Answer
$30 \pi$
Work Step by Step
When $C$ is in counterclockwise direction then $A=\oint_{C} x dy=-\oint_{C} y dx$ and when $C$ is in clockwise direction then $A=-\oint_{C} x dy=\oint_{C} dx$
Now, $x =5 \cos t-\cos 5t; dx=[-5 \sin t +5 \sin (5t) ] dt$ and $y=5 \sin t -\sin 5t; dy=[5\cos t -5 \cos 5t] dt$
Let us consider that $F=(\dfrac{-y}{2}, \dfrac{x}{2})$
Now, $A=\oint_{C} F dr=\oint_{C} \dfrac{-y}{2} dx+\dfrac{x}{2} dy$
$= \int_{0}^{2 \pi} [-(5 \sin t -\sin 5t)( [-5 \sin t +5 \sin (5t) ])] + (5 \cos t-\cos 5t) ([-5 \sin t +5 \sin (5t) ] dt) \times \dfrac{1}{2}$
$=\int_{0}^{2 \pi} 30-30 \sin t \times \sin 5t -30 \cos t \cos (5t) \times \dfrac{1}{2}$
$=\int_{0}^{2 \pi} 30 \ dt + \int_{0}^{2 \pi} (-15) \times [\cos (-4t) -\cos 6t] -15 [\cos (-4t) +\cos 6t) dt \times \dfrac{1}{2}$
$= 30 \pi+[\dfrac{-15 \sin 6t}{4}]_0^{2 \pi}$
$=30 \pi$
