Step 1

We have to prove that,

\(\displaystyle\oint_{{C}}{F}.{n}{d}{\sin{{t}}}\int_{{D}}\div{F}{\left({x},{y}\right)}{d}{A}\)

Step 2

Consider Vector field ,

F=Pi+Qj

\(\displaystyle\div{\left({F}\right)}=\frac{{\partial{P}}}{{\partial{x}}}+\frac{{\partial{Q}}}{{\partial{y}}}\)

Step 3

If C is given by the vector equation ,

r(t)=x(t)i+y(t)j \(\displaystyle{a}\le{t}\le{b}\)

Then unit tangent vector is

\(\displaystyle{T}{\left({t}\right)}=\frac{{{x}'{\left({t}\right)}}}{{{\left|{r}\right|}{\left({r}'{\left({t}\right)}\right)}}}{i}+\frac{{{y}'{\left({t}\right)}}}{{{\left|{{r}'{\left({t}\right)}}\right|}}}{j}\)

n(t) is the outward unit normal vector C is given by,

\(\displaystyle{n}{\left({t}\right)}=\frac{{{y}'{\left({t}\right)}}}{{{\left|{r}\right|}{\left({r}'{\left({t}\right)}\right)}}}{i}+\frac{{{x}'{\left({t}\right)}}}{{{\left|{{r}'{\left({t}\right)}}\right|}}}{j}\)

As we know

Step 4

\(\displaystyle\oint_{{C}}{F}.{n}{d}{s}={\int_{{a}}^{{b}}}{\left({F}.{n}\right)}{\left({t}\right)}{\left|{{r}'{\left({t}\right)}}\right|}{\left.{d}{t}\right.}\)

\(\displaystyle={\int_{{a}}^{{b}}}{\left[\frac{{{P}{\left({x}{\left({t}\right)},{y}{\left({t}\right)}\right)}{y}'{\left({t}\right)}}}{{{\left|{{r}'{\left({t}\right)}}\right|}}}-\frac{{{Q}{\left({x}{\left({t}\right)},{y}{\left({t}\right)}\right)}{x}'{\left({t}\right)}}}{{{\left|{{r}'{\left({t}\right)}}\right|}}}\right]}{\left|{{r}'{\left({t}\right)}}\right|}{\left.{d}{t}\right.}\)

\(\displaystyle={\int_{{a}}^{{b}}}{P}{\left({x}{\left({t}\right)},{y}{\left({t}\right)}\right)}{y}'{\left({t}\right)}{\left.{d}{t}\right.}-{Q}{\left({x}{\left({t}\right)},{y}{\left({t}\right)}\right)}{x}'{\left({t}\right)}{\left.{d}{t}\right.}\)

\(\displaystyle=\int_{{C}}{P}{\left.{d}{y}\right.}-{Q}{\left.{d}{x}\right.}\)

\(\displaystyle=\int\int_{{D}}{\left(\frac{{\partial{P}}}{{\partial{x}}}+\frac{{\partial{Q}}}{{\partial{y}}}\right)}{d}{A}\)

Put Div(F) value in the above equation

Step 5

\(\displaystyle\oint_{{C}}{F}.{n}{d}{s}=\int\int_{{D}}{D}{i}{v}{\left({F}\right)}{d}{A}\)

We have to prove that,

\(\displaystyle\oint_{{C}}{F}.{n}{d}{\sin{{t}}}\int_{{D}}\div{F}{\left({x},{y}\right)}{d}{A}\)

Step 2

Consider Vector field ,

F=Pi+Qj

\(\displaystyle\div{\left({F}\right)}=\frac{{\partial{P}}}{{\partial{x}}}+\frac{{\partial{Q}}}{{\partial{y}}}\)

Step 3

If C is given by the vector equation ,

r(t)=x(t)i+y(t)j \(\displaystyle{a}\le{t}\le{b}\)

Then unit tangent vector is

\(\displaystyle{T}{\left({t}\right)}=\frac{{{x}'{\left({t}\right)}}}{{{\left|{r}\right|}{\left({r}'{\left({t}\right)}\right)}}}{i}+\frac{{{y}'{\left({t}\right)}}}{{{\left|{{r}'{\left({t}\right)}}\right|}}}{j}\)

n(t) is the outward unit normal vector C is given by,

\(\displaystyle{n}{\left({t}\right)}=\frac{{{y}'{\left({t}\right)}}}{{{\left|{r}\right|}{\left({r}'{\left({t}\right)}\right)}}}{i}+\frac{{{x}'{\left({t}\right)}}}{{{\left|{{r}'{\left({t}\right)}}\right|}}}{j}\)

As we know

Step 4

\(\displaystyle\oint_{{C}}{F}.{n}{d}{s}={\int_{{a}}^{{b}}}{\left({F}.{n}\right)}{\left({t}\right)}{\left|{{r}'{\left({t}\right)}}\right|}{\left.{d}{t}\right.}\)

\(\displaystyle={\int_{{a}}^{{b}}}{\left[\frac{{{P}{\left({x}{\left({t}\right)},{y}{\left({t}\right)}\right)}{y}'{\left({t}\right)}}}{{{\left|{{r}'{\left({t}\right)}}\right|}}}-\frac{{{Q}{\left({x}{\left({t}\right)},{y}{\left({t}\right)}\right)}{x}'{\left({t}\right)}}}{{{\left|{{r}'{\left({t}\right)}}\right|}}}\right]}{\left|{{r}'{\left({t}\right)}}\right|}{\left.{d}{t}\right.}\)

\(\displaystyle={\int_{{a}}^{{b}}}{P}{\left({x}{\left({t}\right)},{y}{\left({t}\right)}\right)}{y}'{\left({t}\right)}{\left.{d}{t}\right.}-{Q}{\left({x}{\left({t}\right)},{y}{\left({t}\right)}\right)}{x}'{\left({t}\right)}{\left.{d}{t}\right.}\)

\(\displaystyle=\int_{{C}}{P}{\left.{d}{y}\right.}-{Q}{\left.{d}{x}\right.}\)

\(\displaystyle=\int\int_{{D}}{\left(\frac{{\partial{P}}}{{\partial{x}}}+\frac{{\partial{Q}}}{{\partial{y}}}\right)}{d}{A}\)

Put Div(F) value in the above equation

Step 5

\(\displaystyle\oint_{{C}}{F}.{n}{d}{s}=\int\int_{{D}}{D}{i}{v}{\left({F}\right)}{d}{A}\)