Chapter 5.2, Problem 5.48P

5.48 and *5.49 Determine by direct integration the centroid of the area shown.

Fig. P5.48

To determine

The centroid of shaded area in Fig. P5.48 by method of direct integration.

Answer to Problem 5.48P

Centroid is located at $\left(-9.27a,3.09a\right)$.

Explanation of Solution

Refer the figure P5.48 and figure given below.

Write the equation for curve $r$.

$r=a{e}^{\theta }$

Here, $y$ is the radial coordinate, $a$ is an unknown constant, and $\theta$ is the polar angle.

In the figure given above, black dot denotes the center of mass of triangular shaped differential area element.

Write the expression for the x-coordinate of center of mass of triangular shaped differential area element.

${\overline{x}}_{EL}=\frac{2}{3}r\cos \theta$

Here, ${\overline{x}}_{EL}$ is the center of mass of element $dA$.

Rewrite the above relation by substituting $a{e}^{\theta }$ for $r$ .

${\overline{x}}_{EL}=\frac{2}{3}a{e}^{\theta }\cos \theta$

Write the expression for the y-coordinate of center of mass of triangular shaped differential area element.

${\overline{y}}_{EL}=\frac{2}{3}r\sin \theta$

Here, ${\overline{y}}_{EL}$ is the y-coordinate of center of mass of element $dA$.

Rewrite the above relation by substituting $a{e}^{\theta }$ for $r$ .

${\overline{y}}_{EL}=\frac{2}{3}a{e}^{\theta }\sin \theta$

Write the expression to calculate the triangular differential area element.

$dA=\frac{1}{2}\left(r\right)\left(rd\theta \right)$

Here, $dA$ is the differential area element and $d\theta$ denotes a small change in $\theta$.

Rewrite the above relation by substituting $a{e}^{\theta }$ for $r$ .

$\begin{array}{c}dA=\frac{1}{2}\left(a{e}^{\theta }\right)\left(a{e}^{\theta }d\theta \right)\\ =\frac{1}{2}{a}^2{e}^{2\theta }d\theta \end{array}$

Write the expression to calculate the total area of shaded region in P5.48.

$A={\displaystyle \int dA}$

Here, $A$ is the area of shaded region in P5.48.

Rewrite the above equation by substituting $\frac{1}{2}{a}^2{e}^{2\theta }d\theta$ for $dA$ and integrate it from $\theta =0$ to $\theta =\pi$.

$\begin{array}{c}A={\displaystyle \underset{0}{\overset{\pi }{\int }}\frac{1}{2}{a}^2{e}^{2\theta }d\theta }\\ =\frac{1}{2}{a}^2{\left[\frac{1}{2}{e}^{2\theta }\right]}_0^{2L}\\ =\frac{1}{4}{a}^2\left(e^{2\pi }-1\right)\\ =133.623a^2\end{array}$

Calculate ${\displaystyle \int {\overline{x}}_{EL}dA}$ by substituting $\frac{1}{2}{a}^2{e}^{2\theta }d\theta$ for $dA$ and $\frac{2}{3}a{e}^{\theta }\cos \theta$ for ${\overline{x}}_{EL}$.

$\begin{array}{c}{\displaystyle \int {\overline{x}}_{EL}dA}={\displaystyle \underset{0}{\overset{\pi }{\int }}\frac{2}{3}a{e}^{\theta }\cos \theta }\left(\frac{1}{2}{a}^2{e}^{2\theta }d\theta \right)\\ =\frac{a^3}{3}{\displaystyle \underset{0}{\overset{\pi }{\int }}{e}^{3\theta }\cos \theta }d\theta \end{array}$ (I)

Apply the integration by parts method to solve the above integral. The required formula is given below.

