Find the mass of a fluid of constant mass density $\rho$ flowing across the cube enclosed by the planes $x=0$, $x=1$, $y=0$, $y=1$, $z=0$, and $z=1$ in a unit of time, in the direction of the outer unit normal vectors if the velocity of the fluid at any point on the cube is $\mathbf{F}=\mathbf{F} (x,y,z)=4xz\mathbf{i}-y^{2}\mathbf{j}+yz\mathbf{k}$. (This is Example 8 on page 1035.)

Solution Suppose $E$ is the solid cube and $S$ is its surface. The mass of the fluid is $\iint\limits_S \rho \mathbf{F}\,{\bf\cdot}\, \mathbf{n} \,dS$, where $\mathbf{n}$ is the outer unit normal vector of $S$. Since the velocity of the fluid is $\mathbf{F}=4xz\mathbf{i}-y^{2}\mathbf{j}+yz\mathbf{ k}$, we have ${\rm{div}}\mathbf{F}=4z-2y+y=4z-y$. Now we use the Divergence Theorem. $$\begin{eqnarray*} \iint\limits_{S}\rho \mathbf{F}\,{\bf\cdot}\, \mathbf{n}\,dS &=&\rho \iint\limits_S \mathbf{F}\,{\bf\cdot}\, \mathbf{n}\,dS=\rho \iiint\limits_E {\rm{div}}\mathbf{F} \,dV=\rho \int_{0}^{1}\int_{0}^{1}\int_{0}^{1}(4z-y)\,dz\,dy\,dx \\ &=&\rho \int_{0}^{1}\int_{0}^{1}(2-y)\,dy\,dx=\rho \int_{0}^{1}\dfrac{3}{2} \,dx=\dfrac{3}{2}\rho \end{eqnarray*}$$

which is the same as the result found in Example 8, Section 15.7.