91Ó°ÊÓ

The given information is,$\mathrm{V}=(2−\mathrm{y})\mathrm{i}+\mathrm{xzj}+\mathrm{xyzk}$

The divergence theorem, often known as Gauss' theorem or Ostrogradsky's theorem, is a theorem that connects the flow of a vector field across a closed surface to the field's divergence in the volume enclosed.

Use the divergence theorem ${âˆ\neg }_{{∫}_{\mathrm{T}} } \mathrm{∇}×\mathrm{VdT}={âˆ\neg }_{\mathrm{∂}\mathrm{T}} \mathrm{V}×\mathrm{²Ô»åσ},$, where is the surface area that encloses the volume $\mathrm{T}$.

$\mathrm{∇}×(\mathrm{∇}×\mathrm{V})=0$

But the divergence of any curve is zero. Use the triplet scale product.

${âˆ\neg }_{\mathrm{∂}\mathrm{T}} (\mathrm{∇}×\mathrm{V})×\mathrm{²Ô»åσ}={∭}_{\mathrm{T}} \mathrm{∇}×(\mathrm{∇}×\mathrm{V})\mathrm{dT}$

$=0$

The value of the integral over the surface in the $\mathrm{xy}$-plane cancels the sum of the values of the integral over the rest of the cube. It is enough to evaluate the integral over the face of the cube in the $\mathrm{xy}$-plane.

It is given that

$\mathrm{z}=0$

$\mathrm{n}=-\mathrm{k}$

$\mathrm{»åσ}=\mathrm{dxdy}$

$\mathrm{∇}×\mathrm{V}=\left|\begin{array}{ccc}\mathrm{i}& \mathrm{j}& \mathrm{k}\\ \frac{\mathrm{∂}}{\mathrm{∂}\mathrm{x}}& \frac{\mathrm{∂}}{\mathrm{∂}\mathrm{y}}& \frac{\mathrm{∂}}{\mathrm{∂}\mathrm{z}}\\ 2−\mathrm{y}& \mathrm{xz}& \mathrm{xyz}\end{array}\right|$

$=(\mathrm{xz}−\mathrm{x})\mathrm{i}−\left(\mathrm{yz}\right)\mathrm{j}+(\mathrm{z}+1)\mathrm{k}$

${âˆ\neg }_{\mathrm{xy}−\mathrm{plane}} \mathrm{V}×\mathrm{²Ô»åσ}={âˆ\neg }_{\mathrm{xy}−\text{tace}} −\mathrm{dxdy}−(−2)\left(2\right)$

$=-4$

Hence, the solution is ${âˆ\neg }_{\mathrm{xy}−\mathrm{plane}} \mathrm{V}×\mathrm{²Ô»åσ}=−4$.