Suppose that the plane region D, its boundary curve C, and the functions P and Q satisfy the hypothesis of Green's Theorem. Considering the vector field F = Pi+Qj, prove the vector form of Green's Theorem oint_C F*n ds = int int_D div F(x,y) dA where n(t) is the outward unit normal vector to C.

Suppose that the plane region D, its boundary curve C, and the functions P and Q satisfy the hypothesis of Green's Theorem. Considering the vector field F = Pi+Qj, prove the vector form of Green's Theorem oint_C F*n ds = int int_D div F(x,y) dA where n(t) is the outward unit normal vector to C.

Question
Suppose that the plane region D, its boundary curve C, and the functions P and Q satisfy the hypothesis of Green's Theorem. Considering the vector field F = Pi+Qj, prove the vector form of Green's Theorem \(\displaystyle\oint_{{C}}{F}\cdot{n}{d}{s}=\int\int_{{D}}\div{F}{\left({x},{y}\right)}{d}{A}\)
where n(t) is the outward unit normal vector to C.

Answers (1)

2020-11-03
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
\(\displaystyle\oint_{{C}}{F}.{n}{d}{s}=\int\int_{{D}}{D}{i}{v}{\left({F}\right)}{d}{A}\)
0

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