Question

Complete Example 2 by showing that

$$\begin{gathered} {\int }_{-\pi /4}^{\pi /4}\frac{k}{3}\left(8{\cos }^3\theta -{\sec }^3\theta \right)d\theta \\ =\frac{k}{9}(17\sqrt{2}+3\ln (\sqrt{2}-1\left)\right) \end{gathered}$$

Short answer

The mass of semicircular lamina is

$m=\frac{k}{9}(17\sqrt{2}+3\ln (\sqrt{2}-1\left)\right)$

Step-by-step solution

Given information

The expression is

$$\begin{gathered} {\int }_{-\pi /4}^{\pi /4}\frac{k}{3}\left(8{\cos }^3\theta -{\sec }^3\theta \right)d\theta \\ =\frac{k}{9}(17\sqrt{2}+3\ln (\sqrt{2}-1\left)\right) \end{gathered}$$

Calculation

Density $\rho \left(r\right)=kr$

Equation of circle

$r=2\cos \theta$

Equation of line $x=1$in polar form $r=\sec \theta$

Mass of the plate

$$\begin{gathered} m={\iint }_{\Omega }\rho (x,y)dA \\ m={\int }_{-\pi /4}^{\pi /4}{\int }_{\sec \theta }^{2\cos \theta }k{r}^{2}drd\theta \end{gathered}$$

Integrate the inner integral with respect to $r$first.

$$\begin{gathered} m=k{\int }_{-\pi /4}^{\pi /4}{\left[\frac{r^3}{3}\right]}_{\sec \theta }^{2\cos \theta }d\theta \\ m=k{\int }_{-n/4}^{\pi /4}\left[\frac{8{\cos }^3\theta -{\sec }^3\theta }{3}\right]d\theta \end{gathered}$$

Substitute the limits

$m=\frac{k}{3}{\int }_{-\pi /4}^{\pi /4}\left[8{\cos }^3\theta -{\sec }^3\theta \right]d\theta$

$m=\frac{k}{3}\left[\frac{8}{12}\{9\sin \theta +\sin 3\theta \}-\frac{1}{2}{\left\{\begin{array}{l}\tan \theta \sec \theta -\ln \left(\cos \left(\frac{\theta }{2}\right)-\sin \left(\frac{\theta }{2}\right)\right)+\\ \ln \left(\cos \left(\frac{\theta }{2}\right)+\sin \left(\frac{\theta }{2}\right)\right)\end{array}\right]}_{-\pi /4}^{\pi /4}\right.$

$m=\frac{k}{3}\left[\begin{array}{l}\frac{8}{12}\left\{9\sin \left(\frac{\pi }{4}\right)+\sin \left(\frac{3\pi }{4}\right)\right\}-\frac{1}{2}\left\{\begin{array}{l}\tan \left(\frac{\pi }{4}\right)\sec \left(\frac{\pi }{4}\right)-\ln \left(\cos \left(\frac{\pi }{8}\right)-\sin \left(\frac{\pi }{8}\right)\right)+\\ \ln \left(\cos \left(\frac{\pi }{8}\right)+\sin \left(\frac{\pi }{8}\right)\right)d\end{array}\right\}\\ -\frac{8}{12}\left\{9\sin \left(-\frac{\pi }{4}\right)+\sin \left(-\frac{3\pi }{4}\right)\right\}+\frac{1}{2}\left\{\begin{array}{l}\tan \left(-\frac{\pi }{4}\right)\sec \left(-\frac{\pi }{4}\right)-\ln \left(\begin{array}{l}\cos \left(-\frac{\pi }{8}\right)\\ \ln \left(\cos \left(-\frac{\pi }{8}\right)+\sin \left(-\frac{\pi }{8}\right)\right)\end{array}\right]+3\\ \ln \left(-\frac{\pi }{8}\right)\end{array}\right\}\end{array}\right]$

$m=\frac{k}{9}(17\sqrt{2}+3\ln (\sqrt{2}-1\left)\right)$

Thus, the mass of semicircular lamina is

$m=\frac{k}{9}(17\sqrt{2}+3\ln (\sqrt{2}-1\left)\right)$