Solution:
we are given that [tex]r = \frac{3}{2} sin(2\theta)[/tex]
when [tex] r=0, \theta=0, \pi/2 [/tex] which corresponds to one loop.
Therefore area of the region enclosed by this curve is
[tex]A=\frac{9}{4}* \frac{1}{2} \int_0^ {\pi/2} sin^2 2\theta d\theta\\
\\ A=\frac{9}{8}\int_0^ {\pi/2} \frac{1-cos 4\theta}{2} d \theta\\
\\
A=\frac{9}{8} [\theta-\frac{sin 4\theta}{2} ]_0^{\pi/2}\\
\\
A=\frac{9}{8}*\pi/2\\
\\
A=\frac{9 \pi}{16}\\[/tex]
Area of the required loop is [tex] \frac{9 \pi}{16} )[/tex] only.
