Chapter H.2, Problem 25E

To determine

To calculate:

The area of the region made by the following two curves:

  $\begin{array}{l}r=\sin 2\theta \\ r=\cos 2\theta \end{array}$

Answer to Problem 25E

The area of the region enclosed by two curves is $\frac{\pi }{2}-1$ .

Explanation of Solution

Given information:

  $\begin{array}{l}r=\sin 2\theta \\ r=\cos 2\theta \end{array}$

Calculation:

The equation of the region is

  $\begin{array}{l}r=\sin 2\theta \\ r=\cos 2\theta \end{array}$

So, the curves are plotted as below:

  

Now, need to find the interval in which the inner loop exists:

  $\begin{array}{l}\sin 2\theta =\cos 2\theta \\ 2\theta =\frac{\pi }{4}\\ \theta =\frac{\pi }{8}\end{array}$

The total area of two curves region is sixteen times of area region from $0$ to $\frac{\pi }{8}$

The area of given equation of the polar region is:

  $\begin{array}{l}A=16{\displaystyle \underset{0}{\overset{\frac{\pi }{8}}{\int }}}{\left(\sin 2\theta \right)}^{2}d\theta \\ A=16{\displaystyle \underset{0}{\overset{\frac{\pi }{8}}{\int }}}\left(\frac{1-\cos 4\theta }{2}\right)d\theta \left[\because {\sin }^{2}x=\frac{1-\cos 2x}{2}\right]\\ A=8{\displaystyle \underset{0}{\overset{\frac{\pi }{8}}{\int }}}\left(1-\cos 4\theta \right)d\theta \\ A=8{\left(1-\sin 4\theta \right)}_0^{\frac{\pi }{8}}\\ A=8\left(\frac{\pi }{8}-\frac{\sin 4\left(\frac{\pi }{8}\right)}{4}-0+\sin \left(0\right)\right)\\ A=8\left(\frac{\pi }{8}-\frac{1}{4}\right)\\ A=\frac{\pi }{2}-1\end{array}$