## Thomas' Calculus 13th Edition

$\frac{16}{3}$
$\int^2_0 \int^{\sqrt{4-y^2}}_{\sqrt{4-y^2}} \int^{2x+y}_0$ dz dx dy =$\int^2_0 \int^{\sqrt{4-y^2}}_{-{\sqrt{4-y^2}}}(2x+y)$ dx dy =$\int^2_0[x^2+xy]^{\sqrt{4-y^2}}_{-{\sqrt{4-y^2}}}dy$ =$\int^2_0 (4-y^2)^{1/2}(2y)dy$ =$[-\frac{2}{3}(4-y^2)^{3/2}]^2_0$ =$\frac{2}{3}(4)^{3/2}$ =$\frac{16}{3}$