${\displaystyle \int u\left(\theta \right)v\left(\theta \right)}d\theta =u\left(\theta \right){\displaystyle \int v\left(\theta \right)}d\theta -{\displaystyle \int \left[\left(\frac{du\left(\theta \right)}{d\theta }\right){\displaystyle \int v\left(\theta \right)}d\theta \right]}$

Here, $u\left(\theta \right)$ and $v\left(\theta \right)$ are functions of $\theta$.

Substitute $e^{3\theta }$ for $u\left(\theta \right)$ and $\cos \theta$ for $v\left(\theta \right)$ and solve integral.

$\begin{array}{c}{\displaystyle \int e^{3\theta }\cos \theta }d\theta =e^{3\theta }{\displaystyle \int \cos \theta }d\theta -{\displaystyle \int \left[\left(\frac{d{e}^{3\theta }}{d\theta }\right){\displaystyle \int \cos \theta }d\theta \right]}\\ =e^{3\theta }\sin \theta -{\displaystyle \int \left(3e^{3\theta }\sin \theta \right)}\\ =e^{3\theta }\sin \theta -3\left[e^{3\theta }\left(-\cos \theta \right)-{\displaystyle \int 3e^{3\theta }\left(-\cos \theta \right)d\theta }\right]\\ =e^{3\theta }\sin \theta +3e^{3\theta }\cos \theta -9{\displaystyle \int e^{3\theta }\cos \theta d\theta }\\ {\displaystyle \int e^{3\theta }\cos \theta }d\theta +9{\displaystyle \int e^{3\theta }\cos \theta d\theta }=e^{3\theta }\sin \theta +3e^{3\theta }\cos \theta \\ 10{\displaystyle \int e^{3\theta }\cos \theta }d\theta =e^{3\theta }\sin \theta +3e^{3\theta }\cos \theta \end{array}$

Rewrite the above equation in terms of ${\displaystyle \int e^{3\theta }\cos \theta }d\theta$.

${\displaystyle \int e^{3\theta }\cos \theta }d\theta =\frac{e^{3\theta }}{10}\left(\sin \theta +3\cos \theta \right)$

Calculate $\frac{a^3}{3}{\displaystyle \underset{0}{\overset{\pi }{\int }}{e}^{3\theta }\cos \theta }d\theta$ by using the above result.

$\begin{array}{c}\frac{a^3}{3}{\displaystyle \underset{0}{\overset{\pi }{\int }}{e}^{3\theta }\cos \theta }d\theta =\frac{a^3}{3}{\left[\frac{e^{3\theta }}{10}\left(\sin \theta +3\cos \theta \right)\right]}_0^{\pi }\\ =\frac{a^3}{3}\left[\frac{e^{3\pi }}{10}\left(\sin \pi +3\cos \pi \right)-\frac{e^0}{10}\left(\sin 0+3\cos 0\right)\right]\\ =\frac{a^3}{3}\left[\frac{e^{3\pi }}{10}\left(-3\right)-\frac{3}{10}\right]\\ =\frac{a^3}{30}\left(-3e^{3\pi }-3\right)\\ =-1239.26a^3\end{array}$

Rewrite equation (I) by substituting the above result.

${\displaystyle \int {\overline{x}}_{EL}dA}=-1239.26a^3$

\

Calculate ${\displaystyle \int {\overline{y}}_{EL}dA}$ by substituting $\frac{1}{2}{a}^2{e}^{2\theta }d\theta$ for $dA$ and $\frac{2}{3}a{e}^{\theta }\sin \theta$ for ${\overline{y}}_{EL}$.

$\begin{array}{c}{\displaystyle \int {\overline{y}}_{EL}dA}={\displaystyle \underset{0}{\overset{\pi }{\int }}\frac{2}{3}a{e}^{\theta }\sin \theta }\left(\frac{1}{2}{a}^2{e}^{2\theta }d\theta \right)\\ =\frac{a^3}{3}{\displaystyle \underset{0}{\overset{\pi }{\int }}{e}^{3\theta }\sin \theta }d\theta \end{array}$ (II)

Apply the integration by parts method to solve the above integral. The required formula is given below.

