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A weird thing about Stokes’ theorem is that if $S_1$ and $S_2$ are two oriented surfaces in $R^3$ that share the same boundary $∂S$ with the same induced orientation on $∂S$, then $\iint_{S_1} \operatorname{curl}F \cdot dS = \iint_{S_2} \operatorname{curl}F \cdot dS$ for any $C^1$ vector field $F$.

How do I check that $S_1$ and $S_2$ induce the same orientation on $∂S$, and how does this change if the induced orientations are not the same?

Also, when they induce the same orientation, why do the two integrals have to be the same in that case?

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How do I check that $S_1$ and $S_2$ induce the same orientation on $∂S$, and how does this change if the induced orientations are not the same?

An orientation on a curve is a “forward” direction. The induced orientation on $\partial S_1$ is the one that keeps $S_1$ on the left as you walk around it with your head pointed in the direction of the normal vector on $S_1$. So if both $S_1$ and $S_2$ are on the left as you walk around their shared boundary curve, the surfaces induce the same orientation. Otherwise, they induce opposite orientations, and the integrals differ by a factor of $-1$.

Also, when they induce the same orientation, why do the two integrals have to be the same in that case?

The reason for that is Stokes's Theorem. We have both \begin{align*} \iint_{S_1} \operatorname{curl}\vec F \cdot d\vec S &= \oint_{\partial S_1} \vec F \cdot d\vec s \\ \iint_{S_2} \operatorname{curl}F \cdot dS &= \oint_{\partial S_2} \vec F \cdot d\vec s \end{align*} If $\partial S_1 = \partial S_2$ as oriented curves, the integrals on the right are the same. So the integrals on the left are equal.

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