# Stokes' theorem on surface with boundary being a union of closed, simple, piecewise smooth curves

According to Stokes' theorem, we need an oriented smooth surface $$S$$ whose boundary is a simple, closed, piecewise smooth curve $$C=\partial S$$. Is the theorem valid when the oriented smooth surface $$S$$ has a boundary that is a union of simple, closed, piecewise smooth curves $$C_1,\ldots,C_n$$?

Attempt. I believe the answer is yes. Let's, for example, take the part $$S=\{(x,y,z):z^2=x^2+y^2,~1\leqslant z\leqslant 2\}$$ of the standard cone $$z^2=x^2+y^2$$. Then $$\partial S=C_1\cup C_2$$, where $$C_i$$ is the flat disc of radius $$i$$. If $$\vec{n}$$ is the outer unit normal vector, then $$C_1,\,C_2$$ should be oriented anticlockwise and clockwise, respectively. Take: $$S'=\{(x,y,z):z^2=x^2+y^2,~0\leqslant z\leqslant 1\}.$$ Then we may apply Stokes' theorem on $$S',\,S+S'$$, whose boundaries are $$C_1,\,C_2,$$ respectively. Since $$\iint_{S+S'}= \iint_{S}+\iint_{S'}$$ we get: $$\iint_{S}=\iint_{S+S'}-\iint_{S'}=\oint_{C_2}-\oint_{C_1}.$$

By the same method, of $$\textit{closing}$$ the surface properly, we may derive in general the above claim.

Is the procedure correct?

Thanks in advance.

## 1 Answer

What you did is correct. It is true that Stokes' theorem is often proved just for disc like surfaces having a single closed boundary curve. But in fact this theorem is valid for any orientable surface $$S$$ with boundary $$\partial S$$, whereby the boundary is assumed to consist of finitely many piecewise smooth closed curves $$C_i$$. Orientable means that on $$S$$ can be defined a continuous unit normal $$x\mapsto {\bf n}(x)$$. For Stokes' formula to hold it is then further assumed that the $$C_i$$ are properly directed, namely such that near each point $$x\in C_i$$ the interior of $$S$$, when viewed from the tip of $${\bf n}(x)$$, is to the left of the arrow $${\bf t}(x)$$ marking the tangent direction of $$C_i$$ at $$x$$.