91Ó°ÊÓ

$\mathrm{V}=\mathrm{curl}(\mathrm{yi}−\mathrm{xj})$

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 Divergence theorem.

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

Where$\mathrm{∂}\mathrm{T}$ is the surface area that encloses the volume $\mathrm{T}$.

$\mathrm{∇}×\mathrm{V}=\mathrm{∇}×(\mathrm{∇}×(\mathrm{yi}−\mathrm{xj}\left)\right)$

But the divergence of any curve is zero.

Use the triplet scale product.

$âˆ\neg \mathrm{V}×\mathrm{²Ô»åσ}={âˆ\neg }_{\text{cunved}} \mathrm{V}×\mathrm{²Ô»åσ}+{âˆ\neg }_{\mathrm{plane}} \mathrm{V}×\mathrm{²Ô»åσ}$

$=0$

By simplifying it is enough to calculate the integral over the plane part.

${âˆ\neg }_{\text{plane}} \mathrm{V}×\mathrm{²Ô»åσ}={âˆ\neg }_{\text{disk}} 2\mathrm{dxdy}−2\mathrm{Ï€}(3{)}^2$

$=18\mathrm{Ï€}$

Thus, the integrated part in this question is

${âˆ\neg }_{\text{curved}} \mathrm{V}×\mathrm{²Ô»åσ}=−18\mathrm{Ï€}$

Hence, The Solution to the problem is${âˆ\neg }_{\text{curved}} \mathrm{V}×\mathrm{²Ô»åσ}=−18\mathrm{Ï€}$