Question

If \({\rm{a}}\) is a constant vector,\({\rm{r = xi + yj + zk,}}\) and \({\rm{S}}\) is an oriented, smooth surface with a simple, closed, smooth,positively oriented boundary curve \({\rm{C}}\), show that

\(\iint_{S}{2a\cdot }dS=\int_{C}{\left( a*r \right)\cdot dr}\)

Short answer

Proved \(\iint_{S}{2a\cdot }dS=\int_{C}{\left( a*r \right)\cdot dr}\)

Step-by-step solution

Expand the equation.

Let

\({\rm{a = }}{{\rm{a}}_{\rm{1}}}{\rm{i + }}{{\rm{a}}_{\rm{2}}}{\rm{j + }}{{\rm{a}}_{\rm{3}}}{\rm{k}}\)

Where \({{\rm{a}}_{\rm{1}}}{\rm{,}}{{\rm{a}}_{\rm{2}}}\) and \({{\rm{a}}_{\rm{3}}}\)are constants.

Known \({\rm{F = a * r}}\)

It is necessary to prove this.

\(\int_{C}{F}\cdot dr=\iint_{S}{2}a\cdot dS\)

Can write using Stoke's theorem

\(\int_{C}{F}\cdot dr=\iint_{S}{curl}F\cdot dS\)

It is sufficient to prove that \({\rm{curlF = 2a}}\).To prove the provided item.

\(\begin{array}{l}{\rm{F = a * r = }}\left( {{{\rm{a}}_{\rm{1}}}{\rm{i + }}{{\rm{a}}_{\rm{2}}}{\rm{j + }}{{\rm{a}}_{\rm{3}}}{\rm{k}}} \right){\rm{ * (xi + yj + zk)}}\\{\rm{F = }}\left| {\begin{array}{*{20}{c}}{\rm{i}}&{\rm{j}}&{\rm{k}}\\{{{\rm{a}}_{\rm{1}}}}&{{{\rm{a}}_{\rm{2}}}}&{{{\rm{a}}_{\rm{3}}}}\\{\rm{x}}&{\rm{y}}&{\rm{z}}\end{array}} \right|\\{\rm{F = i}}\left| {\begin{array}{*{20}{c}}{{{\rm{a}}_{\rm{2}}}}&{{{\rm{a}}_{\rm{3}}}}\\{\rm{y}}&{\rm{z}}\end{array}} \right|{\rm{ - j}}\left| {\begin{array}{*{20}{c}}{{{\rm{a}}_{\rm{1}}}}&{{{\rm{a}}_{\rm{3}}}}\\{\rm{x}}&{\rm{z}}\end{array}} \right|{\rm{ + k}}\left| {\begin{array}{*{20}{c}}{{{\rm{a}}_{\rm{1}}}}&{{{\rm{a}}_{\rm{2}}}}\\{\rm{x}}&{\rm{y}}\end{array}} \right|\\{\rm{F = }}\left( {{{\rm{a}}_{\rm{2}}}{\rm{z - }}{{\rm{a}}_{\rm{3}}}{\rm{y}}} \right){\rm{i + }}\left( {{{\rm{a}}_{\rm{3}}}{\rm{x - }}{{\rm{a}}_{\rm{1}}}{\rm{z}}} \right){\rm{j + }}\left( {{{\rm{a}}_{\rm{1}}}{\rm{y - }}{{\rm{a}}_{\rm{2}}}{\rm{x}}} \right){\rm{k}}\end{array}\)

To check.

Recall that

\({\rm{curlF = }}\left( {\frac{{\partial R}}{{\partial y}}{\rm{ - }}\frac{{\partial Q}}{{\partial z}}} \right){\rm{i + }}\left( {\frac{{\partial P}}{{\partial z}}{\rm{ - }}\frac{{\partial R}}{{\partial x}}} \right){\rm{j + }}\left( {\frac{{\partial Q}}{{\partial x}}{\rm{ - }}\frac{{\partial P}}{{\partial y}}} \right){\rm{k}}\)

Where \({\rm{F = Pi + Qj + Rk}}\)

In this case,\({\rm{F = }}\left( {{{\rm{a}}_{\rm{2}}}{\rm{z - }}{{\rm{a}}_{\rm{3}}}{\rm{y}}} \right){\rm{i + }}\left( {{{\rm{a}}_{\rm{3}}}{\rm{x - }}{{\rm{a}}_{\rm{1}}}{\rm{z}}} \right){\rm{j + }}\left( {{{\rm{a}}_{\rm{1}}}{\rm{y - }}{{\rm{a}}_{\rm{2}}}{\rm{x}}} \right){\rm{k}}\)

Therefore,

\(\begin{aligned} \rm curlF &= \left( {{{\rm{a}}_{\rm{1}}}{\rm{ - ( - }}{{\rm{a}}_{\rm{1}}}{\rm{)}}} \right){\rm{i + }}\left( {{{\rm{a}}_{\rm{2}}}{\rm{ - ( - }}{{\rm{a}}_{\rm{2}}}{\rm{)}}} \right){\rm{j + }}\left( {{{\rm{a}}_{\rm{3}}}{\rm{ - ( - }}{{\rm{a}}_{\rm{3}}}{\rm{)}}} \right){\rm{k}}\\\rm curlF &= 2({{\rm{a}}_{\rm{1}}}{\rm{i + }}{{\rm{a}}_{\rm{2}}}{\rm{j + }}{{\rm{a}}_{\rm{3}}}{\rm{k)}}\\\rm &= 2a\end{aligned}\)

Hence proved \(\iint_{S}{2}a\cdot dS=\int_{C}{(a*r)}\cdot dr\)..