Verify Stokes' Theorem for this vector field on a surface

In summary, the parametrization for the ellipse described in the problem should be ##F=\{(x,y,z)\in \mathbb{R}^3: (\frac{x}{3})^2+(\frac{y}{2})^2\leq 1, z=0\}##.
  • #1
Elder1994
6
1
Homework Statement
Hello, my H.W states: Verify the Stokes Theorem for the vector field A=[(4x/3 - 2y), (3y - x), 0] and the surface F:= {(x/3)^2 + (y/2)^2 =< 1, z = 0}
Relevant Equations
∫∫∇×A⋅dS=∫A⋅dr
I do not understand how can I parameterize the surface and area and line differentials.
 
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  • #2
You must post some try or thought for us to help you. In any case, I would say that parametrize a surface or a line is to assign one or more parameters to each point on the surface or the line, for example, imagine the parabola defined in the ##x-y## plane by the equation
$$y=\frac{x(1-x)}{2}$$ you can parametrize it like $$y=-1+3t-2t^2, x=2-2t, \qquad \text{ with } -\infty < t <\infty$$
Notice that ##t## is a parameter, and that, for any choice of ##t## you get a point ##(x,y)## that lies in the desired parabola. In general, to parametrize a curve you will need one parameter and to parametrize a surface you will need two parameters.

Post some tries and we will be able to help you more.
 
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  • #3
What did you get for ##\nabla \times \vec A##? Think about parameterizing the surface with ##x = 3\cos\theta,~y=2\sin\theta## and similarly for the boundary curve.
 
  • #4
LCKurtz said:
Think about parameterizing the surface with ##x = 3\cos\theta,~y=2\sin\theta## and similarly for the boundary curve.

Would this not be the parameterization for the boundary and not the surface?
 
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  • #5
Hello, I've read all your comments, thanks for answering! I've tried to solve the problem and for the line integral, I used de parametrization: ## x = 3\cos\theta, ~y=2\sin\theta ##. I also defined my ##d\vec r## as ##(-\sin\theta , \cos\theta , 0) d\theta##, after this the line integral (from 0 to ##2\pi##) gave me ## \pi## as result. Then for the surface integral the ## \nabla \times \vec A ## , I obtained ## (0,0,1)## and the normal vector ##(\frac {\partial r(u,v)} {\partial u} \times \frac {\partial r(u,v} {\partial v})## ( where ##r(u,v) = (u,v,0)##) is ##(0,0,1)## as well. With all this, and using a change of variables my surface integral is: ## \int_{0}^{2\pi} \int_{0}^1 r \, dr \, d\theta##
This way, both results are ##\pi##, but I'm not sure if I chose the right limits and the right parametization.
 
  • #6
Elder1994 said:
I also defined my ##d\vec r## as ##(-\sin\theta , \cos\theta , 0) d\theta##, after this the line integral (from 0 to ##2\pi##) gave me ##\pi## as result.

If you want the differential of the parametrization ##\vec r(\theta)## you have defined, you will have to use it:

##\vec r(\theta)=(x(\theta),y(\theta),z(\theta))##

You know what ##x(\theta)##, ##y(\theta)##, and ##z(\theta)## are because they have been given to you explicitly. If you wish to find the differential ##d\vec r##, you will have to differentiate each coordinate of the vector function with respect to your independent variable ##\theta##:

##\frac{d\vec r}{d\theta}=(\frac{d}{d\theta}x(\theta),\frac{d}{d\theta}y(\theta),\frac{d}{d\theta}z(\theta))##

This is your ##d\vec r##; just "append" ##d\theta## to the right sides of the entire expression and "cancel" them out when necessary. Someone can explain this better than me, I am sure.

Elder1994 said:
##r(u,v) = (u,v,0)##

This is not the parametrization of the ellipse described in the problem:

##F=\{(x,y,z)\in \mathbb{R}^3: (\frac{x}{3})^2+(\frac{y}{2})^2\leq 1, z=0\}##

Try to formulate one using polar coordinates. It should be very similar to the parametrization for the boundary given earlier, as already stated. Try drawing a picture of the thing, because it will help. You should not need to draw the whole 3-D space, by the way.
 
Last edited:
  • #7
Eclair_de_XII said:
Would this not be the parameterization for the boundary and not the surface?
Yes. I inadvertently left out the ##r## going from ##0## to ##1##. And for the OP's benefit, I think he should get ##6\pi## for the answer to both.
 
  • #8
Hello, I reviewed the problem and I found indeed that the limits that I used were wrong, changing that the answer for both integrals is ## 6\pi ## ! Thank you.
 

Related to Verify Stokes' Theorem for this vector field on a surface

1. What is Stokes' Theorem and how is it related to vector fields on surfaces?

Stokes' Theorem is a fundamental theorem in vector calculus that relates the integral of a vector field over a surface to the line integral of the vector field along the boundary of the surface. It essentially states that the surface integral of a vector field is equal to the line integral of the same vector field along the boundary of the surface, taking into account the direction of the boundary.

2. How is Stokes' Theorem used in scientific research?

Stokes' Theorem is used in various fields of science, such as fluid dynamics, electromagnetism, and thermodynamics, to calculate flux and circulation integrals over surfaces. It allows scientists to relate the behavior of a vector field on a surface to the behavior of the same vector field along the boundary of the surface.

3. What is the process for verifying Stokes' Theorem for a given vector field on a surface?

To verify Stokes' Theorem for a given vector field on a surface, one must first calculate the line integral of the vector field along the boundary of the surface. Then, the surface integral of the same vector field must be calculated over the surface. If the two values are equal, then Stokes' Theorem is verified for that vector field on that surface.

4. Are there any limitations to Stokes' Theorem?

Stokes' Theorem is a powerful tool for calculating surface integrals, but it does have limitations. It can only be applied to smooth surfaces and vector fields that are continuously differentiable. Additionally, the surface and boundary must be orientable, meaning that a consistent direction can be assigned to the surface and boundary.

5. What are some real-world applications of Stokes' Theorem?

Stokes' Theorem has various applications in science and engineering, such as calculating the flow of fluids around a surface, determining the electric field around a conducting wire, and analyzing the circulation of air in a weather system. It is also used in computer graphics and animation to create realistic simulations of fluid and smoke movement over surfaces.

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