Answer
$\dfrac {1}{2}\left( \sqrt {3}-\dfrac {\pi }{6}-\dfrac {1}{\sqrt {3}}\right) \approx 0.315 $
Work Step by Step
$A=\int ^{\pi /3}_{\pi /6}\dfrac {1}{2}\left[ r^{2}\right] d\theta =\int ^{\pi /3}_{\pi /6}\dfrac {1}{2}\tan ^{2}\theta d\theta =\dfrac {1}{2}\int ^{\pi /3}_{\pi /6}\left( {sec^2\theta }-1\right) d\theta =\dfrac {1}{2}\left[ \tan \theta -\theta \right] ^{\pi /3}_{\pi /6}=\dfrac {1}{2}\left( \sqrt {3}-\dfrac {\pi }{3}-\left( \dfrac {1}{\sqrt {3}}-\dfrac {\pi }{6}\right) \right) =\dfrac {1}{2}\left( \sqrt {3}-\dfrac {\pi }{6}-\dfrac {1}{\sqrt {3}}\right) \approx 0.315 $