${\displaystyle \int u\left(\theta \right)v\left(\theta \right)}d\theta =u\left(\theta \right){\displaystyle \int v\left(\theta \right)}d\theta -{\displaystyle \int \left[\left(\frac{du\left(\theta \right)}{d\theta }\right){\displaystyle \int v\left(\theta \right)}d\theta \right]}$

Here, $u\left(\theta \right)$ and $v\left(\theta \right)$ are functions of $\theta$.

Substitute $e^{3\theta }$ for $u\left(\theta \right)$ and $\sin \theta$ for $v\left(\theta \right)$ and solve integral.

$\begin{array}{c}{\displaystyle \int e^{3\theta }\sin }\theta d\theta =e^{3\theta }{\displaystyle \int \sin \theta }d\theta -{\displaystyle \int \left[\left(\frac{d{e}^{3\theta }}{d\theta }\right){\displaystyle \int \sin \theta }d\theta \right]}\\ =-e^{3\theta }\cos \theta -{\displaystyle \int \left(3e^{3\theta }\left(-\cos \theta \right)\right)}\\ =-e^{3\theta }\cos \theta +3\left[e^{3\theta }\sin \theta -{\displaystyle \int 3e^{3\theta }\sin \theta d\theta }\right]\\ =-e^{3\theta }\cos \theta +3e^{3\theta }\sin \theta -9{\displaystyle \int e^{3\theta }\sin \theta d\theta }\\ {\displaystyle \int e^{3\theta }\sin }\theta d\theta +9{\displaystyle \int e^{3\theta }\sin \theta d\theta }=-e^{3\theta }\cos \theta +3e^{3\theta }\sin \theta \\ 10{\displaystyle \int e^{3\theta }\sin \theta }d\theta =e^{3\theta }\left(-\cos \theta +3\sin \theta \right)\end{array}$

Rewrite the above equation in terms of ${\displaystyle \int e^{3\theta }\sin \theta }d\theta$.

${\displaystyle \int e^{3\theta }\sin \theta }d\theta =\frac{e^{3\theta }}{10}\left(-\cos \theta +3\sin \theta \right)$

Calculate $\frac{a^3}{3}{\displaystyle \underset{0}{\overset{\pi }{\int }}{e}^{3\theta }\sin \theta }d\theta$ by using the above result.

$\begin{array}{c}\frac{a^3}{3}{\displaystyle \underset{0}{\overset{\pi }{\int }}{e}^{3\theta }\cos \theta }d\theta =\frac{a^3}{3}{\left[\frac{e^{3\theta }}{10}\left(-\cos \theta +3\sin \theta \right)\right]}_0^{\pi }\\ =\frac{a^3}{3}\left[\frac{e^{3\pi }}{10}\left(-\cos \pi +3\sin \pi \right)-\frac{e^0}{10}\left(-\cos 0+3\sin 0\right)\right]\\ =\frac{a^3}{3}\left[\frac{e^{3\pi }}{10}+\frac{1}{10}\right]\\ =\frac{a^3}{30}\left(e^{3\pi }+1\right)\\ =413.09a^3\end{array}$

Rewrite equation (I) by substituting the above result.

${\displaystyle \int {\overline{y}}_{EL}dA}=413.09a^3$

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 $133.623a^2$ for $A$ and $-1239.26a^3$ for ${\displaystyle \int {\overline{x}}_{EL}dA}$.

$\overline{x}\left(133.623a^2\right)=-1239.26a^3$

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

$\begin{array}{c}\overline{x}=\frac{-1239.26a^3}{133.623a^2}\\ =-9.27a\end{array}$

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 $133.623a^2$ for $A$ and $413.09a^3$ for ${\displaystyle \int {\overline{y}}_{EL}dA}$.

$\begin{array}{c}\overline{y}=\frac{413.09a^3}{133.623a^2}\\ =3.09a\end{array}$

Therefore, the centroid is located at $\left(-9.27a,3.09a\right)$.