${\iint }_{S}curl\mathbf{F}(x,y,z)·\mathbf{n}dS$, where $S$is the portion of the plane$x+y-z=0$with upwards-pointing normal vector and$\mathbf{F}(x,y,z)=yz{e}^{xyz}\mathbf{i}+xz{e}^{xyz}\mathbf{j}+xy{e}^{xyz}\mathbf{k}.$

There is a vector field in which every point has a specific direction. $\mathbf{F}(x,y,z)=yz{e}^{v*}\mathbf{i}+xz{e}^{vz}\mathbf{j}+xy{e}^{xs}\mathbf{k}$

The purpose is to evaluate the integral${\iint }_{s}curl\mathbf{F}(x,y,z)·\mathbf{n}dS$, where the surface is defined as follows:

The surface $S$is the region of the plane$x+y-z=0$that has the normal vector pointing upwards.

Find the curl of the vector field $\mathbf{F}(x,y,z)=yz{e}^{0v}\mathbf{i}+xz{e}^{vz}\mathbf{j}+xy{e}^{vx}\mathbf{k}$

The direction of a vector field is determined by its curl $\mathbf{F}(x,y,z)=F_1(x,y,z)\mathbf{i}+F_2(x,y,z)\mathbf{j}+F_3(x,y,z)\mathbf{k}$defined as:

$curl\mathbf{F}(x,y,z)=\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ F_1(x,y,z)& F_2(x,y,z)& F_3(x,y,z)\end{array}\right|$

$=\left[\frac{\partial F_3}{\partial y}-\frac{\partial F_2}{\partial z}\right]\mathbf{i}-\left[\frac{\partial F_3}{\partial x}-\frac{\partial F_1}{\partial z}\right]\mathbf{j}+\left[\frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y}\right]\mathbf{k}$

Then the curl of the vector field $\mathbf{F}(x,y,z)=yz{e}^{n\pi }\mathbf{i}+xz{e}^{nz}\mathbf{j}+xy{e}^{nn}\mathbf{k}$will be,

$curl\mathbf{F}(x,y,z)=\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ yz{e}^{n*}& xz{e}^{n*}& xy{e}^{nt}\end{array}\right|$

$=\left[\frac{\partial \left(xy{e}^{yz}\right)}{\partial y}-\frac{\partial \left(xz{e}^{yz}\right)}{\partial z}\right]\mathbf{i}-\left[\frac{\partial \left(xy{e}^{yz}\right)}{\partial x}-\frac{\partial \left(yz{e}^{vz}\right)}{\partial z}\right]\mathbf{j}$

$+\left[\frac{\partial \left(xz{e}^{vz}\right)}{\partial x}-\frac{\partial \left(yz{e}^{vz}\right)}{\partial y}\right]\mathbf{k}$ localid="1650710450011" $=\left[(1+xyz)x{e}^{xz}-(1+xyz)x{e}^{yz}\right]\mathbf{i}\mathbf{}-\left[(1+xyz)y{e}^{yz}-(1+xyz)y{e}^{yz}\right]\mathbf{j}$

$+\left[(1+xyz)z{e}^{yz}-(1+xyz)z{e}^{xz}\right]\mathbf{k}$

$$\begin{gathered} =0i-0j+0k \\ =0 \end{gathered}$$

Then, the value of the curl$\mathbf{F}(x,y,z)$' $\mathbf{n}$will,

$curl\mathbf{F}(x,y,z)·\mathbf{n}=\mathbf{0}·\mathbf{n}$
$=0$

Then the necessary integral、

${\iint }_{S}curl\mathbf{F}(x,y,z)·\mathbf{n}dS={\iint }_{S}0dS$

$=0$

Therefore, the required integral is equal to${\iint }_{s}curl\mathbf{F}(x,y,z)·\mathbf{n}dS=0$.