Chapter 10.3, Problem 60E

(a)

To determine

The integral expression to find out the area of R.

Answer to Problem 60E

The integral expression is $A={\displaystyle \underset{0}{\overset{\frac{3}{4}}{\int }}\left(\sqrt{1+y^2}-\frac{5}{3}y\right)dy}$

Explanation of Solution

Given:

The line given is $x=\frac{5}{3}y$ and the curve is $x=\sqrt{1+y^2}$ .

Calculation:

First find the intersection point in order to find the area:

As, $x=\frac{5}{3}y$ and $x=\sqrt{1+y^2}$

Therefore,

  $\begin{array}{l}\frac{5}{3}y=\sqrt{1+y^2}\\ \frac{25}{9}{y}^2=1+y^2\\ \frac{16}{9}{y}^2=1\\ y=\frac{3}{4}\end{array}$

Now, area can be find out as:

  $A={\displaystyle \underset{0}{\overset{\frac{3}{4}}{\int }}\left(\sqrt{1+y^2}-\frac{5}{3}y\right)dy}$

Conclusion:

The integral is $A={\displaystyle \underset{0}{\overset{\frac{3}{4}}{\int }}\left(\sqrt{1+y^2}-\frac{5}{3}y\right)dy}$

(b)

To determine

The curve $x=\sqrt{1+y^2}$ can be described in polar coordinates $r^2=\frac{1}{{\cos }^2\theta -{\sin }^2\theta }$

Answer to Problem 60E

The curve is described in polar coordinates.

Explanation of Solution

Given:

The line given is $x=\frac{5}{3}y$ and the curve is $x=\sqrt{1+y^2}$ .

Calculation:

  $x=r\cos \theta \text{ ; y = r}\sin \theta$

Now,

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

Conclusion:

The curve is described in polar coordinates.

(c)

To determine

The integral expression with respect to $\theta$ in order to find out the area of R.

Answer to Problem 60E

The integral expression is $A={\displaystyle \underset{0}{\overset{\frac{3}{4}}{\int }}\left(\sqrt{1+y^2}-\frac{5}{3}y\right)dy}$

Explanation of Solution

Given:

The line given is $x=\frac{5}{3}y$ and the curve is $x=\sqrt{1+y^2}$ .

Calculation:

Let the origin be O and the intersection point be P. The angle formed by the segment OP with x-axis be $\alpha$

  $\begin{array}{l}\tan \alpha =\frac{y}{x}=\frac{\frac{3}{4}}{\frac{5}{4}}=\frac{3}{5}\\ A={\displaystyle \underset{0}{\overset{{\tan }^{-1}\left(\frac{3}{5}\right)}{\int }}\frac{1}{{\cos }^2\theta -{\sin }^2\theta }}d\theta \end{array}$

Conclusion:

The integral is $A={\displaystyle \underset{0}{\overset{\frac{3}{4}}{\int }}\left(\sqrt{1+y^2}-\frac{5}{3}y\right)dy}$