Chapter 5.2, Problem 5.43P

5.43 and 5.44

Determine by direct integration the centroid of the area shown. Express your answer in terms of a and b.

Fig. P5.43

To determine

The centroid of shaded area in Fig. P5.43 by method of direct integration.

Answer to Problem 5.43P

Centroid is located at $\left(\frac{17}{130}a,\frac{11}{26}b\right)$.

Explanation of Solution

Refer the figure P5.43 and figure given below.

Write the equation for curve $x$.

$x=k{y}^2$ (I)

Here, $y$ is the y co-ordinate, $x$ is the x-coordinate, and $k$ is an unknown constant.

Consider the point $\left(a,b\right)$.

Here, $a$ is a length on x-axis and $b$  is a length parallel to y axis.

Rewrite equation (I) by substituting $a$ for $x$ and $b$ for $y_1$.

$a=k{b}^2$

Rewrite the above equation in terms of $k$

$k=\frac{b}{a^2}$

Rewrite equation (I) by substituting the above relation for $k$.

$x=\frac{b}{a^2}{y}^2$

Rewrite the above equation in terms of $y$.

$y=\frac{b}{\sqrt{a}}\sqrt{x}$

Divide the shaded region in P5.43 into two parts for the purpose of integration. Region $0\le x\le \frac{a}{2}$ and   $\frac{a}{2}\le x\le a$.

Consider the region $0\le x\le \frac{a}{2}$ in P5.43.

Consider a rectangular differential area element in the region. Write the expression for the x-coordinate of center of mass of differential area element.

${\overline{x}}_{EL1}=x$

Here, ${\overline{x}}_{EL1}$ is the center of mass of differential area element in region $0\le x\le \frac{a}{2}$

Write the expression for the y-coordinate of center of mass of differential area element in region $0\le x\le \frac{a}{2}$ in P5.43.

${\overline{y}}_{EL1}=\frac{y}{2}$

Here, ${\overline{y}}_{E{L}_1}$ is the y-coordinate of differential area element

Rewrite the above relation by substituting $\frac{b}{\sqrt{a}}\sqrt{x}$ for $y$ in the above equation.

${\overline{y}}_{EL1}=\frac{b}{2\sqrt{a}}\sqrt{x}$ (II)

Write the expression to calculate the differential area element in $0\le x\le \frac{a}{2}$.

$d{A}_1=ydx$

Here, $d{A}_1$ is the differential area element in $0\le x\le \frac{a}{2}$ and $dx$ denotes a small change in $x$.

Rewrite the above relation by substituting $\frac{b}{\sqrt{a}}\sqrt{x}$ for $y$ in the above equation.

$d{A}_1=\frac{b}{\sqrt{a}}\sqrt{x}dx$ (III)

Consider the region $\frac{a}{2}\le x\le a$ in in P5.43.

Write the expression for the x-coordinate of center of mass of differential area element in region $\frac{a}{2}\le x\le a$.

${\overline{x}}_{EL2}=x$

Here, ${\overline{x}}_{EL2}$ is the center of mass of differential area element in region $\frac{a}{2}\le x\le a$

Write the expression for the y-coordinate of center of mass of differential area element in region $\frac{a}{2}\le x\le a$.

${\overline{y}}_{EL2}=\frac{1}{2}\left(y+y_1\right)$ (IV)

Here, ${\overline{y}}_{EL}$ is the y-coordinate of center of mass of differential area element in region $\frac{a}{2}\le x\le a$ and $y_1$ is the straight line in $\frac{a}{2}\le x\le a$.

Calculate the slope of $y_1$ using points $\left(\frac{a}{2},0\right)$ and $\left(a,\frac{b}{2}\right)$.

$\begin{array}{c}m=\frac{\frac{b}{2}-0}{a-\frac{a}{2}}\\ =\frac{b}{a}\end{array}$

Here, $m$ is the slope of $y_1$.

Write the equation of $y_1$ using slope-point form of equation for straight line. Use the point $\left(\frac{a}{2},0\right)$ for finding the equation.

