Example

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

Solution Figure 66 shows the cube and its six faces. The mass per unit time of fluid flowing across the cube in the direction of the outer unit normal vector $\mathbf{n}$ is $\iint\limits_{\kern-13ptS}\rho \mathbf{F}\,{\cdot}\, \mathbf{n}\,dS=\rho \iint\limits_{\kern-8ptS} \mathbf{F}\,{\cdot}\, \mathbf{n}\,dS$

We decompose $S$ into its six faces and find the surface integral over each one, as shown in Table 1.

TABLE 1

Face		$\mathbf{n}$	$\mathbf{F}$	${\bf F} \,{\cdot}\, {\bf n}$	$\rho \iint\limits_{\kern-8ptS}\mathbf{F}\,{\cdot}\, \mathbf{n}\,dS$
$S_{1}={\it ABCO}$	$z=0$	$-\mathbf{k}$	$-y^{2}\mathbf{j}$	0	0
$S_{2}={\it OAEG}$	$y=0$	$-\mathbf{j}$	$4xz\mathbf{i}$	0	0
$S_{3}={\it OCDG}$	$x=0$	$-\mathbf{i}$	$-y^{2}\mathbf{j}+yz\mathbf{k}$	0	0
$S_{4}={\it ABFE}$	$x=1$	$\mathbf{i}$	$4z\mathbf{i}-y^{2}\mathbf{j}+yz\mathbf{k}$	$4z$	$\rho \int_{0}^{1}\int_{0}^{1}4z\,dy\,dz=2\rho$
$S_{5}={\it BCDF}$	$y=1$	$\mathbf{j}$	$4xz\mathbf{i}-\mathbf{j}+z\mathbf{k}$	$-1$	$-\rho \int_{0}^{1}\int_{0}^{1}\,dx\,dz=-\rho$
$S_{6}={\it DFEG}$	$z=1$	$\mathbf{k}$	$4x\mathbf{i}-y^{2}\mathbf{j}+y\mathbf{k}$	$y$	$\rho\int_{0}^{1}\int_{0}^{1}y\,dx\,dy=\dfrac{1}{2}\rho$

The mass per unit time of fluid flowing across the cube is $0+0+0+2\rho -\rho +\dfrac{1}{2}\rho = \dfrac{3}{2}\rho$