Question

Prove each identity, assuming that \({\rm{S}}\) and \({\rm{E}}\)satisfy the conditions of the Divergence Theorem and the scalar function and components of the vector fields have continuous second-order partial derivatives. \(\iint_{\text{S}} {{{\text{D}}_{\text{n}}}{\text{ f}} \times {\text{dS = }}}\iiint_{\text{E}} {{\nabla ^{\text{2}}}{\text{fdV}}}\).

Short answer

It is proved, \(\iint_{\text{S}} {{{\text{D}}_{\text{n}}}{\text{ f}} \times {\text{dS = }}}\iiint_{\text{E}} {{\nabla ^{\text{2}}}{\text{fdV}}}\).

Step-by-step solution

Define vector

A vector is a number that specifies not only the size of an item, but also its movement or location with respect to another point or object.

Explanation

As defined by definition,

\({{\rm{D}}_{\rm{n}}}{\rm{f = }}\left( {\nabla {\rm{f}}} \right){\rm{ \times n}}\)

Hence,

\(\begin{gathered}\iint_{\text{S}} {{{\text{D}}_{\text{n}}}{\text{fdS}}}{\text{ = }}\iint_{\text{S}} {\left( {\nabla {\text{f}}} \right)} \times {\text{ndS}} \\ {\text{ = }}\iint_{\text{S}} {\left( {\nabla {\text{f}}} \right)} \times {\text{dS}} \\ \end{gathered} \)

The divergence theorem states that,

\(\iint_{\text{S}} {{\text{F}} \times {\text{dS = }}\iiint_{\text{E}} {{\text{div(F)dV}}}}\)

Where, \({\rm{S}}\) = closed surface

\({\rm{E}}\) = inside that surface's region

\(\begin{gathered}\iint_{\text{S}} {\left( {\nabla {\text{f}}} \right)} \times {\text{dS = }}\iiint_{\text{E}} {{\text{div}}}\left( {\nabla {\text{f}}} \right){\text{dV}} \\ {\text{ = }}\iiint_{\text{E}} \nabla \left( {\nabla {\text{f}}} \right){\text{dV}} \\ {\text{ = }}{\iiint_{\text{E}} \nabla ^{\text{2}}}{\text{fdV}} \\ \end{gathered} \)

Therefore, \(\iint_{\text{S}} {{{\text{D}}_{\text{n}}}{\text{fdS}}}{\text{ = }}{\iiint_{\text{E}} \nabla ^{\text{2}}}{\text{fdV}}\).