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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.

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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 $x$.

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