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I'm confused as to what the expression $\iint_D \dfrac{\partial N}{\partial x} - \dfrac{\partial M}{\partial y} \ dA$ in Green's theorem is supposed to be. Is this supposed to be the surface integral? In which case, $\iint_D \dfrac{\partial{N}}{\partial{x}} - \dfrac{\partial M}{\partial y} \ dA = \iint_S \mathbf{F} \cdot \hat{n} \ dS$ ? I was under the impression that Green's theorem is the relationship between the surface integral of a surface and the line integral around the aforementioned surface: $$\oint_C M \ dx + N \ dy = \iint_D \dfrac{\partial N}{\partial x} - \dfrac{\partial M}{\partial{y}} \ dA$$

Is this correct? I'm confused as to whether I am misunderstanding Green's theorem, because I know the surface integral to be $\iint_S \mathbf{F} \cdot \hat{n} \ dS$, or whether these two expressions are equivalent representations of the surface integral.

I would greatly appreciate it if people could please take the time to clarify this.

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  • $\begingroup$ It is the Riemann "$dx \ dy$" integral of a scalar function/surface $z=f(x,y)$ of two variables. $\endgroup$
    – Randall
    Aug 17, 2017 at 19:00
  • $\begingroup$ Green's theorem in the plane allows you to take a line integral to a double integral. See 2.2 of my paper/project thing here: math.tamu.edu/~szroberson/VCalc_SZR.pdf $\endgroup$ Aug 17, 2017 at 19:00

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Green's Theorem allows us to compute line integrals on simple closed curves by computing a double integral on the plane region which is enclosed by the aforementioned curve, and vice-versa. $$\iint_D\left(\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}\right)dA$$ represents a double integral over the surface $D$ which is enclosed by the curve $C$ in the $xy$-plane. This gives you an alternative way to compute the line integral $$\oint_C\left(M~dx+N~dy\right)$$ by using a double integral. A double integral is analogous to an integral over the real line (e.g. $\int_a^bf(x)dx$). Where an integral $\int_a^bf(x)dx$ is the area under $f$ from $a$ to $b$, the integral $\iint_Df(x,y)dA$ is the volume under $f$ on the 2D-region $D$.

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  • $\begingroup$ Thanks for the response. I just found out that $\iint_D \dfrac{\partial{N}}{\partial{x}} - \dfrac{\partial{M}}{\partial{y}} \ dA$ is the AREA of the surface -- not the surface integral, right? But I know the formula for surface area is $\iint_R |\mathbf{r_u} \times \mathbf{r_v}| \ dA$. So how is the expression $\iint_D \dfrac{\partial{N}}{\partial{x}} - \dfrac{\partial{M}}{\partial{y}} \ dA$ representing surface area? It makes sense that it is a double integral, but where did the $\dfrac{\partial{N}}{\partial{x}} - \dfrac{\partial{M}}{\partial{y}}$ come from? $\endgroup$ Aug 17, 2017 at 19:29
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    $\begingroup$ Check out this link: en.wikipedia.org/wiki/Green%27s_theorem. The proof of Green's Theorem is included, and you can see where the $\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}$ comes from. $\endgroup$
    – Dave
    Aug 17, 2017 at 19:34
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Green's Theorem in the plane is simply Stokes' Theorem applied to a planar surface $D$ bounded by a closed-contour $C$. To see this, recall that we have from Stokes' Theorem

$$\oint_{\partial S}\vec F\cdot d\vec \ell=\int_S \nabla \times \vec F\cdot \hat n\,dS$$

Now, let the surface $S=D$, with boundary $\partial S=C$, lie in the $xy$-plane so that $\hat n \,dS=\hat z\,dx\,dy$ and $d\vec \ell=\hat x\,dx+\hat y\,dy$. Furthermore, let $\vec F=\hat x M+\hat y N$. Then, Stokes' Theorem reduces to

$$\begin{align} \oint_{C}(M\,dx+N\,dy)&=\iint_{S} \hat z \cdot \nabla \times (\hat x M+\hat y N)\,dx\,dy\\\\ &=\iint_D \left(\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}\right)\,dx\,dy\\\\ &=\iint_D \left(\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}\right)\,dA \end{align}$$

which is Green's Theorem in the plane.

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  • $\begingroup$ Thanks for the response. I just found out that $\iint_D \dfrac{\partial{N}}{\partial{x}} - \dfrac{\partial{M}}{\partial{y}} \ dA$ is the AREA of the surface -- not the surface integral, right? But I know the formula for surface area is $\iint_R |\mathbf{r_u} \times \mathbf{r_v}| \ dA$. So how is the expression $\iint_D \dfrac{\partial{N}}{\partial{x}} - \dfrac{\partial{M}}{\partial{y}} \ dA$ representing surface area? It makes sense that it is a double integral, but where did the $\dfrac{\partial{N}}{\partial{x}} - \dfrac{\partial{M}}{\partial{y}}$ come from? $\endgroup$ Aug 17, 2017 at 19:24
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    $\begingroup$ You're welcome; my pleasure. To answer your last question, we have $$\hat z\cdot \nabla \times (\hat x M+\hat y N)=\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}$$As far as the double integral, it is taken in the $xy$-plane. And $$\iint_D \left(\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}\right)\,dA$$is not the area of $D$ in general. The area of $D$ is given by $\iint_D (1)\,dA$. $\endgroup$
    – Mark Viola
    Aug 17, 2017 at 19:33

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