$y_1-0=m\left(x-\frac{a}{2}\right)$

Rewrite the above equation by substituting $\frac{b}{a}$ for $m$ to find

$y_1=\frac{b}{a}\left(x-\frac{a}{2}\right)$

Rewrite equation (IV) by substituting the above relation for $y_1$ and $\frac{b}{\sqrt{a}}\sqrt{x}$ for $y$.

$\begin{array}{c}{\overline{y}}_{EL2}=\frac{1}{2}\left(\frac{b}{\sqrt{a}}\sqrt{x}+\frac{b}{a}\left(x-\frac{a}{2}\right)\right)\\ =\frac{b}{2}\left(\frac{\sqrt{x}}{\sqrt{a}}+\frac{x}{a}-\frac{1}{2}\right)\end{array}$ (V)

Write the expression for $d{A}_2$.

$d{A}_2=\left(y-y_1\right)dx$

Rewrite the above relation by substituting $\frac{b}{\sqrt{a}}\sqrt{x}$ for $y$ and $\frac{b}{a}\left(x-\frac{a}{2}\right)$ for $y_1$.

$\begin{array}{c}d{A}_2=\left(\frac{b}{\sqrt{a}}\sqrt{x}-\frac{b}{a}\left(x-\frac{a}{2}\right)\right)dx\\ =b\left(\frac{\sqrt{x}}{\sqrt{a}}-\frac{x}{a}+\frac{1}{2}\right)dx\end{array}$ (VI)

Write the equation to calculate the total area of shaded region in P5.43.

$A={\displaystyle \underset{0}{\overset{a/2}{\int }}d{A}_1+}{\displaystyle \underset{a/2}{\overset{a}{\int }}d{A}_2}$

Here, $A$ is the area shaded region in P5.43.

Rewrite the above equation by substituting equation (III) and (V).

$\begin{array}{c}A={\displaystyle \underset{0}{\overset{a/2}{\int }}\frac{b}{\sqrt{a}}\sqrt{x}dx+}{\displaystyle \underset{a/2}{\overset{a}{\int }}b\left(\frac{\sqrt{x}}{\sqrt{a}}-\frac{x}{a}+\frac{1}{2}\right)dx}\\ =\frac{b}{\sqrt{a}}{\left[\frac{2}{3}{x}^{3/2}\right]}_0^{a/2}+b{\left[\frac{2}{3}\frac{x^{3/2}}{\sqrt{a}}-\frac{x^2}{2a}+\frac{x}{2}\right]}_{a/2}^a\\ =\frac{2b}{3\sqrt{a}}\left[{\left(\frac{a}{2}\right)}^{3/2}+a^{3/2}-{\left(\frac{a}{2}\right)}^{3/2}\right]+b\left[-\frac{1}{2a}\left(a^2-{\left(\frac{a}{2}\right)}^2\right)+\frac{1}{2}\left(a-\frac{a}{2}\right)\right]\\ =\frac{13}{24}ab\end{array}$

Write the expression for ${\displaystyle \int {\overline{x}}_{EL}dA}$.

$\begin{array}{c}{\displaystyle \int {\overline{x}}_{EL}dA}={\displaystyle \underset{0}{\overset{a/2}{\int }}{\overline{x}}_{EL1}d{A}_1+}{\displaystyle \underset{a/2}{\overset{a}{\int }}{\overline{x}}_{EL2}d{A}_2}\\ ={\displaystyle \underset{0}{\overset{a/2}{\int }}xd{A}_1+}{\displaystyle \underset{a/2}{\overset{a}{\int }}xd{A}_2}\end{array}$

Rewrite the above equation by substituting equation (III) and (V).

