## Thomas' Calculus 13th Edition

$\frac{1}{6}$
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: =$\frac{1}{6}$