Answer
$\frac{1}{6}$
Work Step by Step
We integrate the triple integral as follows:
$\int^{1}_0 \int^{1-x}_0 \int^1_{x+z} F(x,y,z)dydzdx $
=$\int^{1}_0 \int^{1-x}_0 \int^{1}_{x+z} du dz dx $
=$\int^{1}_0 \int^{1-x}_0(1-x-z)dzdx $
=$\int^1_0[(1-x)-x(1-x)-\frac{(1-x)^2}{2}]dx $
=$\int^1_0\frac{(1-x)^2}{2}dx $
=$[-\frac{(1-x)^3}{6}]^1_0$
We plug in $1$ and $0$ and subtract to obtain:
=$\frac{1}{6}$
