Answer
$\dfrac{1}{6}$
Work Step by Step
We have the triple integral:
$\int^{1}_0 \int^{1-x}_0 \int^1_{x+z} F(x,y,z)dy \space dz \space dx=\int^{1}_0 \int^{1-x}_0 \int^{1}_{x+z} du \space dz \space dx $
or, =$\int^{1}_0 \int^{1-x}_0(1-x-z)dz \space dx $
or, =$\int^1_0[(1-x)-x(1-x)-\dfrac{(1-x)^2}{2}]dx $
or, =$\int^1_0\dfrac{(1-x)^2}{2}dx $
or, =$-\dfrac{(1-x)^3}{6}^1_0=\frac{1}{6}$
so, $\int^{1}_0 \int^{1-x}_0 \int^1_{x+z} F(x,y,z)dydzdx=\dfrac{1}{6}$
