What exactly does it mean for a surface to span a contour?

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Homework Statement


What exactly does it mean for a surface to span a contour/curve? For example, which surface(s) span the contour/curve defined by the equations x^2 + y^2 = 1 and z = y^2?

Homework Equations


N/A

The Attempt at a Solution


I'm not sure if I'm saying something nonsensical, but does any surface that covers at least the contour/curve in question span it? Would a cylinder as well as a sphere each span the contour curve defined by the equations x^2 + y^2 = 1 and z = y^2?
 

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  • #2
haruspex
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Homework Statement


What exactly does it mean for a surface to span a contour/curve? For example, which surface(s) span the contour/curve defined by the equations x^2 + y^2 = 1 and z = y^2?

Homework Equations


N/A

The Attempt at a Solution


I'm not sure if I'm saying something nonsensical, but does any surface that covers at least the contour/curve in question span it? Would a cylinder as well as a sphere each span the contour curve defined by the equations x^2 + y^2 = 1 and z = y^2?
Just a guess, but I'd say it was looking for a description of a surface with a boundary along that curve. So it's more of a verbal description of some possibilities than any standard shape. Can you picture the curve? Can you find two solutions for which the surface could be produced by curving a sheet of paper (i.e. at every point, one of the principal curvatures is zero - I don't know what the technical term for that is)?
 
  • #3
pasmith
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A surface spans a curve if that curve is the boundary of the surface and the surface is bounded.

Thus a surface which spans the given curve might consist of that portion of the cylinder [itex]x^2 + y^2 = 1[/itex] having [itex]0 \leq z \leq y^2[/itex], together with the circle [itex]x^2 + y^2 \leq 1[/itex] in the [itex]z = 0[/itex] plane. Another example, this time smooth, is [itex]z = r\sin^2 \theta[/itex] where [itex](x,y,z) = (r \cos \theta, r \sin \theta, z)[/itex] with [itex]0 \leq r \leq 1[/itex] and [itex]0 \leq \theta \leq 2\pi[/itex].

However a cylinder doesn't span that curve (because a cylinder either isn;t bounded or has a boundary consisting of two disjoint simple closed curves), and a spherical cap wouldn't span that curve because the curve doesn't lie on a sphere: you can parametrize it as [itex](\cos t, \sin t, \sin^2 t)[/itex], and the distance of the curve from the origin is then [itex]\sqrt{1 + \sin^4 t}[/itex] which is not constant.
 
  • #4
haruspex
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I suspect that the main thing the question is after is for you to spot an interesting symmetry the curve has. This symmetry means that there is a third surface, also a quadratic in x, y and z, which contains the curve. Using that, you can obtain another spanning surface analogous to pasmith's ##z = r\sin^2 \theta## solution.
 
  • #5
s3a
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pasmith:
Shouldn't z = r sin^2 θ be z = r^2 sin^2 θ?

Also, isn't (x,y,z) = (r cosθ, r sinθ, r^2 sin^2 θ) the same thing as (x,y,z) = (cos t, sin t, sin^2 t) except that the r is assumed to be 1 (with θ = t)? Wouldn't it be wrong to assume that r = 1?

And, is there a need for knowing the distance to the curve from the origin?

haruspex:
Actually, the problem that I'm currently working (which is the problem attached in this post) does not require that I work with any surface, because of the fact that the curl of the vector field in question is 0. (but I just wanted to know what I would have to do algebraically if the curl of the vector field in question were not 0).

Everyone:
In the problem attached in this post, both integrals over the contour C should be F ⋅ dr instead of what the solution gives, right?
 

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  • #6
pasmith
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pasmith:
Shouldn't z = r sin^2 θ be z = r^2 sin^2 θ?

[itex]z = r^n \sin^2 \theta[/itex] for any [itex]n > 0[/itex] works, since [itex]1^n = 1[/itex].
 
  • #7
haruspex
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haruspex:
Actually, the problem that I'm currently working (which is the problem attached in this post) does not require that I work with any surface, because of the fact that the curl of the vector field in question is 0. (but I just wanted to know what I would have to do algebraically if the curl of the vector field in question were not 0).
OK. But out of interest, the curve has an unexpected symmetry: invert it (##z\rightarrow 1-z##) and rotate 90 degrees about the z axis. Thus ##1-z = x^2## is another surface through it.
 
  • #8
s3a
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pasmith (and others):
Okay, so, is what is within the ExtraWork.pdf correct? I suspect that I made at least one mistake, since I still don't feel like I've mastered this problem (the problem in the TheProblemAndSolution.jpg attachment).

haruspex (and others):
As for the symmetry argument, first, is this ( http://www.wolframalpha.com/input/?i=parametric plot (1 * cos t, 1 * sin t, 1 * sin^2 t) ) the correct graph?

How am I supposed to notice the symmetry? How did you notice the symmetry?
 

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  • #9
haruspex
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