Answer
$$\dfrac{2}{3}$$
Work Step by Step
$$Volume =\int_{0}^{2} \int_{0}^{y}
int_0^{(2-y/2)} dz dx dy \\=\int_{0}^{2} \int_{0}^{y} (1-\dfrac{y}{2}) dx dy \\=\\=\int_{0}^{2} \int_{0}^{y} (1-\dfrac{y}{2}) dx dy\\=\int_0^2 y dy - \dfrac{1}{2} \times \int_0^2 y^2 dy\\=\dfrac{2}{3}$$
