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For Stoke's Theorem we are tasked with picking the boundary of a surface. What happens if there'smore than one boundary?
e.g. in
$$\int_{S}(\nabla \times {\textbf{F}})\cdot dS$$
where ${\textbf{F}} = (y,z,x^2y^2)$ and $S$ is the surface given by $z=x^2 + y^2$ and $0\leq z \leq 5$.
The surface looks like it has a boundary at $z=0$ with the circle $x^2 + y^2 = 1$ and also the same boundary except at $z=5$. How do we pick the correct one?

What if instead of the boundary $0 \leq z \leq 5$, we have $0 \leq z \leq 4x+3y+5$ or something like that? (There will also be a boundary with a slanted plane).

OneGapLater
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  • You have to use the whole boundary. – Mariano Suárez-Álvarez May 30 '17 at 06:21
  • I'm not sure what that means, sorry. I'm quite new to Stoke's Theorem and the context of a boundary wasn't actually defined for me. / Do you mean I'd have to take the sum or osmething alike – OneGapLater May 30 '17 at 06:23
  • You have a wrong idea about boundary in your mind. boundary at given point , doesn't make any sense. – Red shoes May 30 '17 at 06:26
  • I mean, that the boundary of the surface $S$ would be the circle with radius 1, centre 0. This, occuring at both heights $z=0,5$ (by my given intuition of a boundary, but never formally or mathematically defined) – OneGapLater May 30 '17 at 06:27
  • Note that at height $z=5$ the radius of the boundary-circle is $\sqrt{5}$, not $1$. However at height $z=0$ there is no "boundary-circle", the surface is perfectly smooth there. – Zubzub May 30 '17 at 07:16
  • I think the question was for surfaces with multiple boundaries, like the cylinder $x^2+y^2=4$ from $0\leq z\leq 1$. What do you do in this case, would the surface integral be equal to the sum of the line integrals for each contour? – oriont Dec 12 '22 at 15:24

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