Question

$\mathbf{\iint }\mathbf{V}\mathbf{\times }\mathbf{nd\sigma }$over the closed surface of the tin can bounded by if ${\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{=}\mathbf{9}\mathbf{,}\mathbf{z}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{z}\mathbf{=}\mathbf{5}$if$\mathbf{V}\mathbf{=}\mathbf{2}\mathit{x}\mathit{y}\mathit{i}\mathbf{-}{\mathbf{y}}^{\mathbf{2}}\mathbf{j}\mathbf{+}\mathbf{(}\mathbf{z}\mathbf{+}\mathit{x}\mathit{y}\mathbf{)}\mathbf{k}$.

Short answer

The solution derived is$\mathrm{I}=45\mathrm{\pi }$.

Step-by-step solution

Given Information.

The given expression is${\mathrm{x}}^2+{\mathrm{y}}^2=9,\mathrm{z}=0,\mathrm{z}=5\mathrm{,}$if$\mathrm{V}=2\mathrm{xyi}-{\mathrm{y}}^2\mathrm{j}+(\mathrm{z}+\mathrm{xy})\mathrm{k}$.

Definition of vector.

A quantity that has magnitude as well as direction is called a vector. It is typically denoted by an arrow in which the head determines the direction of a vector and the length determines its magnitude.

Apply the divergence theorem.

Apply the divergence theorem.

${\text{∮}}_{+\mathrm{\partial }\mathrm{V}}\mathrm{V}\times \mathrm{nd\sigma }={\int }_{\mathrm{V}} \mathrm{\nabla }\times \mathrm{Vd\tau }$

But$\mathrm{\nabla }\times \mathrm{V}=2\mathrm{y}-2\mathrm{y}+1=1$then the integral becomes as shown below.

$\mathrm{I}={\int }_{\mathrm{V}} \mathrm{d\tau }=\mathrm{\pi }\left(3^2\right)\left(5\right)$

$=45\mathrm{\pi }$

Hence, the solution derived is $\mathrm{I}=45\mathrm{\pi }$.