Q. 23 - Calculus

Question

Complete Example 3 by showing that

$$\begin{gathered} {\int }_{-\pi /4}^{\pi /4}{\int }_{\sec \theta }^{2\cos \theta }k{r}^3\cos \theta drd\theta \\ =\frac{k}{60}(157\sqrt{2}+15\ln (\sqrt{2}-1\left)\right). \end{gathered}$$

Step-by-Step Solution

Answer

The first moment of the mass about y-axis is 

$M_y=k{\int }_{-\pi /4}^{\pi /4}{\int }_{\sec \theta }^{2\cos \theta }{r}^3\cos \theta drd\theta =\frac{k}{60}(157\sqrt{2}+15\ln (\sqrt{2}-1\left)\right)$

Step 1: Given information

The expression is 

$$\begin{gathered} {\int }_{-\pi /4}^{\pi /4}{\int }_{\sec \theta }^{2\cos \theta }k{r}^3\cos \theta drd\theta \\ =\frac{k}{60}(157\sqrt{2}+15\ln (\sqrt{2}-1\left)\right). \end{gathered}$$

Step 2: Calculation

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

Equation of circle

$r=2\cos \theta$

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

First moment of the mass about $y$ - axis is

$$\begin{gathered} M_y={\iint }_{\Omega }x\rho (x,y)dA \\ {M}_y={\iint }_{\Omega }k{r}^{3}drd\theta \\ {M}_y=k{\int }_{-\pi /4}^{n/4}{\int }_{\sec \theta }^{2\cos \theta }{r}^3\cos \theta drd\theta \end{gathered}$$

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

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

Substitute the limits

$M_y=\frac{k}{4}{\int }_{-\pi /4}^{\pi /4}\left[16{\cos }^4\theta {\sec }^4\theta \right]d\theta$

Integrate with respect to $\theta .$

$$\begin{gathered} M_y=\frac{k}{4}{\left[\frac{16}{32}\{12\theta +8\sin 2\theta +\sin 4\theta \}-\frac{1}{3}\left\{(\cos 2\theta +2)\tan \theta {\sec }^2\theta \right\}\right]}_{-\pi /4}^{\pi /4} \\ {M}_y=\frac{k}{4}\left[\frac{1}{2}\left\{12\right(\pi /4)+8\sin (\pi /2)+\sin (\pi \left)\right\}-\frac{1}{3}\left\{\left(\cos \right(\pi /2)+2)\tan (\pi /4){\sec }^2(\pi /4)\right\}\right] \\ -\frac{k}{4}\left[\frac{1}{2}\left\{12\right(-\pi /4)+8\sin (-\pi /2)+\sin (-\pi \left)\right\}-\frac{1}{3}\left\{\left(\cos \right(-\pi /2)+2)\tan (-\pi /4){\sec }^2(-\pi /4)\right\}\right] \\ {M}_y=\frac{k}{60}(157\sqrt{2}+15\ln (\sqrt{2}-1\left)\right) \end{gathered}$$

Thus, the first moment of the mass about $y$ - axis is

$M_y=k{\int }_{-\pi /4}^{\pi /4}{\int }_{\sec \theta }^{2eos\theta }{r}^3\cos \theta drd\theta =\frac{k}{60}(157\sqrt{2}+15\ln (\sqrt{2}-1\left)\right)$