Answer
$\frac{16}{3}$
Work Step by Step
The volume under the surface $z=f(x,y)$ and above the region D in the xy-plane can be defined as:
$$Volume=\int\int_Df(x,y)dA$$
We can express the region D in the triangular region as follows:
$$D=\{(x,y)|0\leq x\leq2,0\leq y\leq -2x+4\}$$
Therefore,
$$Volume=\int\int_Df(x,y)dA
\\=\int_0^2\int_0^{-2x+4}(-2x-y+4)dydx
\\=\int_0^2\biggr[-2xy-\frac{1}{2}y^2+4y\biggr]_{y=0}^{y=-2x+4}dx
\\=\int_0^2\biggr[-2x(-2x+4)-\frac{1}{2}(-2x+4)^2+4(-2x+4)\biggr]dx
\\=\int_0^2(2x^2-8x+8)dx
\\=\int_0^22(x-2)^2dx
\\=\frac{2}{3}(x-2)^3\biggr|_0^2
\\=\frac{16}{3}$$
