Chapter 8, Problem 8.74P

To determine

Centroid of the surface generated by revolving the parabola about y axis

Answer to Problem 8.74P

The centroid $\left(\overline{x},\overline{y},\overline{z}\right)$ of the volume is $\left(0,1.64,0\right)m$.

Explanation of Solution

Given information:

The centroid of the surface is defined as:

  $\begin{array}{l}\overline{x}=\frac{Q_{yz}}{A}=\frac{{\displaystyle \underset{A}{\int }xdA}}{{\displaystyle \underset{A}{\int }dA}}\\ \overline{y}=\frac{Q_{xz}}{A}=\frac{{\displaystyle \underset{A}{\int }ydA}}{{\displaystyle \underset{A}{\int }dA}}\\ \overline{z}=\frac{Q_{xy}}{A}=\frac{{\displaystyle \underset{A}{\int }zdA}}{{\displaystyle \underset{A}{\int }dA}}\end{array}$

Calculation:

Consider the surface element, a ring obtained by rotating the line of length $ds$ about y axis.

We know that

  $y=\frac{3}{16}{x}^2$

Differentiate

  $\frac{dy}{dx}=\frac{3}{8}x$

  $\begin{array}{l}\frac{ds}{dx}=\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}\\ =\sqrt{1+{\left(\frac{3}{8}x\right)}^2}\end{array}$

The differential area $dA$

  $\begin{array}{l}dA=2\pi xds\\ =2\pi x\frac{ds}{dx}dx\\ =2\pi x\sqrt{1+{\left(\frac{3}{8}x\right)}^2}dx\end{array}$

The centroidal coordinate $\overline{y_{el}}$ of element

  $\overline{y_{el}}=y$

The area A of the element

  $\begin{array}{l}A={\displaystyle \underset{A}{\int }dA}\\ =2\pi {\displaystyle {\int }_0^{4}x\sqrt{1+{\left(\frac{3}{8}x\right)}^2}dx}\end{array}$

The first moment $Q_{xz}$ of element

  $\begin{array}{l}{Q}_{xz}={\displaystyle \underset{A}{\int }\overline{y_{el}}dA}\\ =2\pi {\displaystyle {\int }_0^{4}xy\sqrt{1+{\left(\frac{3}{8}x\right)}^2}dx}\end{array}$

Apply Simpson's rule to evaluate above integrals.

x (m)	y (m)	$x\sqrt{1+{\left(\frac{3}{8}x\right)}^2}$	$xy\sqrt{1+{\left(\frac{3}{8}x\right)}^2}$
0	0	0	0
1	0.1875	1.068	0.20025
2	0.75	2.5	1.875
3	1.6875	4.5155	7.62
4	3.0	7.211	21.633

The Area A of the element

  $\begin{array}{l}A=2\pi \frac{1}{3}\left[0+4\left(1.068\right)+2\left(2.5\right)+4\left(4.5155\right)+7.211\right]\\ =72.35m^2\end{array}$

The first moment $Q_{xz}$ of element

  $\begin{array}{l}{Q}_{xz}=2\pi \frac{1}{3}\left[0+4\left(0.20055\right)+2\left(1.875\right)+4\left(7.62\right)+21.633\right]\\ =118.68m^3\end{array}$

The point of action

  $\overline{y}$ along y axis

  $\overline{y}=\frac{Q_{xz}}{A}=\frac{118.68m^3}{72.35m^2}=1.64m$

Due to symmetry

  $\overline{x}=\overline{z}=0$

Conclusion:

The centroid $\left(\overline{x},\overline{y},\overline{z}\right)$ of the volume is $\left(0,1.64,0\right)m$.