## Area of the Leaf of a Four Leaf Rose

The equation of a four leaf rose is
$r=sin \: 2 \theta$
.

According to the polar form of Green's Theorem, the area enclosed by a closed curve is
$A= \frac{1}{2} \oint r^2 \: d \theta$

Hence the area enclosed by the outline of a leaf of a four leaf rose is
\begin{aligned} A &= \frac{1}{2} \int^{ \pi /2}_0 r^2 d \theta \\ &= \frac{1}{2} \int^{ \pi /2}_0 sin^2 \: 2 \theta d \theta \\ &= \frac{1}{2} \int^{ \pi /2}_0 \frac{1}{2} - \frac{ cos \: 2 \theta}{2} d \theta \\ &= \frac{1}{2} [ \frac{ \theta}{2} - \frac{ sin \: 2 \theta}{4}]^{ \pi /2}_0 \\ &= \frac{1}{2} (( \frac{\pi}{4} -0-(0-0)) \\ &= \frac{\pi}{8} \end{aligned}