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# Conditions for stokes theorem

Understanding when you can use Stokes. Piecewise-smooth lines and surfaces. Created by Sal Khan.

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• In R^3 the same boundary will be shared by many surfaces. For instance a circular loop can be the boundary of a circle, a hemisphere, a paraboloid, etc. Does Stoke's theorem imply that the flux over all of these surfaces is the same? • Is it also a condition for F to be well defined and continuous differentiable at S (class C1)? • Can I use stokes to solve flux problems? How do you know which theorem out of stokes,divergence,green to use?
(1 vote) • As Maclaurin is to Taylor, Green is to Stokes, that is Green (2 dimensions) is a special case of Stokes (3 dimensions).

Stokes' Theorem equates the single integral of a function f along the boundary of a surface with the double integral of some kind of derivative of f along the surface itself.

Gauss's Theorem (a.k.a. the Divergence Theorem) equates the double integral of a function along a closed surface which is the boundary of a three-dimensional region with the triple integral of some kind of derivative of f along the region itself.

Thus the situation in Gauss's Theorem is "one dimension up" from the situation in Stokes's Theorem, so it should be easy to figure out which of these results applies. If you see a three dimensional region bounded by a closed surface, or if you see a triple integral, it must be Gauss's Theorem that you want. Conversely, if you see a two dimensional region bounded by a closed curve, or if you see a single integral (really a line integral), then it must be Stokes' Theorem that you want.

This is only two theorems: what about Green's Theorem? Green's Theorem is in fact the special case of Stokes's Theorem in which the surface lies entirely in the plane. Thus when you are applying Green's Theorem you are technically applying Stokes's Theorem as well, however in a case which leads to some simplifications in the formulas. Especially, when you have a vector field in the plane, the curl of the vector field is always a purely vertical vector, so it makes sense to identify this with a scalar quantity, and this scalar quantity is precisely the "derivative" appearing in the double integral in Green's Theorem.

All are used for flux calculations.
• Wikipedia says that the divergence theorem is a more generalized form of Stokes' theorem. Is this true? • At , I think Sal means they're not smooth at the corners, not edges. • Can you give an example of a curve that is simple & closed, but not piece-wise smooth? Thanks! • Hi bkmurthy99, here's an example. Keep in mind we need to get a little beyond what most multivariable courses cover in terms of rigorous definitions. Recall smooth means differentiable in this context, and piecewise-smooth means we can at least split up our function into finitely many pieces that are themselves smooth. To violate this condition, we'd need a function everywhere undifferentiable. There's a pathological (meaning it's wild!) function called the Weierstrass function, basically a fractal curve, that is continuous everywhere but differentiable nowhere. Since we can't break it up into finitely many smooth pieces (or any smooth pieces actually), it's not piecewise-smooth. Normally it's a function from R to R, but to make it a simple, closed curve, we just have to connect up two ends of the graph to make a jagged circle-ish boundary. It's simple because it doesn't intersect itself (which I haven't rigorously shown, but is probably true), and closed because the Weierstrass function is symmetric, so we can definitely join up two ends of it into that jagged circle. See this image for a visual of the function. Imagine picking up the graph between x = -1 and x = 1, and forming a circle with that strand of graph. https://i.stack.imgur.com/RgN7d.png

Hope this example clears up how it's theoretically possible to have a simple, closed, not piecewise-smooth curve. In practice, since the real world is constrained by the Planck-length and quantum mechanical stuff, every actual curve is piecewise smooth in real life. Hope this helps!
(1 vote)
• So could a normal surface, with continuous derivatives everywhere (gradually changing slope), also be called a piecewise smooth surface? Even though it's not really "piecewise" per se? because it's not necessarily smooth in pieces...

(1 vote) • Hi Karun, yes, even perfectly smooth surfaces can be called piecewise smooth as well. The reason comes down to the adjective piecewise smooth being more general than just smooth. The definition of piecewise smooth is that it's possible to break up the surface into a finite number of sections, each of which is itself smooth. For your regular, smooth surface, you can "break" it into just one piece, namely itself, to satisfy the definition. Similarly, we could say a circle is piecewise smooth, even though it's also perfectly smooth on its own. Hope this helps :)
(1 vote)
• why we use green's theorem,stoke,s theorem and divergence theorem?reply must
(1 vote) • I still don't see why the integration of the boundary is the same as the integration of the total surface area of the boundary. This is counter-intuitive because it's assumed in our everyday thinking that the surface area of an objection is larger than the mere boundary line of that object. ? I wish this part could be made clear (unless I'm understanding the theorem's conclusion wrong)?
(1 vote) • I believe you also need orientability, right?
(1 vote) 