Hello,
\(Area= \int\limits^a_b {\rho^2} \, d\theta\)
\(\frac{1}{2} \int\limits^{\frac{\pi}{5}} _0 {(3\ sin(5\theta))^2} \, d\theta\\\\
=\frac{9}{4} \int\limits^{\frac{\pi}{5}} _0 {(1-cos(10\theta))} \, d\theta\\\\
=\frac{9}{4} [\theta-\frac{sin(10\theta)}{10}]^\frac{\pi}{5}} _0 \\\\
=\frac{9\pi}{20}\)
≈1.41371669412
