Given,
{x}^{2}+{(y-2)}^{2}\leq 4&{x}^{2}\geq 2y
Now on solving, {x}^{2}+{(y-2)}^{2}=4&{x}^{2}=2y we get intersecting point of the curves as (±2,2),
Now plotting the diagram we get,

So, from above diagram area of required region will be,
=2[41(π×22)−∫022⋅ydy]
=2[π−[232⋅y23]02]
=2[π−322⋅22]
=2[π−38]
=2π−316