Chapter 11.3, Problem 4QQ

(a)

To determine

To find: The area bounded by the curve and the x axis.

Answer to Problem 4QQ

The required area is $4.382$ .

Explanation of Solution

Given Data:

The given equation for the polar curve is,

  $r=\theta +\sin 2\theta$

The range is,

  $0\le \theta \le \pi$

Calculation:

Consider that for the given polar curve $r=f\left(\theta \right)$ is given then the area of the region between the origin and the curve $r=f\left(\theta \right)$ for $\alpha \le \theta \le \beta$ is given by,

  $\begin{array}{c}A={\displaystyle {\int }_{\alpha }^{\beta }\frac{1}{2}{r}^{3}d\theta }\\ ={\displaystyle {\int }_{\alpha }^{\beta }{\left(f\left(\theta \right)\right)}^2}d\theta \\ ={\displaystyle {\int }_0^{\pi }\frac{1}{2}{\left(\theta +\sin 2\theta \right)}^2}d\theta \\ =4.382\end{array}$

(b)

To determine

To find: The angle $\theta$ that corresponds to the point on the curve where $x=-2$ .

Answer to Problem 4QQ

The value of $\theta =2.786°$ .

Explanation of Solution

Consider for any point on the curved in polar coordinates system the $x=r\cos \theta$ and $x=-2$ then,

  $\begin{array}{l}r\cos \theta =-2\\ \left(\theta +\sin 2\theta \right)=-2\\ \theta =2.786°\end{array}$

(c)

To determine

To find: The way in which the $\frac{\pi }{3}<\theta <\frac{2\pi }{3}$ and $\frac{dr}{d\theta }$ is negative and is seen from graph.

Answer to Problem 4QQ

The $\frac{\pi }{3}<\theta <\frac{2\pi }{3}$ is negative then the graph is getting closer to the origin as $\theta$ and it increases from $\frac{\pi }{3}$ to $\frac{2\pi }{3}$ .

Explanation of Solution

Consider for $\frac{\pi }{3}<\theta <\frac{2\pi }{3}$ is negative then the graph is getting closer to the origin as $\theta$ and it increases from $\frac{\pi }{3}$ to $\frac{2\pi }{3}$ .

(d)

To determine

To find: The angle $\theta$ in the interval $0\le \theta \le \frac{\pi }{2}$ is the curve farthest away from the origin.

Answer to Problem 4QQ

The curve is farthest away from the origin at $\theta =\frac{\pi }{3}$ .

Explanation of Solution

In order to obtain the angle $\theta$ at which the curve is farthest away from the origin the value of $r=\theta +\sin 2\theta$ for $0\le \theta \le \frac{\pi }{2}$ then,

  $\frac{dr}{d\theta }=1+2\cos \theta$

For maximum value,

  $\begin{array}{l}0=1+2\cos \theta \\ \theta =\frac{\pi }{3}\end{array}$

Therefore,

  $\frac{dr}{d\theta }=1+2\cos 2\theta >0\text{for}0<\theta <\frac{\pi }{3}$

Also,

  $\frac{dr}{d\theta }=1+2\cos 2\theta <0\text{for}\frac{\pi }{3}<\theta <\frac{\pi }{2}$

This shows that the curve is farthest away from the origin at $\theta =\frac{\pi }{3}$ .