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Green's theorem is usually stated as follows:

Let $U \subseteq \mathbb{R}^2$ be an open bounded set. Suppose its boundary $\partial U$ is the range of a closed, simple, piecewise $C^1$, positively oriented curve $\phi: [0,1] \to \mathbb{R}^2$ with $\phi(t) = (x(t),y(t))$. Let $f,g: \overline{U} \to \mathbb{R}$ be continuous with continuous, bounded partial derivatives in $U$. Then $$ \int_U \left(\frac{\partial f}{\partial x}(x,y) - \frac{\partial g}{\partial y}(x,y)\right) dx dy = \int_{[0,1]} f(\phi(t))y'(t) dt + \int_{[0,1]} g(\phi(t))x'(t) dt. $$

Is there a complete rigorous proof of this theorem somewhere? Most texts (Rudin, Munkres, Spivak) proves the generalized Stokes' theorem first, and shows Green's theorem as a corollary. However, the "piecewise" criteria is never addressed, since the version of Stokes' proven in the texts above do not work for manifolds with corners. Hence the version of Green's theorem proven is always with full $C^1$ assumption.

Because Green's theorem is only on the plane, I'm wondering if there is an easy way to obtain the "piecewise" generalization from just the $C^1$ assumption. Spivak mentions

Green's theorem is true for a square ... can be proved by approximating the square ... by manifolds with boundary.

Is this easy to do in the context of just $\mathbb{R}^2$?

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It is easy to do by approximation -- just replace each corner of the square by an arc of a small circle, and show that the errors in both the area integral and the line integral approach zero as the radius of the circle approaches zero.

Alternatively, you can prove directly that Stokes's theorem applies to manifolds with corners. I do this in my Introduction to Smooth Manifolds (2nd ed.), Theorem 16.25. See also this MathOverflow question and its answers for more references.

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  • $\begingroup$ Thanks for the book reference, I will definitely look at that someday! About the approximation, would it be as easy for generic piecewise $C^1$ curves though? I'm not sure how I would describe the arc to connect lines that are not "straight"... $\endgroup$ – Wakaka Dec 24 '16 at 17:40
  • $\begingroup$ Just do it in local coordinates near the vertex. It shouldn't be hard to find an arc of a hyperbola that matches up with your curves and their tangent vectors at two points. $\endgroup$ – Jack Lee Dec 24 '16 at 21:11

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