# 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 Greens
You can still ask an expert for help

## Want to know more about Green's, Stokes', and the divergence theorem?

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Luvottoq
Step 1
We have to prove that,
${\oint }_{C}F.nd\mathrm{sin}t{\int }_{D}÷F\left(x,y\right)dA$
Step 2
Consider Vector field ,
F=Pi+Qj
$÷\left(F\right)=\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}$
Step 3
If C is given by the vector equation ,
r(t)=x(t)i+y(t)j $a\le t\le b$
Then unit tangent vector is
$T\left(t\right)=\frac{{x}^{\prime }\left(t\right)}{|r|\left({r}^{\prime }\left(t\right)\right)}i+\frac{{y}^{\prime }\left(t\right)}{|{r}^{\prime }\left(t\right)|}j$
n(t) is the outward unit normal vector C is given by,
$n\left(t\right)=\frac{{y}^{\prime }\left(t\right)}{|r|\left({r}^{\prime }\left(t\right)\right)}i+\frac{{x}^{\prime }\left(t\right)}{|{r}^{\prime }\left(t\right)|}j$
As we know
Step 4
${\oint }_{C}F.nds={\int }_{a}^{b}\left(F.n\right)\left(t\right)|{r}^{\prime }\left(t\right)|dt$
$={\int }_{a}^{b}\left[\frac{P\left(x\left(t\right),y\left(t\right)\right){y}^{\prime }\left(t\right)}{|{r}^{\prime }\left(t\right)|}-\frac{Q\left(x\left(t\right),y\left(t\right)\right){x}^{\prime }\left(t\right)}{|{r}^{\prime }\left(t\right)|}\right]|{r}^{\prime }\left(t\right)|dt$
$={\int }_{a}^{b}P\left(x\left(t\right),y\left(t\right)\right){y}^{\prime }\left(t\right)dt-Q\left(x\left(t\right),y\left(t\right)\right){x}^{\prime }\left(t\right)dt$
$={\int }_{C}Pdy-Qdx$
$=\int {\int }_{D}\left(\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}\right)dA$
Put Div(F) value in the above equation
Step 5
${\oint }_{C}F.nds=\int {\int }_{D}Div\left(F\right)dA$