Answer
$\dfrac{4}{3}$
Work Step by Step
$\int^0_{-2} \int^{4-x^2}_{2x+4} dy \ dx =\int^0_{-2}(-x^2-2x)dx $
or, $=[-\dfrac{x^3}{3}-x^2]^0_{-2}$
or, $=-(\dfrac{8}{3}-4)$
or, $=\dfrac{4}{3}$
Now, $\int^4_0 \int^{(y-4)/2}_{-\sqrt{4-y}}dx \ dy =\int^4_0 (\frac{y-4}{2}+\sqrt{4-y})dy $
or, $=[\dfrac{y^2}{2}-2y-\dfrac{2}{3}(4-y)^{3/2}]^{4}_0$
or, $=4-8+\dfrac{2}{3} \cdot 4^{3/2}$
or $=-4+\dfrac{16}{3}$
or, $=\dfrac{4}{3}$
