I was wondering if what my teacher said was correct and complete in that in Stokes' Theorem the "boundary curve" of a surface can be defined as the mapping along the boundary of the two-dimensional region from which the surface is parameterized. For example, if a surface $\mathbf r(u,v)$ is a mapping from the u-v plane to x-y-z space, then the boundary of $\mathbf r(u,v)$ is the curve that you get from mapping from the u-v plane to x-y-z along the boundary of the domain region in the u-v plane. Thanks in advance!

Edit: I was discussing with my teacher and we think the above is the case (although I would really like someone to please confirm) but interestingly depending on how you parameterize a surface you may get similar but not exactly the same results for the boundary curve by following the above method, although you do get equivalent results if you account for direction, and the same answer via two parametrization makes me think I'm on the right track. To demonstrate what I mean let me use an example. Call the surface S the part of the sphere $x^2+y^2+z^2=4$ above the plane $z=1$. The boundary curve is obviously the intersection of the plane and the sphere, that is the circle $x^2+y^2=3$ (and $z=1$). It seems from the fact that we can get this result using the method that traces the out the boundary curve by mapping to the surface along the domain using two different parametrizations that this definition of a boundary curve is correct (though note that the spherical one requires that two parts cancel each other out). Example:

Cartesian Coordinates

We can write the surface S as $\mathbf r(x,y)= \big<x,y,\sqrt{4-x^2-y^2}\big>$ (where we take the positive root since $z\geq1 \rightarrow z\geq0$). Now, the surfaces is a mapping of all $(x,y)$ from the "filled-in circle" $x^2+y^2\leq3$ to it via $z=\sqrt{4-x^2-y^2}$, so tracing along the boundary of the planar domain region $x^2+y^2\leq3$, that is, $x^2+y^2=3$, and mapping to the surface gives the boundary curve $x^2+y^2=3$ with $z=1$ (the first part of which is given by the fact that the domain of the curve we are tracing along the surface is $x^2+y^2=3$ and the $z=1$ part we got from plugging in). Now in spherical coordinates we get a similar result for the boundary curve.

Spherical Coordinates

We can write S as $\mathbf r(\phi,\theta) = \big<2\sin\phi\cos\theta,2\sin\phi\sin\theta,2\cos\phi\big> $ where $0\leq\theta\leq2\pi$ and $0\leq\phi\leq\frac{\pi}{3}$. This gives us a rectangular domain in the "$\phi$-$\theta$" plane (where $\phi$ is where the x-axis would be and $\theta$ where the y-axis would be and the surface is a mapping from the rectangle into x-y-z space) bounded by the coordinate axis and the lines $\theta = 2\pi$ and $\phi=\frac{\pi}{3}$. Update: Here's an image: Phi-Theta Domain Going along the bottom ($\theta =0, \phi$ varying to $\frac{\pi}{3}$) gives us a downward arch along the sphere from the top to the bounding circle. Similarly, going along the right gives us the bounding circle, going along the top gives the arch but in the opposite direction (so they cancel -- I presume) and at $\phi=0$ moving down the last side of the rectangle (where $\theta$ is varying from $2\pi$ back to $0$) all we get is a point which doesn't contribute to the curve leaving us with the same boundary curve to S. To recap, as we map to the surface alone the boundary of the domain, we go from the "north pole", down a longitude to the 60-degree latitude, around the latitude 360 degrees and back up the same longitude to the north pole, gibing us the boundary circle.

In both cases you get the "boundary ring" by mapping to the surface along the boundary of the planar domain region. I was wondering if this method can be a general definition of the boundary of a surface. If someone could confirm or correct that would be great.

  • $\begingroup$ It depends on your definition of a psrameterized surface. If you assume the parameterization to be 1-1 on the planar region including boundary then the interpretation you gave is correct. Otherwise, it is incorrect. $\endgroup$ Commented Jan 15, 2015 at 21:01
  • $\begingroup$ @studiosus Why, what happens if it's not one-to-one? $\endgroup$
    – Gabriel
    Commented Jan 16, 2015 at 2:05
  • $\begingroup$ Then your notion of boundary is wrong. $\endgroup$ Commented Jan 16, 2015 at 18:01
  • $\begingroup$ @studiosus could you maybe give an example? $\endgroup$
    – Gabriel
    Commented Jan 16, 2015 at 18:30
  • $\begingroup$ You can map unit disk onto the unit sphere by a smooth map which is 1-1 in the open disk and sends the unit circle to a single point. I can write down an explicit formula but it is a bit tedious. In such example, the parameterized surface has empty boundary. $\endgroup$ Commented Jan 16, 2015 at 18:50

1 Answer 1


You are asking a very good question. The problem, however, that a proper answer to it lies beyond the class you are taking. You should consider reading, say, Spivak's Calculus on Manifolds or the book by Guillemin and Pollack Differential Topology.

Here is the upshot: There is a concept of an (oriented) differential manifold with boundary, it is a bit abstract, but the key is that it allows you to define the notion of boundary which is independent of the parameterization that you use. A 2-dimensional manifold is simply a surface. Instead of trying to "parameterize" a surface $S$ (or, more generally, a manifold $M$) via a single map of a planar region, you consider a collection of maps from the closed upper half-plane $\{(x,y): y\ge 0\}$ to your surface. These maps $f_i$ (called charts) are required to be 1-1, which allows you to consider the compositions $f_i^{-1} \circ f_j$. These compositions are called transition maps. Each transition map is required to be differentiable (say, infinitely-differentiable, just to be on the safe side). The boundary points of the $S$ are defined as points which are images of points of the x-axis under charts. One verifies that if $p$ is in the image of the x-axis under one chart, then the same is true for any other chart whose image contains $p$.

Another useful thing that you learn by reading either of these books is that Stokes' theorem holds not only for surfaces with boundary but for manifolds with boundary of arbitrary dimension and in this form it includes all fundamental theorems of multivariable calculus (Stokes, Green, Gauss) and can be written as $$ \int_{M} d\omega = \int_{\partial M} \omega, $$ where $M$ is a compact $k$-dimensional manifold with boundary $\partial M$, $\omega$ is a differential form of degree $k-1$ on $M$ (this is a replacement of vector fields which you see in the multivariable calculus), while $d\omega$ is the exterior derivative of $\omega$, which is the proper replacement of Curl and divergence.

  • $\begingroup$ Are you sure you can't just use the boundary of the domain thing, even for non-one-to-one mappings? I looked in Buck's "Advanced Calculus" and it says: "Let Σ be a continuous surface whose domain D is such that its boundary is the trace of a simple closed curve $\gamma$ defined for $a\leq t\leq b$. The image of $\gamma$ under $\Sigma$ is a closed curve...which is called the boundary or edge of $\Sigma$." It seems as if as long as D is composed of a finite number of simple closed regions, this works. He even includes a picture in his discussion of boundary of a non-one-to surface (I can attach) $\endgroup$
    – Gabriel
    Commented Jan 18, 2015 at 4:36
  • 1
    $\begingroup$ @Gabriel For maps that are one-to-one on the domain including boundary, then the boundary of the surface is indeed the image of the boundary of the domain under this map. And this is justified by the theory of differentiable manifolds(essentially by the inverse function theorem). $\endgroup$
    – AaronS
    Commented Jun 2, 2016 at 0:18

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