Chapter 9.1, Problem 9.18P

To determine

Find the moment of inertia and radius of gyration of the shaded area with respect to y axis by using direct integration.

Answer to Problem 9.18P

The moment of inertia of the shaded area with respect to y axis by using direct integration is $\underset{\_}{\frac{1}{8}\pi a^{3}b}$.

The radius of gyration of the shaded area with respect to y axis by using direct integration is $\underset{\_}{\frac{a}{2}}$.

Explanation of Solution

Given information:

The curve equation is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.

Calculation:

Sketch the vertical strip shaded portion as shown in Figure 1.

Write the curve Equation as follows:

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ (1)

Modify Equation (1).

$\begin{array}{l}\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\\ y=b\sqrt{1-\left(\frac{x^2}{a^2}\right)}\end{array}$

Determine the area of the strip element $dA$ as shown in below:

$dA=ydx$

Determine the moment of inertia $\left(d{I}_x\right)$ with respect to x axis of a rectangular strip:

$d{I}_y=x^{2}dA$ (2)

Substitute $ydx$ for $dA$ in Equation (2).

$\begin{array}{c}d{I}_y=x^2\left(ydx\right)\\ I_y={\displaystyle \underset{-a}{\overset{a}{\int }}{x}^2\left(ydx\right)}\\ ={\displaystyle \underset{-a}{\overset{a}{\int }}{x}^2\left(b\sqrt{1-\left(\frac{x^2}{a^2}\right)}\right)}dx\\ =b{\displaystyle \underset{-a}{\overset{a}{\int }}{x}^2\left(\sqrt{1-\left(\frac{x^2}{a^2}\right)}\right)}dx\end{array}$ (3)

Consider $x=a\sin \theta$

Differentiate both sides of the Equation.

$dx=a\cos \theta d\theta$

Find the moment of inertia of the shaded area with respect to y axis by using direct integration.

Substitute $a\sin \theta$ for x and $a\cos \theta d\theta$ for $dx$ and apply the limits in Equation (3).

$\begin{array}{c}{I}_y=b{\displaystyle \underset{-a}{\overset{a}{\int }}{a}^2{\sin }^2\theta \sqrt{1-\left(\frac{a^2{\sin }^2\theta }{a^2}\right)}}a\cos \theta d\theta \\ =b{\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}{a}^2{\sin }^2\theta \sqrt{1-{\sin }^2\theta }a\cos \theta d\theta }\\ =a^{3}b{\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}{\sin }^2\theta {\cos }^2\theta d\theta }\\ =a^{3}b{\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}\frac{1}{4}{\sin }^{2}2\theta d\theta }\end{array}$

$\begin{array}{c}=\frac{1}{4}{a}^{3}b{\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}\frac{1}{2}\left(1-\cos 4\theta \right)d\theta }\\ =\frac{1}{8}{a}^{3}b{\left[\left(\theta -\frac{1}{4}\sin 4\theta \right)\right]}_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\\ =\frac{1}{8}{a}^{3}b\left[\left(\frac{\pi }{2}-\frac{1}{4}\sin 4\left(\frac{\pi }{2}\right)\right)-\left(-\frac{\pi }{2}-\frac{1}{4}\sin 4\left(-\frac{\pi }{2}\right)\right)\right]\\ =\frac{1}{8}{a}^{3}b\left(\frac{\pi }{2}+\frac{\pi }{2}\right)\end{array}$

$\begin{array}{c}=\frac{1}{8}{a}^{3}b\left(\frac{2\pi }{2}\right)\\ I_y=\frac{1}{8}\pi a^{3}b\end{array}$

Thus, the moment of inertia of the shaded area with respect to y axis by using direct integration is $\underset{\_}{\frac{1}{8}\pi a^{3}b}$.

Find the area of shaded portion (A) as shown below:

$A={\displaystyle \int dA}$

Substitute $2xdy$ for $dA$ in Equation (4).

$\begin{array}{c}A={\displaystyle \int 2xdy}\\ =2a{\displaystyle \underset{0}{\overset{b}{\int }}\sqrt{1-\left(\frac{y^2}{b^2}\right)dy}}\end{array}$

Substitute $b\sin \theta$ for y and $b\cos \theta d\theta$ for $dy$ and apply the limit.

$\begin{array}{c}A={\displaystyle \int 2xdy}\\ =2a{\displaystyle \underset{0}{\overset{b}{\int }}\sqrt{1-\left(\frac{b^2{\sin }^2\theta }{b^2}\right)b\cos \theta d\theta }}\\ =2ab{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}{\cos }^2\theta d\theta }\\ =2ab{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{1}{2}\left(1+\cos 2\theta \right)d\theta }\end{array}$

$\begin{array}{c}A=ab{\left[\theta +\frac{1}{2}\sin 2\theta \right]}_0^{\frac{\pi }{2}}\\ =ab\left[\frac{\pi }{2}+\frac{1}{2}\sin 2\left(\frac{\pi }{2}\right)\right]\\ =\frac{1}{2}\pi ab\end{array}$

Find the radius of gyration of the shaded area with respect to y axis by using direct integration.

$k_y^2=\frac{I_y}{A}$ (4)

Here, A is are of shaded portion and $I_y$ is moment of inertia about y axis.

Substitute $\frac{1}{2}\pi ab$ for A and $\frac{1}{8}\pi a^{3}b$ for $I_y$ in Equation (4).

$\begin{array}{c}{k}_y^2=\frac{\frac{1}{8}\pi a^{3}b}{\frac{1}{2}\pi ab}\\ =\frac{1}{4}{a}^2\\ k_y=\frac{1}{2}a\end{array}$

Thus, the radius of gyration of the shaded area with respect to y axis by using direct integration is $\underset{\_}{\frac{a}{2}}$.