$\mathbf{\iint }\left(\nabla \times V\right)\mathbf{.}\mathit{n}\mathit{d}\mathit{\sigma }$over the surface consisting of the four slanting faces of a pyramid whose base is the square in the (x,y) plane with corners at $\left(0,0\right)\mathbf{,}\left(0,2\right)\mathbf{,}\left(2,0\right)\mathbf{,}\left(2,2\right)$, and whose top vertex is at (1,1,2) where$\mathit{V}\mathbf{=}\left(x^{2}z-2\right)\mathit{i}\mathbf{+}\left(x+y+z\right)\mathit{j}\mathbf{-}\mathit{x}\mathit{y}\mathit{z}\mathit{k}$.

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.

${\iint }_{\mathit{\sigma }}\left(\nabla \times A\right).nd\sigma ={\oint }_{\partial \sigma }A.dr$

$\sigma$ 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

${\oint }_{\partial \sigma }V.dx=\int \left(x^{2}z-2\right)dx+\left(x+y-z\right)dy-xyzdz$

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

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

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

dx = 0,

x = 2

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

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

dy = 0

y = 2,

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

iv) From

dx = 0

x = 0,

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

Hence, the Solution to the problem islocalid="1659239276120" ${\iint }_{\mathit{\sigma }}\left(\nabla \times V\right).nd\sigma =4$