Question

Find the area of the region that is bounded by the given curve and lies in the specified sector.

\({\rm{r = cos\theta ,0}} \le {\rm{\theta }}{{ \le {\rm{\pi }}} \mathord{\left/

{\vphantom {{ \le {\rm{\pi }}} {\rm{6}}}} \right.

\kern-\nulldelimiterspace} {\rm{6}}}\)

Short answer

The area of the region is\(\frac{{2\pi + 3\sqrt 3 }}{{48}} \approx 0.239\).

Step-by-step solution

The region to be discovered is defined by the pink portion of the curve.

\({\rm{r = cos\theta ,0}} \le {\rm{\theta }}{{ \le {\rm{\pi }}} \mathord{\left/

{\vphantom {{ \le {\rm{\pi }}} {\rm{6}}}} \right.

\kern-\nulldelimiterspace} {\rm{6}}}\)

This graph is shaped like a circle.

Find the area of the region.

Make use of the formula for polar area and power reduction.

\(\begin{aligned}{c}{\cos ^2}x = \frac{1}{2}(1 + \cos 2x)\\A = \frac{1}{2}\int_a^b {{r^2}} d\theta \\ = \frac{1}{2}\int_0^{\pi /6} {{{(\cos \theta )}^2}} d\theta \\ = \frac{1}{2}\int_0^{\pi /6} {{{\cos }^2}} \theta d\theta \\ = \frac{1}{2}\int_0^{\pi /6} {\frac{1}{2}} (1 + \cos 2\theta )d\theta \\ = \frac{1}{4}\left( {\theta + \frac{1}{2}\sin 2\theta } \right)_0^{\pi /6}\\ = \frac{1}{4}\left( {\frac{\pi }{6} + \frac{1}{2}\sin \frac{\pi }{3} - \left( {0 + \frac{1}{2}\sin 0} \right)} \right)\\ = \frac{1}{4}\left( {\frac{\pi }{6} + \frac{{\sqrt 3 }}{4} - 0} \right)\\ = \frac{\pi }{4} + \frac{{\sqrt 3 }}{{16}}\\ = \frac{{2\pi + 3\sqrt 3 }}{{48}}\\ = 0.239\end{aligned}\)

Therefore, the area of the region is\(\frac{{2\pi + 3\sqrt 3 }}{{48}} \approx 0.239\).