Find the area of the region bounded by $x^2 = 4y, y = 2, y = 4$ and the $y$ - axis in the first quadrant.

Question

$\begin{array}{1 1} \frac{32-8\sqrt 2}{3}\;sq.units. \\ \frac{16-8\sqrt 2}{3}sq.units. \\ \frac{32-4\sqrt 2}{3}sq.units. \\ \frac{16-4\sqrt 2}{3}\;sq.units\end{array}$