Chapter 9, Problem 9.92RP

Use integration to find $I_x,I_y,$ and $I_{xy}$ for the shaded region shown.

To determine

  $I_x,I_{y}and{I}_{xy}$ for the shaded region.

Answer to Problem 9.92RP

  $I_x=21.9i{n}^4$

  $I_y=21.9i{n}^4$

  $I_{xy}=21.3i{n}^4$

Explanation of Solution

Given information:

The shaded region:

Calculations:

Choosing a small differential area element as shown in the figure.

Using double integration:

  $\begin{array}{l}\text{The moment of inertia with respect to x-axis is defined as:}\\ I_x={\displaystyle \underset{A}{\int }{y}^2}dA\\ \therefore I_x={\displaystyle {\int }_0^4}{\displaystyle {\int }_{x^2/4}^{2\sqrt{x}}{y}^2}dydx={\displaystyle {\int }_0^4{\left[\frac{y^3}{3}\right]}_{x^2/4}^{2\sqrt{x}}}dx\\ =\frac{1}{3}{\displaystyle {\int }_0^4\left(8x^{3/2}-\frac{x^6}{64}\right)}dx=\frac{1}{3}{\left[\frac{16x^{5/2}}{5}-\frac{x^7}{448}\right]}_0^4\\ \Rightarrow I_x=21.9i{n}^4\\ \text{And, the moment of inertia with respect to y-axis:}\\ I_y={\displaystyle \underset{A}{\int }{x}^2}dA\\ I_y={\displaystyle {\int }_0^4{\displaystyle {\int }_{x^2/4}^{2\sqrt{x}}{x}^2}dydx={\displaystyle {\int }_0^4{x}^2{\left[y\right]}_{x^2/4}^{2\sqrt{x}}}dx}\\ ={\displaystyle {\int }_0^4\left(2x^{5/2}-\frac{x^4}{4}\right)}dx={\left[\frac{4}{7}{x}^{7/2}-\frac{x^5}{20}\right]}_0^4\\ \Rightarrow I_y=21.9i{n}^4\end{array}$

  $\begin{array}{l}Notethat{I}_x=I_y,whichwas\exp ectedbecauseofsymmetry.\\ \text{The product of inertia is calculated as:}\\ I_{xy}={\displaystyle \underset{A}{\int }xy}dA\\ \therefore I_{xy}={\displaystyle {\int }_0^4{\displaystyle {\int }_{x^2/4}^{2\sqrt{x}}xy}dydx={\displaystyle {\int }_0^{4}x{\left[\frac{y^2}{2}\right]}_{x^2/4}^{2\sqrt{x}}}dx={\displaystyle {\int }_0^4\left(2x^2-\frac{x^5}{32}\right)dx}}\\ ={\left[\frac{2}{3}{x}^2-\frac{x^6}{192}\right]}_0^4\\ \Rightarrow I_{xy}=21.3i{n}^4\end{array}$

Conclusion:

For the shaded region shown, $I_x=21.9i{n}^4$, $I_y=21.9i{n}^4$ and $I_{xy}=21.3i{n}^4$.