Chapter 6, Problem 1RP

Locate the centroid $\bar{x}$ of the area.

To determine

Find the centroid $\overline{x}$ of the shaded area.

Answer to Problem 1RP

The centroid $\overline{x}$ of the shaded area is $\underset{\_}{\frac{b-a}{\ln \frac{b}{a}}}$.

Explanation of Solution

Show the area of the differential element as in Figure 1.

Using Figure 1,

Express the parabolic value, y.

$xy=c^2$

Rearrange Equation to get the value of y.

$y=\frac{c^2}{x}$

Compute the area of the differential element.

$dA=ydx$

Substitute $\frac{c^2}{x}$ for y.

$dA=\left(\frac{c^2}{x}\right)dx$

Compute the centroid of the differential element along the x-axis using the formula.

$\tilde{x}=x$

Determine the location of the centre of gravity along the x-axis $\left(\overline{x}\right)$ using the relation.

$\overline{x}=\frac{{\displaystyle {\int }_A\tilde{x}dA}}{{\displaystyle {\int }_{A}dA}}$

Apply the limits from $a$ to $b$; substitute $x$ for $\tilde{x}$ and $\left(\frac{c^2}{x}\right)dx$ for dA.

$\begin{array}{c}\overline{x}=\frac{{\displaystyle {\int }_a^b\left(x\right)}\left(\frac{c^2}{x}\right)dx}{{\displaystyle {\int }_a^b\left(\frac{c^2}{x}\right)dx}}\\ =\frac{{\displaystyle {\int }_a^b{c}^{2}dx}}{c^2\ln \frac{b}{a}}\\ =\frac{c^2{\left[x\right]}_a^b}{c^2\ln \frac{b}{a}}\\ =\frac{b-a}{\ln \frac{b}{a}}\end{array}$

Thus, the centroid $\overline{x}$ of the shaded area is $\underset{\_}{\frac{b-a}{\ln \frac{b}{a}}}$.