# Stokes' theorem confusion

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?

## 2 Answers

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.

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.

• 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? – JustWandering Jun 12 at 11:07
• 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. – Adam Latosiński Jun 12 at 15:46
• Ahh you are right sorry I misread the question, my apologies! – JustWandering Jun 12 at 17:34