Question

Find the area of the region using double integral.

Short answer

The area of the region is \(\frac{\pi }{{12}}\).

Step-by-step solution

Formula used

If\(f\)is a polar rectangle\(R\)given by\(0 \le a \le r \le b,\alpha \le \theta \le \beta \), where\(0 \le \beta - \alpha \le 2\pi \), then,

Substitute \(x = r\cos \theta \) and \(y = r\sin \theta \) in the equation (1) 

Integrate with respect to \(r\) apply the limit as shown below.

\(\begin{aligned}{l}\int_{ - \frac{x}{6}}^{\frac{\pi }{6}} {\int_0^{\cos 3\theta } r } drd\theta = \int_{ - \frac{x}{6}}^{\frac{\pi }{6}} {\left( {\frac{{{r^2}}}{2}} \right)_0^{\cos 3\theta }} d\theta \\ = \int_{ - \frac{\pi }{6}}^{\frac{\pi }{6}} {\left( {\frac{{{{(\cos 3\theta )}^2}}}{2} - \frac{{{0^2}}}{2}} \right)} d\theta \\ = \frac{1}{2}\int_{ - \frac{\pi }{6}}^{\frac{\pi }{6}} {{{\cos }^2}} 3\theta d\theta \end{aligned}\)

\( = \frac{1}{2}\int_{ - \frac{\pi }{6}}^{\frac{\pi }{6}} {\frac{1}{2}} (1 + \cos 6\theta )d\theta \)

Integrate with respect to \(\theta \) apply the limit for \(\theta \)

\(\int_{ - \frac{\pi }{6}}^{\frac{\pi }{6}} {\int_0^{\cos 3\theta } r } drd\theta = \frac{1}{4}\left( {\theta + \frac{1}{6}\sin 6\theta } \right)_{ - \frac{\pi }{6}}^{\frac{\pi }{6}}\)

\( = \frac{1}{4}\left( {\left( {\frac{\pi }{6} + \frac{1}{6}\sin \frac{{6\pi }}{6}} \right) - \left( {\frac{{ - \pi }}{6} + \frac{1}{2}\sin \left( { - \frac{{6\pi }}{6}} \right)} \right)} \right)\)

\( = \frac{1}{4}\left( {\left( {\frac{\pi }{6} + \frac{1}{2}(0)} \right) + \left( {\frac{\pi }{6} + \frac{1}{2}(0)} \right)} \right)\)

\( = \frac{1}{4}\frac{\pi }{3} + 0\)

Thus, the area of the region is \(\frac{\pi }{{12}}\).