Stokes theorem and line integral

In summary, the problem is to prove that 2A is equal to the line integral of \vec{r}\times d\vec{r}. Two methods were suggested, one using Stokes theorem and the other using a vector field and its curl. After some clarification, it was determined that the formula only applies to a planar surface. Using this information, the solution was found by setting z=0 and evaluating the vector field and its curl, resulting in the correct value of 2A.
  • #1
rado5
73
0

Homework Statement



Prove that [tex]2A=\oint \vec{r}\times d\vec{r}[/tex]

Homework Equations





The Attempt at a Solution



From stokes theorem we have [tex]\oint d\vec{r}\times \vec{r}=\int _{s}(d\vec{s}\times \nabla)\times \vec{r}= \int _{s}(2ds\frac{\partial f}{\partial x},-ds+ds\frac{\partial f}{\partial y},-2ds)[/tex]. I'm stuck here and I don't know how to continue. Please help me.

Another method is [tex]\oint \vec{r}\times d\vec{r}=\oint (ydz-zdy,zdx-xdz,xdy-ydx)[/tex]. Please help me solve it.
 
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  • #2
Maybe they didn't state it very clearly, but that formula is only true for a planar surface, say a region in the xy plane. And I'm not sure what formulation you are using of Stokes theorem. If you are using the vector form, that means a unit normal points in the z direction, say k. To get the area you want to integrate 1*dx*dy. Now if you define the vector field F=(-y/2,x/2,0) what is curl(F)?
 
  • #3
Dick said:
Maybe they didn't state it very clearly, but that formula is only true for a planar surface, say a region in the xy plane.

So in this case z=0 and then we have: [tex]\oint \vec{r}\times d\vec{r}=\oint (ydz-zdy,zdx-xdz,xdy-ydx)=\oint (0,0,-ydx+xdy) [/tex] so [tex]\vec{F}=(-y,x,0) and \vec{n}=(0,0,1) and \nabla \times \vec{F}=(0,0,2) and so (\nabla \times \vec{F}).\vec{n}=(0,0,2)[/tex] so we have [tex]\oint \vec{F}.d\vec{r}=\int_{s}(\nabla\times \vec{F}).\vec{n}ds=\int\int_{D}2dxdy=2A[/tex]
I think my method is now correct. Thank you very much for your kind help.
 

Related to Stokes theorem and line integral

1. What is Stokes theorem and how is it related to line integrals?

Stokes theorem is a fundamental theorem in vector calculus that relates the surface integral of a vector field over a closed surface to the line integral of the same vector field around the boundary of that surface. In other words, it connects the flow of a vector field through a surface to the circulation of that same vector field around the boundary of the surface.

2. What is the significance of Stokes theorem in physics and engineering?

Stokes theorem is used in physics and engineering to solve problems related to fluid flow, electromagnetism, and other fields that involve vector fields. It allows us to relate the behavior of a vector field over a closed surface to the behavior of that same vector field along the boundary of the surface, making it a powerful tool in understanding and analyzing physical systems.

3. Can you provide an example of how to use Stokes theorem to solve a problem?

One example of using Stokes theorem is in calculating the work done by a force field on a moving object. By converting the work integral into a line integral using Stokes theorem, we can simplify the calculation and make use of the fundamental theorem of calculus to find the solution.

4. How is Stokes theorem related to Green's theorem?

Stokes theorem is a higher-dimensional version of Green's theorem, which relates the line integral of a two-dimensional vector field to the double integral of its curl over a region in the plane. Stokes theorem extends this concept to three dimensions, where the line integral is related to the surface integral of the curl of a vector field over a closed surface.

5. Are there any restrictions or assumptions to keep in mind when using Stokes theorem?

Stokes theorem can only be applied to smooth vector fields and surfaces. Additionally, the surface must be oriented in a specific direction and its boundary must be a closed curve. It is also important to keep track of any singularities or discontinuities in the vector field, as they may affect the validity of the theorem.

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