Answer
$3 \pi$
Work Step by Step
The area of a given region can be calculated as: $Area; A=\dfrac{1}{2}\int_C xdy-y dx$
Here, we are given that $x=t-\sin t; y=1-\cos t$
Therefore,
$A=\dfrac{1}{2}\int_C (t-\sin t) \sin t \ dt -(1-\cos t) (1-\cos t) \ dt \\=\dfrac{1}{2}\int_0^{2 \pi} (t \sin t -2+2 \cos t) \ dt \\=\dfrac{1}{2} [-t \cos t +\sin t -2t +2 \sin t]_0^{2 \pi} \\=-3 \pi$
But the area cannot be negative, so we have $Area=3 \pi$