Answer
$$\dfrac{16}{3}$$
Work Step by Step
Our aim is to integrate the integral as follows:
$$\int^2_0 \int^{\sqrt{4-y^2}}_{\sqrt{4-y^2}} \int^{2x+y}_0 dz \space dx \space dy =\int^2_0 \int^{\sqrt{4-y^2}}_{-{\sqrt{4-y^2}}}(2x+y) \space dx \space dy \\= \int^2_0[x^2+xy]^{\sqrt{4-y^2}}_{-{\sqrt{4-y^2}}}dy\\= 2 \int^2_0 (4-y^2)^{1/2} \times y \space dy\\=[-\dfrac{2}{3}(4-y^2)^{(3/2)}]^2_0\\=\dfrac{2}{3}(4)^{3/2}=\dfrac{16}{3}$$
