Answer
$\dfrac{4}{3}$
Work Step by Step
Our aim is to integrate the integral as follows:
$ Area =\iint_D dA $
$\int^{1}_{-1} \int^{y^2-1}_{2y^2-2} \space dx \space dy = \int^{1}_{-1}(y^2-1-2y^2+2) \space dy $
or, $=\int^{1}_{-1}(1-y^2) \space dy $
or, $=[y-\dfrac{y^3}{3}]^1_{-1}$
or, $=[1-1/3]-[-1+1/3]$
or, $ Area=\dfrac{4}{3}$
