Question

$\mathbf{\iint }\mathbf{V}\mathbf{-}\mathbf{nd\sigma }$over the curved part of the hemisphere in Problem $\mathbf{24}$, if role="math" localid="1657355269158" $\mathbf{V}\mathbf{=}\mathbf{curl}\mathbf{(}\mathbf{yi}\mathbf{-}\mathbf{xj}\mathbf{)}$.

Short answer

The Solution to the problem is${\iint }_{\text{curved}} \mathrm{V}\times \mathrm{nd\sigma }=-18\mathrm{\pi }$

Step-by-step solution

Given Information.

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

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 Divergence theorem.

${\iiint }_{\mathrm{T}} \mathrm{\nabla }\times \mathrm{VdT}={\iint }_{\mathrm{\partial }\mathrm{T}} \mathrm{V}\times \mathrm{nd\sigma }$

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

$\mathrm{\nabla }\times \mathrm{V}=\mathrm{\nabla }\times (\mathrm{\nabla }\times (\mathrm{yi}-\mathrm{xj}\left)\right)$

But the divergence of any curve is zero.

Use the triplet scale product.

$\iint \mathrm{V}\times \mathrm{nd\sigma }={\iint }_{\text{cunved}} \mathrm{V}\times \mathrm{nd\sigma }+{\iint }_{\mathrm{plane}} \mathrm{V}\times \mathrm{nd\sigma }$

$=0$

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

${\iint }_{\text{plane}} \mathrm{V}\times \mathrm{nd\sigma }={\iint }_{\text{disk}} 2\mathrm{dxdy}-2\mathrm{\pi }(3{)}^2$

$=18\mathrm{\pi }$

Thus, the integrated part in this question is

${\iint }_{\text{curved}} \mathrm{V}\times \mathrm{nd\sigma }=-18\mathrm{\pi }$

Hence, The Solution to the problem is${\iint }_{\text{curved}} \mathrm{V}\times \mathrm{nd\sigma }=-18\mathrm{\pi }$