$\begin{array}{c}{\displaystyle \int {\overline{x}}_{EL}dA}={\displaystyle \underset{0}{\overset{a/2}{\int }}x\frac{b}{\sqrt{a}}\sqrt{x}dx+}{\displaystyle \underset{a/2}{\overset{a}{\int }}xb\left(\frac{\sqrt{x}}{\sqrt{a}}-\frac{x}{a}+\frac{1}{2}\right)dx}\\ =\frac{b}{\sqrt{a}}{\left[\frac{2}{5}{x}^{5/2}\right]}_0^{a/2}+b{\left[\frac{2}{5}\frac{x^{5/2}}{\sqrt{a}}-\frac{x^3}{3a}+\frac{x^4}{4}\right]}_{a/2}^a\\ =\frac{2b}{5\sqrt{a}}\left[{\left(\frac{a}{2}\right)}^{5/2}+a^{5/2}-{\left(\frac{a}{2}\right)}^{5/2}\right]+b\left[-\frac{1}{3a}\left(a^3-{\left(\frac{a}{2}\right)}^3\right)+\frac{1}{4}\left(a^2-{\left(\frac{a}{2}\right)}^2\right)\right]\\ =\frac{71}{249}{a}^{2}b\end{array}$

Write the expression for ${\displaystyle \int {\overline{y}}_{EL}dA}$.

${\displaystyle \int {\overline{y}}_{EL}dA}={\displaystyle \underset{0}{\overset{a/2}{\int }}{\overline{y}}_{EL1}d{A}_1+}{\displaystyle \underset{a/2}{\overset{a}{\int }}{\overline{y}}_{EL2}d{A}_2}$

Rewrite the above equation by substituting equations (II), (III), (V) and (VI).

$\begin{array}{c}{\displaystyle \int {\overline{y}}_{EL}dA}={\displaystyle \underset{0}{\overset{a/2}{\int }}\frac{b}{2\sqrt{a}}\sqrt{x}\frac{b}{\sqrt{a}}\sqrt{x}dx+}{\displaystyle \underset{a/2}{\overset{a}{\int }}\frac{b}{2}\left(\frac{\sqrt{x}}{\sqrt{a}}+\frac{x}{a}-\frac{1}{2}\right)b\left(\frac{\sqrt{x}}{\sqrt{a}}-\frac{x}{a}+\frac{1}{2}\right)dx}\\ =\frac{b^2}{2a}{\left[\frac{1}{2}{x}^2\right]}_0^{a/2}+\frac{b^2}{2}{\left[\frac{x^2}{2a}-\frac{1}{3a}{\left(\frac{x}{a}-\frac{1}{2}\right)}^3\right]}_{a/2}^a\\ =\frac{b}{4a}\left[{\left(\frac{a}{2}\right)}^2+a^2-{\left(\frac{a}{2}\right)}^2\right]-\frac{b^2}{6a}{\left(\frac{a}{2}-\frac{1}{2}\right)}^3\\ =\frac{11}{48}a{b}^2\end{array}$

Write the expression for first moment of whole area about y-axis.

$\overline{x}A={\displaystyle \int {\overline{x}}_{EL}dA}$

Here, $\overline{x}$ is the x-coordinate of center of mass of total area.

Rewrite the above relation by substituting $\frac{13}{24}ab$ for $A$ and $\frac{71}{249}{a}^{2}b$ for ${\displaystyle \int {\overline{x}}_{EL}dA}$.

$\overline{x}\left(\frac{13}{24}ab\right)=\frac{71}{249}{a}^{2}b$

Rewrite the above relation in terms of $\overline{x}$.

$\begin{array}{c}\overline{x}=\frac{\left(\frac{71}{249}{a}^{2}b\right)}{\left(\frac{13}{24}ab\right)}\\ =\frac{17}{130}a\end{array}$

Write the expression for first moment of whole area about x-axis.

$\overline{y}A={\displaystyle \int {\overline{y}}_{EL}dA}$

Here, $\overline{y}$ is the y-coordinate of center of mass of total area.

Rewrite the above relation in terms of $\overline{y}$ by substituting $\frac{13}{24}ab$ for $A$ and $\frac{11}{48}a{b}^2$ for ${\displaystyle \int {\overline{y}}_{EL}dA}$.

$\begin{array}{c}\overline{y}=\frac{\left(\frac{11}{48}a{b}^2\right)}{\left(\frac{13}{24}ab\right)}\\ =\frac{11}{26}b\end{array}$

Therefore, the centroid is located at $\left(\frac{17}{130}a,\frac{11}{26}b\right)$.