Suppose that an electric field is given by

$\mathbf{E}=2y\mathbf{i}+2xy\mathbf{j}+yz\mathbf{k}$

Compute the flux ${\int }_S \mathbf{E}\cdot \mathbf{n}dA$ of the field through the unit cube $[0,1]\times [0,1]\times [0,1]$.

given, $\mathbf{E}=2y\mathbf{i}+2xy\mathbf{j}+yz\mathbf{k}$

More explicitly, suppose that a cube's faces are portions of the planes$x=0,x=1,y=0,y=1,z=0,z=1$.Then the cube's faces are parameterized as follows:
$\begin{array}{l}{\mathbf{r}}_1(u,v)=(0,u,v);\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathbf{r}}_2(u,v)=(1,u,v);\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathbf{r}}_3(u,v)=(u,0,v);\\ {\mathbf{r}}_4(u,v)=(u,1,v);\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathbf{r}}_5(u,v)=(u,v,0);\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathbf{r}}_6(u,v)=(u,v,1).\end{array}$

In this case, the flux of this field through the unit cube will be,

localid="1650478851021" $$\begin{gathered} {\int }_S \mathbf{E}\cdot \mathbf{n}dA={\int }_{S_1} \mathbf{E}\cdot \mathbf{n}dA+{\int }_{S_2} \mathbf{E}\cdot \mathbf{n}dA+{\int }_{S_3} \mathbf{E}\cdot \mathbf{n}dA+{\int }_{S_4} \mathbf{E}\cdot \mathbf{n}dA+ \\ {\int }_{S_3} \mathbf{E}\cdot \mathbf{n}dA+{\int }_{S_6} \mathbf{E}\cdot \mathbf{n}dA.....\left(1\right) \end{gathered}$$

For the faces $S_1,S_2,S_3,S_4,S_{5}and{S}_6$, the unit nominal vectors will be,

${\mathbf{n}}_1=-\mathbf{i},{\mathbf{n}}_2=\mathbf{i},{\mathbf{n}}_3=-\mathbf{j},{\mathbf{n}}_4=\mathbf{j},{\mathbf{n}}_5=-\mathbf{k},{\mathbf{n}}_6=\mathbf{k}$,

respectively.

For each, the region of integration is described as follows,

$D=\left\{\right(u,v)\mid 0\le u\le 1,0\le v\le 1\}$

$\begin{array}{r}{\int }_S \mathbf{E}\cdot \mathbf{n}dA\\ ={\int }_{S_1} \mathbf{E}\cdot \mathbf{n}dA+{\int }_{S_2} \mathbf{E}\cdot \mathbf{n}dA+{\int }_{S_3} \mathbf{E}\cdot \mathbf{n}dA+{\int }_{S_4} \mathbf{E}\cdot \mathbf{n}dA+{\int }_{S_3} \mathbf{E}\cdot \mathbf{n}dA+{\int }_{S_6} \mathbf{E}\cdot \mathbf{n}dA\\ ={\int }_{S_1} (2y\mathbf{i}+2xy\mathbf{j}+yz\mathbf{k})\cdot (-\mathbf{i})dA+{\int }_{S_2} (2y\mathbf{i}+2xy\mathbf{j}+yz\mathbf{k})\cdot \mathbf{i}dA+{\int }_{S_3} (2y\mathbf{i}+2xy\mathbf{j}+yz\mathbf{k})\cdot (-\mathbf{j})dA\\ +{\int }_{S_4} (2y\mathbf{i}+2xy\mathbf{j}+yz\mathbf{k})\cdot \mathbf{j}dA+{\int }_{S_3} (2y\mathbf{i}+2xy\mathbf{j}+yz\mathbf{k})\cdot (-\mathbf{k})dA+{\int }_{S_6} (2y\mathbf{i}+2xy\mathbf{j}+yz\mathbf{k})\cdot \mathbf{k}dA\\ ={\int }_{S_1} -2ydA+{\int }_{S_2} 2ydA+{\int }_{S_3} -2xydA+{\int }_{S_4} 2xydA+{\int }_{S_5} -yzdA+{\int }_{S_6} yzdA.\end{array}$

$\begin{array}{r}{\int }_S \mathbf{E}\cdot \mathbf{n}dA\\ ={\int }_{S_1} -2ydA+{\int }_{S_2} 2ydA+{\int }_{S_3} -2xydA+{\int }_{S_4} 2xydA+{\int }_{S_5} -yzdA+{\int }_{S_6} yzdA\\ ={\int }_0^1 {\int }_0^1 -2ududv+{\int }_0^1 {\int }_0^1 2ududv+{\int }_0^1 {\int }_0^1 -2u\cdot 0dudv+{\int }_0^1 {\int }_0^1 2u1dudv+{\int }_0^1 {\int }_0^1 -v0dudv+{\int }_0^1 {\int }_0^1 v1dudv\\ =-{\int }_0^1 {\int }_0^1 2ududv+{\int }_0^1 {\int }_0^1 2ududv+0+{\int }_0^1 {\int }_0^1 2u1dudv+0+{\int }_0^1 {\int }_0^1 v1dudv\\ ={\int }_0^1 {\int }_0^1 2ududv+{\int }_0^1 {\int }_0^1 vdudv\\ ={\int }_0^1 \left[{\int }_0^1 2udu\right]dv+{\int }_0^1 \left[{\int }_0^1 vdu\right]dv\\ ={\int }_0^1 {\left[u^2\right]}_0^{1}dv+{\int }_0^1 v[u{]}_0^{1}dv\\ ={\int }_0^1 \left[1^2-0^2\right]dv+{\int }_0^1 v[1-0]dv\\ ={\int }_0^1 1dv+{\int }_0^1 vdv\\ =[v{]}_0^1+{\left[\frac{v^2}{2}\right]}_0^1\\ =[1-0]+\left[\frac{1^2}{2}-\frac{0^2}{2}\right]\\ =\frac{3}{2}.\end{array}$

Therefore, the required flux of the vector field through the surface S is$\frac{3}{2}.$