Chapter 8, Problem 8.73P

To determine

(a)

Centroid of the volume generated by revolving the area shown about x axis

Answer to Problem 8.73P

The centroid $\left(\overline{x},\overline{y},\overline{z}\right)$ of the volume is $\left(8,0,0\right)in$.

Explanation of Solution

Given information:

The centroid of the volume is defined as:

  $\begin{array}{l}\overline{x}=\frac{Q_{yz}}{V}=\frac{{\displaystyle \underset{V}{\int }xdV}}{{\displaystyle \underset{V}{\int }dV}}\\ \overline{y}=\frac{Q_{xz}}{V}=\frac{{\displaystyle \underset{V}{\int }ydV}}{{\displaystyle \underset{V}{\int }dV}}\\ \overline{z}=\frac{Q_{xy}}{V}=\frac{{\displaystyle \underset{V}{\int }zdV}}{{\displaystyle \underset{V}{\int }dV}}\end{array}$

Calculation:

Consider a thin shell element generated by rotating the shaded area about x axis.

The differentiate volume $dV$ of the element

  $dV=\pi y^{2}dx$

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

  $\overline{x_{el}}=x$

The volume V of the element

  $\begin{array}{l}V={\displaystyle \underset{V}{\int }dV}\\ =\pi {\displaystyle {\int }_0^{12}{y}^{2}dx}\end{array}$

The first moment $Q_{yz}$ of element

  $\begin{array}{l}{Q}_{yz}={\displaystyle \underset{V}{\int }\overline{x_{el}}dV}\\ =\pi {\displaystyle {\int }_0^{12}{y}^{2}xdx}\end{array}$

Apply Simpson's rule to evaluate above integrals.

x (in)	y (in)	$y^2$ (mm2)	$y^{2}x$   $\left({10}^{3}m{m}^3\right)$
0	0	0	0
2	4.9	24	48
4	6.93	48	192
6	8.49	72	432.5
8	9.8	96	768.5
10	10.95	120	1199
12	12	144	1728

The volume V of the element

  $\begin{array}{l}V=\pi \frac{2}{3}\left[0+4\left(24\right)+2\left(48\right)+4\left(72\right)+2\left(96\right)+4\left(120\right)+144\right]\\ =2714i{n}^3\end{array}$

The first moment $Q_{yz}$ of element

  $\begin{array}{l}{Q}_{yz}=\pi \frac{2}{3}\left[0+4\left(48\right)+2\left(192\right)+4\left(432.5\right)+2\left(768.5\right)+4\left(1199\right)+1728\right]\\ =21713i{n}^4\end{array}$

The point of action $\overline{x}$ along x axis

  $\overline{x}=\frac{Q_{yz}}{V}=\frac{21713i{n}^4}{2714i{n}^3}=8in$

Due to symmetry

  $\overline{y}=\overline{z}=0$

Conclusion:

The centroid $\left(\overline{x},\overline{y},\overline{z}\right)$ of the volume is $\left(8,0,0\right)in$.

To determine

(b)

Assume curve OB is a parabola and verify the result found on part a

Answer to Problem 8.73P

The results obtained at sub part 'a' is equal to the results in sub part 'b'.

Explanation of Solution

Given information:

According to table 8.3

Paraboloid of revolution:

  $\begin{array}{l}V=\frac{\pi }{2}{r}^{2}h\\ \overline{x}=\frac{2}{3}h\end{array}$

Calculation:

The volume V

  $V=\frac{\pi }{2}{R}^{2}h$

Substitute and solve

  $\begin{array}{l}V=\frac{\pi }{2}{\left(12in\right)}^2\left(12in\right)\\ =2714i{n}^3\end{array}$

The point of action $\overline{x}$ along x axis

  $\begin{array}{l}\overline{x}=\frac{2}{3}h\\ =\frac{2}{3}\left(12in\right)\\ =8in\end{array}$

The results obtained at sub part 'a' is equal to the results in sub part 'b'.

Conclusion:

The results obtained at sub part 'a' is equal to the results in sub part 'b'.