# Why does Green's theorem fail for this exact differential form $\int Mdx + Ndy$

Why does Green's theorem fail for this exact differential form $\int {Mdx + Ndy} = 0$ since it's an exact differential and M and N both are not functions of z ( the case for which i want to ask the question). Green says this $$\int \int {\partial N/\partial X - \partial M/\partial Y}dA$$ which is equal to zero as this is an exact differential. If i do the line integral in some path to some point to some point then it is equal to the change of function between the final and initial points which may not be zero. So my question is, why does not Gauss' theorem become true for this?

• First, your $ds$ needs to be $dA$ (ordinarily $ds$ stands for an arclength integral). But if the form is exact and it's continuously differentiable, that integrand is precisely $0$. I.e., $\partial N/\partial x - \partial M/\partial y = 0$ (assuming the functions are defined and continuously differentiable everywhere inside your curve). What's your issue? – Ted Shifrin Aug 18 '18 at 0:47
• @TedShifrin is it okay now? – user187604 Aug 18 '18 at 0:51
• The integral of an exact form is path-independent. What you wrote in the first line applies only to line integrals around closed paths. – Ted Shifrin Aug 18 '18 at 0:57
• @TedShifrin so when im moving from a point to another which is not a closed integral that also doesn't enclose a surface. So we can't apply gauss theorem here. Is it the answer? – user187604 Aug 18 '18 at 1:21

Let $C$ be a positively oriented, piecewise smooth, simple closed curve in a plane, and let $D$ be the region bounded by $C$. If $M$ and $N$ are functions of $(x, y)$ defined on an open region containing $D$ and have continuous partial derivatives there, then $$\int_C M\,dx + N\, dy = \iint_D \left(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}\right) \, dx\, dy$$ where the path of integration along $C$ is anticlockwise.