Question

$\mathbf{\iint }\mathbf{F}\mathbf{\times }\mathbf{nd\sigma }$where $\mathbf{F}\mathbf{=}\left(y^2-x^2\right)\mathbf{i}\mathbf{+}\mathbf{(}\mathbf{2}\mathbf{xy}\mathbf{-}\mathbf{y}\mathbf{)}\mathbf{j}\mathbf{+}\mathbf{3}\mathbf{zk}$and $\mathrm{\sigma }$is the entire surface of the tin can bounded by the cylinder

role="math" localid="1657353627256" ${\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{=}\mathbf{16}$

role="math" localid="1657353639412" $\mathbf{z}\mathbf{=}\mathbf{3}$

role="math" localid="1657353647648" $\mathbf{z}\mathbf{=}\mathbf{-}\mathbf{3}$

Short answer

The solution is ${\iint }_{\mathrm{\partial }\mathrm{T}} \mathrm{F}\times \mathrm{nd\sigma }=192\mathrm{\pi }$.

Step-by-step solution

Given Information.

$\mathrm{F}=\left({\mathrm{y}}^2-{\mathrm{x}}^2\right)\mathrm{i}+(2\mathrm{xy}-\mathrm{y})\mathrm{j}+3\mathrm{zk}$

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

$\mathrm{\nabla }\times \mathrm{F}=\frac{\mathrm{\partial }{\mathrm{F}}_{\mathrm{x}}}{\mathrm{\partial }\mathrm{x}}+\frac{\mathrm{\partial }{\mathrm{F}}_{\mathrm{y}}}{\mathrm{\partial }\mathrm{y}}+\frac{\mathrm{\partial }{\mathrm{F}}_{\mathrm{z}}}{\mathrm{\partial }\mathrm{z}}$

$=-2\mathrm{x}+(2\mathrm{x}-1)+3$

$=2$$=2$

Here, $\mathrm{\partial }\mathrm{\sigma }$is the entire surface of the tin, and can be bounded by the cylinder.

$x^2+y^2=16$

$\mathrm{z}=3$

$\mathrm{z}=-3$

The tin’s radius and height are $4$and $6$respectively.

${\iint }_{\mathrm{\partial }\mathrm{T}} \mathrm{F}\times \mathrm{nd\sigma }={\iiint }_{\mathrm{T}} \mathrm{\nabla }\times \mathrm{FdT}$

$=2{\iint }_T dT$

$=\left(2\mathrm{\pi }\right)\left(4{)}^2\right(6)$

$=192\mathrm{\pi }$

Hence, the solution is ${\iint }_{\mathrm{\partial }\mathrm{T}} \mathrm{F}\times \mathrm{nd\sigma }=192\mathrm{\pi }$.