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. \({\text{V(E) = }}\frac{{\text{1}}}{{\text{3}}}\iint_{\text{S}} {{\text{F}} \times {\text{dS}}}\) where \({\rm{F(x,y,z) = xi + yj + zk}}\).

Short answer

It is proved, \(\frac{{\text{1}}}{{\text{3}}}\iint_{\text{S}} {{\text{F}} \times {\text{ds}}}{\text{ = V(E)}}\).

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

The divergence theorem states that,

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

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

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

\(\begin{aligned}{c}{\rm{divF = }}\frac{{\partial {\rm{x}}}}{{\partial {\rm{x}}}}{\rm{ + }}\frac{{\partial {\rm{y}}}}{{\partial {\rm{y}}}}{\rm{ + }}\frac{{\partial {\rm{z}}}}{{\partial {\rm{z}}}}\\{\rm{divF = 1 + 1 + 1}}\\{\rm{ = 3}}\end{aligned}\)

So,

\(\begin{gathered} \iint_{\text{S}} {{\text{F}} \times {\text{ds}}}{\text{ = }}\iiint_{\text{E}} {{\text{div(F)dV}}} \\ \iint_{\text{S}} {{\text{F}} \times {\text{ds}}}{\text{ = }}\iiint_{\text{E}} {{\text{3dV}}} \\ \frac{{\text{1}}}{{\text{3}}}\iint_{\text{S}} {{\text{F}} \times {\text{ds}}}{\text{ = }}\iiint_{\text{E}} {{\text{dV}}} \\ \frac{{\text{1}}}{{\text{3}}}\iint_{\text{S}} {{\text{F}} \times {\text{ds}}}{\text{ = V(E)}} \\ \end{gathered} \)

Hence, it is proved.

Therefore, \(\frac{{\text{1}}}{{\text{3}}}\iint_{\text{S}} {{\text{F}} \times {\text{ds}}}{\text{ = V(E)}}\).