Question

Find the moments of inertia about the x - and y-axes for the semicircular lamina described in Example 2. Assume that the density at every point is proportional to the distance of the point from the origin.

Short answer

The moment of inertia about the x and y axes are

$I_x=\frac{k}{5}\left[\frac{821}{210\sqrt{2}}+\frac{1}{8}\ln (3+2\sqrt{2})\right]$

$I_y=\frac{k}{5}\left[\frac{673\sqrt{2}}{35}-2{\tanh }^{-1}\left(\tan \left(\frac{\pi }{8}\right)\right)\right]$

Step-by-step solution

part (a) step 1: Given information

The objective of this problem is to find the moments of inertia about x - axis and y- axis for the semicircular lamina . Density of region is proportional to the distance of the point from origin.

part (a) step 1: calculation

Density $\rho (x,y)=k\sqrt{x^2+y^2}$

Moment of inertia about x - axis

$$\begin{gathered} I_x={\iint }_{\Omega }{y}^2\rho (x,y)dA \\ {I}_x={\iint }_{\Omega }{y}^{2}k\sqrt{x^2+y^2}dydx \end{gathered}$$

Substitute x=r \cos \theta, y=r \sin \theta$ and d x d y=r d r d \theta

Equation of circle

localid="1650635096537" $r=2\cos \theta$

Equation of linelocalid="1650635103420" $x=1inpolarformr=\sec \theta$

localid="1650635109386" $I_x={\int }_{-\pi /4}^{\pi /4}{\int }_{\sec \theta }^{2\cos \theta }k{r}^4{\sin }^2\theta drd\theta$

Integrate the inner integral with respect to r first.

localid="1650635115504" $I_x=k{\int }_{-\pi /4}^{\pi /4}{\left[\frac{r^5}{5}\right]}_{\sec \theta }^{2\cos \theta }{\sin }^2\theta d\theta$

Substitute the limits

localid="1650635122663" $I_x=\frac{k}{5}{\int }_{-\pi /4}^{\pi /4}\left[32{\sin }^2\theta {\cos }^5\theta -{\sin }^2\theta {\sec }^5\theta \right]d\theta$

localid="1650635130893" $I_x=\frac{k}{5}\left[\frac{821}{210\sqrt{2}}+\frac{1}{8}\ln (3+2\sqrt{2})\right]$

Moment of inertia about y - axis

$$\begin{gathered} I_y={\iint }_{\Omega }{x}^2\rho (x,y)dA \\ {I}_y={\iint }_{\Omega }{x}^{2}k\sqrt{x^2+y^2}dydx \end{gathered}$$

Substitute$x=r\backslash \cos \backslash theta,y=r\backslash \sin \backslash theta\$anddxdy=rdrd\backslash theta$

Equation of circle

$x=r\cos \theta$, $y=r\sin \theta anddxdy=rdrd\theta$

Equation of line $x=1inpolarformr=\sec \theta$

$I_y={\int }_{-\pi /4}^{\pi /4}{\int }_{\sec \theta }^{2\cos \theta }k{r}^4{\cos }^2\theta drd\theta$

Integrate the inner integral with respect to r first.

$I_y=k{\int }_{-\pi /4}^{\pi /4}{\left[\frac{r^5}{5}\right]}_{\cos \theta }^{2\cos \theta }{\cos }^2\theta d\theta$

Substitute the limits

$$\begin{gathered} I_y=\frac{k}{5}{\int }_{-\pi /4}^{\pi /4}\left[32{\cos }^2\theta {\cos }^5\theta -{\cos }^2\theta {\sec }^5\theta \right]d\theta \\ {I}_y=\frac{k}{5}{\int }_{-\pi /4}^{\pi /4}\left[32{\cos }^7\theta -{\sec }^3\theta \right]d\theta \\ {I}_y=\frac{k}{5}\left[\frac{673\sqrt{2}}{35}-2{\tanh }^{-1}\left(\tan \left(\frac{\pi }{8}\right)\right)\right] \end{gathered}$$