Question

In Problems 17 to 30, for the curve $\mathit{y}\mathbf{=}\sqrt{\mathbf{x}}$, between role="math" localid="1658830478813" $\mathit{x}\mathbf{=}\mathbf{0}$and $\mathit{x}\mathbf{=}\mathbf{2}$,

find:

The centroid of the surface area.

Short answer

The centroid of the surface area $\left(\overline{x},\overline{y},\overline{z}\right)=\left(\frac{149}{130},0,0\right)$.

Step-by-step solution

Definition of Centroid of a body

The point in a particular body at which the value of the volume is the same on all the sides is defined as the centroid of the body. If the mass is uniform in the body, then center of mass lies on the centeroid.

Representation of the area bounded by the curve

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

Calculation of the value of x 

If the surface be rotated around the x - axis, then only $\overline{x}$remains, and$\overline{y}=\overline{z}=0$.

$$\begin{gathered} \overline{x}=\frac{1}{S}{\int }_0^{2}2\mathrm{\pi xf}\left(\mathrm{x}\right)\mathrm{dl} \\ =\frac{2\mathrm{\pi }}{\mathrm{S}}{\int }_0^2\mathrm{xf}\left(\mathrm{x}\right)\sqrt{1+{\left(\mathrm{f}\text{'}\left(\mathrm{x}\right)\right)}^2\mathrm{dx}} \\ =\frac{2\mathrm{\pi }}{\mathrm{S}}{\int }_0^2\mathrm{x}\sqrt{\mathrm{x}}\sqrt{1+\frac{1}{4\mathrm{x}}\mathrm{dx}} \\ =\frac{2\mathrm{\pi }}{\mathrm{S}}{\int }_0^2\mathrm{x}\sqrt{\mathrm{x}+\frac{1}{4}\mathrm{dx}} \end{gathered}$$

Step 4`: Calculation of centroid of surface area

Evaluate the value,$S=\frac{13\mathrm{\pi }}{3}$, so the integral is as follows,

$$\begin{gathered} S=\frac{6}{13}\left[{\int }_0^2\left(x+\frac{1}{4}\right)\sqrt{1+\frac{1}{4x}}d\left(x+\frac{1}{4}\right)-\frac{1}{4}{\int }_0^2\sqrt{1+\frac{1}{4x}}d\left(x+\frac{1}{4}\right)\right] \\ =\frac{6}{13}{\left[\frac{2}{5}{\left(x+\frac{1}{4}\right)}^{5/2}-\frac{1}{6}{\left(x+\frac{1}{4}\right)}^{3/2}\right]}_0^2 \\ =\frac{149}{130} \end{gathered}$$

Thus, the centroid of the surface area $\left(\overline{x},\overline{y},\overline{z}\right)=\left(\frac{149}{130},0,0\right)$.