Question

$\mathbf{\iint }\mathbf{V}\mathbf{\times }\mathbf{nd\sigma }$over the entire surface of the cube in the first octant with three faces in the three coordinate planes and the other three faces intersecting at $\mathbf{(}\mathbf{2}\mathbf{,}\mathbf{2}\mathbf{,}\mathbf{2}\mathbf{)}$, where $\mathbf{V}\mathbf{=}\mathbf{(}\mathbf{2}\mathbf{-}\mathbf{y}\mathbf{)}\mathbf{i}\mathbf{+}\mathbf{xzj}\mathbf{+}\mathbf{xyzk}$.

Short answer

The Solution is${\iint }_{\mathrm{\partial }\mathrm{T}} (\mathrm{\nabla }\times \mathrm{V})\times \mathrm{nd\sigma }=0.$

Step-by-step solution

Given Information.

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

Definition of Divergence Theorem.

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.

Find the solution.

Use the divergence theorem ${\iiint }_{\mathrm{T}} \mathrm{\nabla }\times \mathrm{VdT}={\iint }_{\mathrm{\partial }\mathrm{T}} \mathrm{V}\times \mathrm{nd\sigma }$, where is the surface area that encloses the volume $\mathrm{T}$.

$\mathrm{\nabla }\times (\mathrm{\nabla }\times \mathrm{V})=0$

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

${\iint }_{\mathrm{\partial }\mathrm{T}} (\mathrm{\nabla }\times \mathrm{V})\times \mathrm{nd\sigma }={\iint }_{\mathrm{J}} \mathrm{\nabla }\times (\mathrm{\nabla }\times \mathrm{V})\mathrm{dT}$

$=0$

Hence, the solution is ${\iint }_{\mathrm{\partial }\mathrm{T}} (\mathrm{\nabla }\times \mathrm{V})\times \mathrm{nd\sigma }=0.$