Para encontrarmos o volume do sólido, realizaremos os cálculos abaixo:

\(\begin{align} & \int_{0}^{2}{\int_{0}^{1}{z}dz=}\int_{0}^{2}{\int_{0}^{1}{(4-x-y)dxdy}} \\ & V=\int_{0}^{2}{\int_{0}^{1}{(4-x-y)dxdy}} \\ & V=\int_{0}^{2}{\left( 4x-\frac{{{x}^{2}}}{2}-xy \right)_{0}^{1}} \\ & V=\int_{0}^{2}{\left( \frac{7}{2}-y \right)dy} \\ & V=\left( \frac{7y}{2}-\frac{{{y}^{2}}}{2} \right)_{0}^{2} \\ & V=\left( \frac{14}{2}-\frac{4}{2} \right)-\left( 0 \right) \\ & V=7-2 \\ & V=5 \\ \end{align}\ \)

Portanto, para esse exercício temos que V=5.