Chapter 11.3, Problem 60E

(a)

To determine

To find: The set up and the evaluate an integral expression with respect to $y$ that gives the area of $R$ .

Answer to Problem 60E

The area of the region is $\frac{1}{2}\ln 2$ .

Explanation of Solution

Given Data:

The given equation for the curve is,

  $\begin{array}{l}x=\frac{5}{3}y\\ x=\sqrt{1+y^2}\end{array}$

The given graph is shown in Figure 1

  

Figure 1

Calculation:

From the given equations,

  $\sqrt{1+y^2}=\frac{5}{3}y$

Solve the above equation as,

  $\begin{array}{l}\frac{25}{9}{y}^2=1+y^2\\ \frac{16}{9}{y}^2=1\\ y=\frac{3}{4}\end{array}$

From the given figure $x=\sqrt{1+y^2}>x=\frac{5}{3}y$ in the interval $\left[0,\frac{3}{4}\right]$ .

Then, the area of the region bounded is,

  $\begin{array}{c}A={\displaystyle {\int }_0^{\frac{3}{4}}\left[\sqrt{1+y^2}-\frac{5y}{3}\right]}dy\\ ={\left[\frac{y}{2}\sqrt{1+y^2}+\frac{1}{2}\ln \left|y+\sqrt{1+y^2}\right|-\frac{5y^2}{6}\right]}_0^{\frac{3}{4}}\\ =\left[\frac{3}{7}\left(\frac{5}{4}\right)+\frac{1}{2}\ln \left|\frac{3}{4}+\frac{5}{4}\right|-\frac{15}{32}\right]\\ =\frac{1}{2}\ln 2\end{array}$

(b)

To determine

To Show: The curve $x=\sqrt{1+y^2}$ is descried in polar coordinates by $r^2=\frac{1}{{\cos }^2\theta -{\sin }^2\theta }$ .

Explanation of Solution

Consider the equation is,

  $\begin{array}{l}x=\sqrt{1+y^2}\\ x^2=1+y^2\\ {\left(r\cos \theta \right)}^2-{\left(r\sin \theta \right)}^2=1\\ r^2=\frac{1}{{\cos }^2\theta -{\sin }^2\theta }\end{array}$

Hence, proved.

(c)

To determine

To find: The polar equation in part (b) to set up the integral expression with respect to $\theta$ that givens the area of R .

Answer to Problem 60E

The required equation is $R={\displaystyle {\int }_0^{{\tan }^{-1}\frac{3}{5}}\frac{1}{2}\left(\frac{1}{{\cos }^2\theta -{\sin }^2\theta }\right)d}\theta$ .

Explanation of Solution

Consider the slope of the line $x=\frac{5}{3}y$ is $\frac{3}{5}$ then,

  $\begin{array}{l}m=\frac{3}{5}\\ \tan \delta =\frac{3}{5}\\ \delta ={\tan }^{-1}\left(\frac{3}{5}\right)\end{array}$

Then the region is bounded between the origin and the curve $r-f\left(\theta \right)$ for $\alpha \le \theta \le \beta$ then the area is,

  $\begin{array}{c}A={\displaystyle {\int }_2^{\beta }{r}^{2}d\theta }\\ ={\displaystyle {\int }_{\alpha }^{\beta }\frac{1}{2}}{\left(f\left(\theta \right)\right)}^{2}d\theta \end{array}$

The area of the region R is,

  $R={\displaystyle {\int }_0^{{\tan }^{-1}\frac{3}{5}}\frac{1}{2}\left(\frac{1}{{\cos }^2\theta -{\sin }^2\theta }\right)d}\theta$