# Use Green's Theorem in the form of this equation to prove Green's first identity, where D and C satisfy the hypothesis of Green's Theorem and the appropriate partial derivatives of f and g exist and are continuous. (The quantity grad g · n = Dng occurs in the line integral. This is the directional derivative in the direction of the normal vector n and is called the normal derivative of g.) oint_c F * n ds = int int_D div F(x,y)dA

Question
Use Green's Theorem in the form of this equation to prove Green's first identity, where D and C satisfy the hypothesis of Green's Theorem and the appropriate partial derivatives of f and g exist and are continuous. (The quantity grad g · n = Dng occurs in the line integral. This is the directional derivative in the direction of the normal vector n and is called the normal derivative of g.)
$$\displaystyle\oint_{{c}}{F}\cdot{n}{d}{s}=\int\int_{{D}}\div{F}{\left({x},{y}\right)}{d}{A}$$

2021-03-05
Step 1
$$\displaystyle\int\int_{{D}}{f}\nabla^{{2}}{g}{d}{A}=\oint_{{C}}{f{{\left(\nabla{g}\right)}}}.{n}{d}{s}-\int\int_{{D}}\nabla{f}.\nabla{g}{d}{A}$$
Add $$\displaystyle\int\int_{{D}}\nabla{f}.\nabla{g}{d}{A}$$ to both sides,
$$\displaystyle\int\int_{{D}}{f}\nabla^{{2}}{g}{d}{A}+\int\int_{{D}}\nabla{f}.\nabla{g}{d}{A}=\oint_{{C}}{f{{\left(\nabla{g}\right)}}}.{n}{d}{s}$$
$$\displaystyle\int\int{\left({f}\nabla^{{2}}{g}+\nabla{f}.\nabla{g}\right)}{d}{A}=\oint_{{C}}{f{{\left(\nabla{g}\right)}}}.{n}{d}{s}$$
Step 2
Using the product rule for derivatives,
we can write $$\displaystyle{f}\nabla^{{2}}{g}+\nabla{f}.\nabla{g}={f}\nabla.{\left(\nabla{g}\right)}+\nabla{f}.\nabla{g}=\nabla.{\left({f}\nabla{g}\right)}$$
$$\displaystyle\int\int_{{D}}\nabla.{f{{\left(\nabla{g}\right)}}}{d}{A}=\oint_{{C}}{f{{\left(\nabla{g}\right)}}}.{n}{d}{s}$$
$$\displaystyle\int\int_{{D}}\div{\left({f}\nabla{g}\right)}{d}{A}=\oint_{{C}}{f{{\left(\nabla{g}\right)}}}.{n}{d}{s}$$

### Relevant Questions

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.
If E(t,x,y,z) and B(t,x,y,z)represent the electric and magnetic fields at point (x,y,z) at time t, a basic principle of electromagnetic theory says that $$\displaystyle\nabla\times{E}=\frac{{-\partial{B}}}{{\partial{t}}}$$. In this expression $$\displaystyle\nabla\times{E}$$ is computed with t held fixed and $$\displaystyle\frac{{\partial{B}}}{{\partial{t}}}$$ is calculated with (x,y,z) fixed.
Use Stokes' Theorem to derive Faraday's law, $$\displaystyle\oint_{{C}}{E}\cdot{d}{r}=-\frac{\partial}{{\partial{t}}}\int\int_{{S}}{B}\cdot{n}{d}\sigma$$,
Use Green's Theorem to evaluate the line integral. Orient the curve counerclockwise.
$$\displaystyle\oint_{{C}}{F}{8}{d}{r}$$, where $$\displaystyle{F}{\left({x},{y}\right)}={\left\langle{x}^{{2}},{x}^{{2}}\right\rangle}$$ and C consists of the arcs $$\displaystyle{y}={x}^{{2}}{\quad\text{and}\quad}{y}={8}{x}{f}{\quad\text{or}\quad}{0}\le{x}\le{8}$$
Use the Divergence Theorem to calculate the surface integral $$\displaystyle\int\int_{{S}}{F}·{d}{S}$$, that is, calculate the flux of F across S.
$$\displaystyle{F}{\left({x},{y},{z}\right)}={\left({\cos{{\left({z}\right)}}}+{x}{y}^{{2}}\right)}{i}+{x}{e}^{{-{{z}}}}{j}+{\left({\sin{{\left({y}\right)}}}+{x}^{{2}}{z}\right)}{k}$$
S is the surface of the solid bounded by the paraboloid $$\displaystyle{z}={x}^{{2}}+{y}^{{2}}$$ and the plane z = 9.
Use the Divergence Theorem to calculate the surface integral $$\displaystyle\int\int_{{S}}{F}·{d}{S}$$, that is, calculate the flux of F across S.
$$\displaystyle{F}{\left({x},{y},{z}\right)}={\left({x}^{{3}}+{y}^{{3}}\right)}{i}+{\left({y}^{{3}}+{z}^{{3}}\right)}{j}+{\left({z}^{{3}}+{x}^{{3}}\right)}{k}$$, S is the sphere with center the origin and radius 2.
Use Green's Theorem to evaluate the line integral
$$\displaystyle\int_{{C}}{\left({y}+{e}^{{x}}\right)}{\left.{d}{x}\right.}+{\left({6}{x}+{\cos{{y}}}\right)}{\left.{d}{y}\right.}$$
where C is triangle with vertices (0,0),(0,2)and(2,2) oriented counterclockwise.
a)6
b)10
c)14
d)4
e)8
f)12
Use Stokes' theorem to evaluate the line integral $$\displaystyle\oint_{{C}}{F}\cdot{d}{r}$$ where A = -yi + xj and C is the boundary of the ellipse $$\displaystyle\frac{{x}^{{2}}}{{a}^{{2}}}+\frac{{y}^{{2}}}{{b}^{{2}}}={1},{z}={0}$$.
Evaluate the line integral $$\displaystyle\oint_{{C}}{x}{y}{\left.{d}{x}\right.}+{x}^{{2}}{\left.{d}{y}\right.}$$, where C is the path going counterclockwise around the boundary of the rectangle with corners (0,0),(2,0),(2,3), and (0,3). You can evaluate directly or use Green's theorem.
Use Stokes' Theorem to evaluate $$\displaystyle\int\int_{{S}}{C}{U}{R}{L}{f}\cdot{d}{S}$$.
$$\displaystyle{F}{\left({x},{y},{z}\right)}={x}^{{2}}{y}^{{3}}{z}{i}+{\sin{{\left({x}{y}{z}\right)}}}{j}+{x}{y}{z}{k}$$,
S is the part of the cone $$\displaystyle{y}^{{2}}={x}^{{2}}+{z}^{{2}}$$ that lies between the planes y = 0 and y = 2, oriented in the direction of the positive y-axis.