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