## Thomas' Calculus 13th Edition

$\frac{4}{3}$
$\int^0_{-2} \int^{4-x^2}_{2x+4}$ dy dx =$\int^0_{-2}(-x^2-2x)dx$ =$[-\frac{x^3}{3}-x^2]^0_{-2}=-(\frac{8}{3}-4)=\frac{4}{3}$ =$\int^4_0 \int^{(y-4)/2}_{-\sqrt{4-y}}dx$ dy =$\int^4_0 (\frac{y-4}{2}+\sqrt{4-y})dy$ =$[\frac{y^2}{2}-2y-\frac{2}{3}(4-y)^{3/2}]^{4}_0$ =$4-8+\frac{2}{3} \cdot 4^{3/2}=-4+\frac{16}{3}=\frac{4}{3}$