm{ucosh}}^2\mathrm{udu} \\ =\frac{\mathrm{\pi }}{4}{\int }_0^{{\sinh }^{-1}\left(2\sqrt{2}\right)}{\sinh }^2{\mathrm{ucosh}}^2\mathrm{udu} \\ =\frac{\mathrm{\pi }}{4}{\int }_0^{{\sinh }^{-1}\left(2\sqrt{2}\right)}\frac{1}{4}\left(\cosh \left(2\mathrm{u}\right)-1\right)\left(\cosh \left(2\mathrm{u}\right)+1\right)\mathrm{du} \\ =\frac{\mathrm{\pi }}{16}{\int }_0^{{\sinh }^{-1}\left(2\sqrt{2}\right)}\frac{1}{4}\left(\cosh \left(2\mathrm{u}\right)-1\right)\mathrm{du} \end{gathered}$$

Further solve the expression.

$$\begin{gathered} S=\frac{\mathrm{\pi }}{16}\left[-{\overline{)u}}_0^{\sin h-1\left(2\sqrt{2}\right)}+\frac{1}{2}{\int }_0^{\sin h-1\left(2\sqrt{2}\right)}\left(\cos h\left(4u\right)+1\right)du\right] \\ =\frac{\mathrm{\pi }}{16}\left[-\frac{1}{2}\sin h^{-1}\left(2\sqrt{2}\right)+\frac{1}{8}\sin h^2{\left[4u\right]}_0^{\sin h^{-1\left(2\sqrt{2}\right)}}\right] \\ =\frac{\mathrm{\pi }}{128}\left[\sin h+\left(4\sin h^{-1}\left(2\sqrt{2}\right)\right)-4\sin h^{-1}\left(2\sqrt{2}\right)\right] \end{gathered}$$

Apply the theorem.

$$\begin{gathered} L\overline{x}=\frac{S}{2\mathrm{\pi }} \\ \approx 2.226 \\ \overline{x}\approx 0.869 \end{gathered}$$

Thus, the centroid is$\left(\overline{x},\overline{y}\right)\approx \left(0.869,0.846\right)$ .