91Ó°ÊÓ

The given information is

$V=\left(x^{2}z-2\right)i+\left(x+y-z\right)j-xyzk$

"The surface integral of the curl of a function over a surface limited by a closed surface is equivalent to the line integral of the particular vector function around that surface," according to Stoke's theorem.

Use Stoke’s theorem.

${âˆ\neg }_{\mathit{σ}}\left(∇×A\right).nd\mathrm{σ}={âˆ\circledR }_{∂\mathrm{σ}}A.dr$

$\mathrm{σ}$ is the surface consisting of the four slanting faces of a pyramid whose base is the square in the xy-plane with corners at $\left(0,0\right)\left(0,2\right)\left(2,0\right)\left(2,2\right)$so is that square

${âˆ\circledR }_{∂\mathrm{σ}}V.dx=∫\left(x^{2}z-2\right)dx+\left(x+y-z\right)dy-xyzdz$

Integrate over each side of the square once at a time in all the four sides

z = dz

= 0.

i) From (0,0) to (2,0)

y = dz

= 0

$$\begin{gathered} ∫\left(x^{2}z-2\right)dx+\left(x+y-z\right)dz-xyzdz={∫}_0^2-2dx \\ =-4 \end{gathered}$$

ii) From (2,0) to (2,2)

dx = 0,

x = 2

$$\begin{gathered} ∫\left(x^{2}z-2\right)dx+\left(x+y-z\right)dz-xyzdz={∫}_0^2\left(2+y\right)dy \\ ={\overline{)\left(2y+\frac{1}{2}{y}^2\right)}}_0^2 \\ =12 \end{gathered}$$

iii) From (2,2) to (0,2)

dy = 0

y = 2,

localid="1659239265407" $$\begin{gathered} ∫\left(x^{2}z-2\right)dx+\left(x+y-z\right)dz-xyzdz={∫}_2^0-2dx \\ =4 \end{gathered}$$

iv) From

dx = 0

x = 0,

localid="1659239271095" $$\begin{gathered} ∫\left(x^{2}z-2\right)dx+\left(x+y-z\right)dy-xyzdz={∫}_2^{0}ydy \\ ={\overline{)\left(\frac{1}{2}{y}^2\right)}}_2^0 \\ =-8 \end{gathered}$$

Hence, the Solution to the problem islocalid="1659239276120" ${âˆ\neg }_{\mathit{σ}}\left(∇×V\right).nd\mathrm{σ}=4$