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Stokes' Theorem states that, "For a smooth oriented region $V$ $\in$ $R^3$ and a smooth vector field defined on $V \cup \partial V$, where $\partial V$ is the boundary curve for $V$, $\int \int _{V} $ curl $\vec F .d\vec S$ = $\int_{\partial V} \vec F . d\vec r$. My question is, do we always take $\partial V$ to be the curve when $V$ is projected onto the $xy$ plane? (So if $V$ is a unit sphere, say, then $\partial V$ is the unit circle). If so, why do we do so?

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See the Math Insight website. In the website there is a toy applet you can play with.

The curve need not lie on a plane, and it is NOT a projection on a plane. It can only be the boundary of an "open" surface, so your should split your "ellipsoid" into two parts and apply the theorem respectively.

In a calculus course, the boundary (curve) is usually on a plane. It's only because of simplicity of calculation. It's not related to the theorem description.

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No. $\partial V$ is a curve located somewhere in $\mathbb R^3$, it's not projected on any plane.

In case when $V$ is a unit sphere it doesn't have a boundary at all and $\partial V = \varnothing$.

If you want $\partial V$ to be a unit circle, you'd need $V$ to be, for example, a half-sphere, or, a unit disk.

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  • $\begingroup$ Hmm, I had an ellipsoid as V for my example and for $\partial V$ they just took the ellipse on xy plane, I am not sure why they did so? $\endgroup$ – JustWandering Jun 12 at 11:07
  • $\begingroup$ Are you sure it was a full ellipsoid? A half-ellipsoid would have the boundary being an allipsoid. Another possibility is that they don't use Stokes' Theorem at all, but they are parametrizing the ellipsoid with the points of a region of a plane bounded by the ellipse. Without knowing what example you're talking about, I can't say. $\endgroup$ – Adam Latosiński Jun 12 at 15:46
  • $\begingroup$ Ahh you are right sorry I misread the question, my apologies! $\endgroup$ – JustWandering Jun 12 at 17:34

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