Stokes' theorem over a tetrahedron

In summary, Stokes' theorem states that the line integral of the curl of a vector field over a closed curve is equal to the surface integral of the vector field over the surface bounded by that curve. In the given example of the function \vec v = y\hat z over a tetrahedron, the right hand side of the equation is computed by finding the line integral from (a,0,0) to (0,2a,0) to (0,0,a) to (a,0,0), which equals a^2. For the surface integral, one can choose any surface as long as it has the triangle as its boundary. In this case, the chosen surface has the area element d\vec
  • #1
Reshma
749
6
Check the Stokes' theorem for the function [tex]\vec v = y\hat z[/tex]
Here it is over a tetrahedron.

Stokes' theorem suggests:
[tex]\int_s {(\nabla\times \vec v).d\vec a = \oint_p\vec v.d\vec r [/tex]

For the right hand side I computed the line integral from (a,0,0)--->(0,2a,0)--->(0,0,a)--->(a,0,0);
which comes out to be [tex]a^2[/tex] (matches with the solution given).

For the surface integral, one needs to obtain the expression for the area element[tex]d\vec a[/tex] of the plane given by the above points, which is where my problem lies. Can someone help me on this?
 
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  • #2
You don't necessarily need the expression for the area element. Remember that you may choose the surface anyway you like as long as it has the triangle as its boundary.
Since the curl point only in the x-direction, choose your surface such that the integration will become trivial.
 
  • #3
Thank you so much for pointing it out to me. Here is how I figured it out!

[tex]\nabla\times\vec v = \hat x[/tex]
so,[tex]d\vec a = dydz\hat x[/tex]
Taking x=a, y-->0 to 2a-2z & z-->0 to a
[tex]\int(\nabla\times\vec v) .d\vec a = \int_{0}^{a}dz\int_{0}^{2a-2z}dy = a^2[/tex]
Since intregration over other surfaces are trivial.
Hence verifying Stokes' law.
 
  • #4
Right. Geometrically, the flux through the parts in the xz-plane and the xy-plane are both zero (curl only points in the x-direction) and the flux through the remaining surface is simple the area of that surface, which is a right triangle with height a and length 2a, thus area a^2.

Always look for ways to exploit geometry, since it will speed up your calculations. Calculating fluxes and line integrals generally involve a lot of work, so every bit helps.
 

1. What is Stokes' theorem over a tetrahedron?

Stokes' theorem over a tetrahedron is a mathematical theorem that relates the surface integral of a vector field over the boundary of a tetrahedron to the volume integral of the curl of the vector field inside the tetrahedron.

2. How is Stokes' theorem over a tetrahedron different from the standard Stokes' theorem?

The standard Stokes' theorem applies to surfaces in three-dimensional space, while Stokes' theorem over a tetrahedron specifically applies to tetrahedrons. It is a special case of the standard Stokes' theorem.

3. What is the significance of using a tetrahedron in Stokes' theorem over a tetrahedron?

A tetrahedron has four triangular faces, making it a simple and easily visualized shape. This allows for an intuitive understanding of the theorem and its applications. Additionally, using a tetrahedron allows for the theorem to be extended to three-dimensional space, whereas the standard Stokes' theorem only applies to two-dimensional surfaces.

4. What are some real-world applications of Stokes' theorem over a tetrahedron?

Stokes' theorem over a tetrahedron has various applications in physics and engineering, such as determining the circulation of a fluid flow or the flux of an electromagnetic field. It can also be used in the study of fluid dynamics, electromagnetism, and other areas of physics.

5. Are there any limitations to using Stokes' theorem over a tetrahedron?

Stokes' theorem over a tetrahedron only applies to vector fields that are continuous and have a well-defined curl. Additionally, the tetrahedron must be a closed surface, meaning that it cannot have any holes or openings. These limitations can make it difficult to apply the theorem in certain scenarios.

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