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.
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 $x$.