Stokes' Theorem, how to apply for this surface?

In summary: My bad.In summary, you are given a parametric equation of a surface. You attempt to solve for the surface using a line integral, but are given trouble because the surface is part of a square and part of a curve.
  • #1
Felipe Lincoln
Gold Member
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Homework Statement


With the stokes' theorem transform the integral ## \iint_\sigma \vec{\nabla}\times\vec{F}\cdot\vec{\mathrm{d}S} ## into a line integral and calculate.
## \vec{F}(x,y,z) = y\hat{i} -x^2\hat{j} +5\hat{k}##
##\sigma(u,v) = (u, v, 1-u^2)##
## v\geq0##, ##u\geq0##, ##u+v\leq1##

Homework Equations


Stokes' Theorem
$$\oint_\gamma\vec{F}\cdot\vec{\mathrm{d}S} = \iint_\sigma\vec{\nabla}\times\vec{F}\cdot\vec{\mathrm{d}S}$$

The Attempt at a Solution


The surface in this case is formed by a surface with several cuts: in the plane x+y=1, x=0 and y=0. So I have no idea on how to apply stokes' theorem here. I know this way: given a closed simple curve with not conservative vectorial field in it, the line integral of this vector field along this curve is equal to the integral of surface in which the surface is bounded by this curve. Thinking this way I can only imagine curves being bounds of a bounded surface, but here we got a surface that is part of a square and part of a curve. How to deal with this case?
 
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  • #2
Your surface is not "part of a square and part of a curve". You are given a parametric equation of the surface ##\sigma(u,v) = \langle u,v,1 - u^2\rangle##. That is ##x = u,~y=v,~z = 1-x^2##. Do you recognize what that is? Plot a picture of it, then look at the part of that surface that satisfies the inequalities. The boundaries of that surface will give you the curves you need for your line integral. Come back when (if) you need more help.
 
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  • #3
Oh, I can see. I was looking more than just the surface, it seemed to be something like a peace of cake hehe, but I saw it's just a slice of a parabola. It's solved now, thank you !
 
  • #4
Just to add a piece of notational advice: Do not use ##d\vec S## for the line element if you are also using it to represent the area element. You risk confusing not only others but also yourself.
 
  • #5
Orodruin said:
Just to add a piece of notational advice: Do not use ##d\vec S## for the line element if you are also using it to represent the area element. You risk confusing not only others but also yourself.
Oh I'm sorry I meant ##\vec{\mathrm{d}r} ## .
 

1. What is Stokes' Theorem and why is it important?

Stokes' Theorem is a fundamental theorem in vector calculus that relates the surface integral of a vector field to the line integral of its curl along the boundary of the surface. It is important because it allows us to easily calculate the surface integral of a vector field, which has many applications in physics and engineering.

2. How do I know when to apply Stokes' Theorem for a given surface?

In order to apply Stokes' Theorem, the given surface must be a closed, orientable surface with a continuous and differentiable boundary curve. Additionally, the vector field must be continuously differentiable on the surface. If these conditions are met, you can use Stokes' Theorem to calculate the surface integral.

3. Can you provide an example of how to apply Stokes' Theorem?

Sure! Let's say we have a closed surface S defined by the curve C, and we want to calculate the surface integral of a vector field F over S. First, we would need to calculate the curl of F and then evaluate the line integral of the curl along C. This line integral would then be equal to the surface integral of F over S, as given by Stokes' Theorem.

4. What are some common mistakes to avoid when applying Stokes' Theorem?

One common mistake is forgetting to check if the surface and vector field meet the necessary conditions for Stokes' Theorem. Another mistake is not being careful with the orientation of the surface and boundary curve, which can result in an incorrect calculation. It is also important to double check your calculations and make sure you are using the correct formula.

5. Are there any limitations to using Stokes' Theorem?

Stokes' Theorem can only be applied to closed surfaces and vector fields that are continuously differentiable. Additionally, the surface and boundary curve must be orientable. If these conditions are not met, Stokes' Theorem cannot be used and another method must be used to calculate the surface integral.

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