## Thomas' Calculus 13th Edition

$3 \pi$
Consider, $A=\oint_{C} \dfrac{x}{2} dy - \dfrac{y}{2} dx$ We have $dx=1- \cos t dt$ and $dy= \sin t dt$ Now $Area=\int_{2 \pi}^{0} \dfrac{1}{2} (t-\sin t) (\sin t dt) -\dfrac{1}{2} (1-\cos t) (1-\cos t \ dt) \\=\dfrac{1}{2} \int_{2 \pi}^{0} t \sin t -\sin^2 t -1-\cos^2 t + 2 \cos t d t \\=\dfrac{1}{2} [\int t \sin t \ dt]_0^{2 \pi} -[t+\sin t]_{2 \pi}^0 \\=\dfrac{1}{2} [ -t \cos t +\int \cos t ]_{2 \pi}^{0} +2 \pi \\=3 \pi$