Chapter 9.6, Problem 66E

To determine

To calculate:

The area of loop of the graph buy using graph $r^2=16\cos 2\theta$ .

Answer to Problem 66E

The area of loop of the graph is $8$ square units.

Explanation of Solution

Given information:

Area of lemiscate $a^2$

lemiscate $r^2=a^2\cos 2\theta$

Calculation:

Draw the graph of $r^2=16\cos 2\theta$ .

Type of curve is lemniscate.

This is the symmetric with respect to the polar axis, $\theta =\frac{\pi }{2}$ and pole.

Maximum value of $\left|r\right|$ is $4$ for $\theta =0,\pi$ .

Zeros of $r:$ here $r=0$ for $\theta =\frac{\pi }{4},-\frac{\pi }{4}$ .

Here $\cos 2\theta \ge 0$ for $-\frac{\pi }{4}\le \theta \le \frac{\pi }{4}$ or $-\frac{5\pi }{4}\le \theta \le \frac{5\pi }{4}$ .

Consider the following values:

$\theta$	$r=\pm 4\sqrt{\cos 2\theta }$
$-\frac{\pi }{4}$	$0$
$-\frac{\pi }{6}$	$\pm 2\sqrt{2}$
$0$	$\pm 4$
$\frac{\pi }{6}$	$\pm 2\sqrt{2}$
$\frac{\pi }{4}$	$0$

Area of lemniscate $r^2=a^2\cos 2\theta$ is $a^2$ .

Area of $r^2=16\cos 2\theta$ is $16$ .

Area of one loop is $\frac{1}{2}\left(16\right)=8$

Hence, the required result is $8$ square units.