Answer
$12$
Work Step by Step
Our aim is to integrate the integral as follows:
$ Area =\iint_D dA $
$\int^{6}_{0} \int^{2y}_{y^2/3} \space dx \space dy = \int^{6}_{0} (2y-\dfrac{y^2}{3}) \space dy $
or, $=[y^2-\dfrac{y^3}{9}^6_0]$
or, $=36-\dfrac{216}{9}$
or, $=12$
