 Solution

The divergence is $y^2+z^2+x^2$. Over the cube $[-1/2,1/2{]}^3$, each squared-coordinate term has integral ${\int }_{-1/2}^{1/2}{s}^{2}ds=1/12$, while the other two coordinate integrals have length one. ThusCompute the surface flux separately. On $x=1/2$, the outward normal is $e_x$ and $F\cdot n=y^2/2$, giving flux $1/24$. On $x=-1/2$, the normal is $-e_x$, so $F\cdot n=-x{y}^2=y^2/2$, again giving $1/24$. The two $y$ faces similarly give $z^2/2$ each, and the two $z$ faces give $x^2/2$ each. All six face integrals therefore giveThe independently evaluated values agree, verifying the divergence theorem for this field and cube.