DS for line integral

  • Thread starter Gauss M.D.
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  • #1
Gauss M.D.
153
1

Homework Statement



Say I have a line integral which I have simplified to:

[itex]\int\int x+y dS[/itex]

Over some surface S, let's say 5x^2 + 3y^2 = 4 or something. Having arrived at this step, how do I determine dS? The formulas and methods we've been taught doesn't really lead to this step all that often but I'd like to know a general approach for determining dS when it is in this form.

Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
szynkasz
115
2
If you have ##dS## it is rather surface integral, not line one.
 
  • #3
13,918
7,866
Here's a discussion on surface integrals:

 
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  • #4
LCKurtz
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Homework Statement



Say I have a line integral which I have simplified to:

[itex]\int\int x+y dS[/itex]

Over some surface S, let's say 5x^2 + 3y^2 = 4 or something. Having arrived at this step, how do I determine dS? The formulas and methods we've been taught doesn't really lead to this step all that often but I'd like to know a general approach for determining dS when it is in this form.

Homework Equations





The Attempt at a Solution


You parameterize the surface with two parameters, say ##u## and ##v##. So you have$$
\vec R(u,v) =\langle x(u,v), y(u,v), z(u,v)\rangle$$Then express your integral in terms of ##u## and ##v## with ##dS = |\vec R_u \times \vec R_v|dudv## with appropriate ##u,v## limits.

In your example you could let ##z## be itself, solve the surface equation for ##x## in terms of ##y## and use ##z## and ##y## for your parameters. The actual integral for your example, which I'm guessing you just made up, might get ugly. If your example was circular instead of elliptical you might use cylindrical coordinates ##z,\theta## for your parameters.
 

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