multivariable calculus determing region and limits of integration

Question

Evaluate $$\iint_R\sqrt{|y-x^2|}dxdy~~~~~~\text{where}~ R=[-1,1]\times[0,2]$$ My effort: The above integral is $$=2\left(\int_0^1\int_{x^2}^2\sqrt{y-x^2}dydx+\int_0^1\int_0^{x^2}\sqrt{x^2-y}dydx\right)$$ $$=2\left(\{\pi/4~~+2/3\}+1/6\right)$$ The first inegral evaluated as follows $$y=x^2sec^2\theta$$ Hence the integral becomes $$=\int_0^1\left(\int_0^{sec^{-1}\left(\sqrt{\dfrac{2}{x^2}} \right)}x.\tan\theta. x^2.2\sec\theta.\sec\theta.\tan\theta d\theta \right)dx$$ $$=\dfrac{2}{3}\int_0^1x^3\tan^3\left(\sec^{-1}\left(\sqrt{\dfrac{2}{x^2}}\right)\right)dx$$ $$=\dfrac{2}{3}\int_0^1x^3\left(\dfrac{\sqrt{2-x^2}}{x}\right)^3dx $$ $$=\dfrac{2}{3}\int_0^1(2-x^2)^{3/2}dx$$ $$=\pi/4~~+2/3$$ Is this the right approach?