91Ó°ÊÓ

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:
$\begin{array}{l}{\mathbf{r}}_1(u,v)=(0,u,v);â€\dots â\P Ä\dots â\P Ä\dots â\P Ä\dots {\mathbf{r}}_2(u,v)=(1,u,v);â€\dots â\P Ä\dots â\P Ä\dots â\P Ä\dots {\mathbf{r}}_3(u,v)=(u,0,v);\\ {\mathbf{r}}_4(u,v)=(u,1,v);â€\dots â\P Ä\dots â\P Ä\dots â\P Ä\dots {\mathbf{r}}_5(u,v)=(u,v,0);â€\dots â\P Ä\dots â\P Ä\dots â\P Ä\dots {\mathbf{r}}_6(u,v)=(u,v,1).\end{array}$

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

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

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,

$D=\left\{\right(u,v)âˆ\pounds 0≤u≤1,0≤v≤1\}$

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

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

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