Compute circulation of vector around the contour

In summary, the conversation discusses how to compute the circulation of a vector around a contour using the direct method and the Stokes Theorem. The equations used are \ointF.dr = \int∇ X F . n dS and \nabla\times \vec{F}= \left(\frac{\partial h}{\partial y}- \frac{\partial g}{\partial z}\right)\vec{i}-\left(\frac{\partial h}{\partial x}- \frac{\partial f}{\partial z}\right)\vec{j}+ \left(\frac{\partial g}{\partial x}- \frac{\partial f}{\partial y}\right)\vec{k}. The conversation also discusses how to find the unit
  • #1
princessme
35
0

Homework Statement



Compute the circulation of the vector a = yi+x2j - zk around the contour L: {x2 +y2 = 4; z = 3}, a) directly and b) via the Stokes Theorem.
Plot the contour and show its orientation.

Homework Equations



Stokes theorem is [itex]\oint[/itex]F.dr = [itex]\int[/itex]∇ X F . n dS

The Attempt at a Solution



For (a) which is solving it directly, i used the left side of the equation [itex]\oint[/itex]F.dr and I obtained 0 as my answer. Having problem with part b. What should my ∇f be? Is it 2x i + 2y j + 0 k?
 
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  • #2
No, it isn't. Don't you know how [itex]nabla\times \vec{F}[/itex] is defined? If [itex]\vec{F}= f(x,y,z)\vec{i}+ g(x,y,z)\vec{j}+ h(x,y,z)\vec{k}[/itex] then
[tex]\nabla\times \vec{F}= \left(\frac{\partial h}{\partial y}- \frac{\partial g}{\partial z}\right)\vec{i}-\left(\frac{\partial h}{\partial x}- \frac{\partial f}{\partial z}\right)\vec{j}+ \left(\frac{\partial g}{\partial x}- \frac{\partial f}{\partial y}\right)\vec{k}[/tex]
 
  • #3
HallsofIvy said:
No, it isn't. Don't you know how [itex]nabla\times \vec{F}[/itex] is defined? If [itex]\vec{F}= f(x,y,z)\vec{i}+ g(x,y,z)\vec{j}+ h(x,y,z)\vec{k}[/itex] then
[tex]\nabla\times \vec{F}= \left(\frac{\partial h}{\partial y}- \frac{\partial g}{\partial z}\right)\vec{i}-\left(\frac{\partial h}{\partial x}- \frac{\partial f}{\partial z}\right)\vec{j}+ \left(\frac{\partial g}{\partial x}- \frac{\partial f}{\partial y}\right)\vec{k}[/tex]

Yes I know that. But to obtain a unit normal vector n, i need to first find [itex]\frac{∇f}{|∇f|}[/itex] , isn't it?

Anyway, I managed to find a mistake I made for the left hand side, and obtained my answer as -4pi. Mind guiding me through for the right hand side?
 
  • #4
How did you get -4pi? Also, did you manage to finish part b?
 
  • #5
princessme said:
Yes I know that. But to obtain a unit normal vector n, i need to first find [itex]\frac{∇f}{|∇f|}[/itex] , isn't it?

No.

The unit normal vector ##\mathbf n## is normal to the surface inside the contour.
The obvious choice for that surface is in the plane z=3, bounded by a circle.
The unit normal vector to that plane is ##\mathbf n = \mathbf k##.

So first you have to determine ##\nabla \times \mathbf a##, and then you have to take the dot product with ##\mathbf n = \mathbf k##.
 
  • #6
How do you know it's "k" and not negative "k"
 
  • #7
dan38 said:
How do you know it's "k" and not negative "k"

You can choose the direction.
The consequence is that for the contour integral you have to follow the contour according to the right hand rule.
 

Related to Compute circulation of vector around the contour

1. What is the definition of circulation of a vector around a contour?

The circulation of a vector around a contour is a measure of the flow of the vector field around a closed path. It takes into account both the magnitude and direction of the vector at each point along the path.

2. How is the circulation of a vector calculated?

The circulation of a vector is calculated by taking the line integral of the vector field around the contour. This involves breaking the contour into small segments and evaluating the vector field at each point along the contour.

3. What does a positive circulation value indicate?

A positive circulation value indicates that the vector field is flowing in a clockwise direction around the contour. This means that the vector is pointing in the same direction as the contour's tangent vector at each point along the path.

4. Can the circulation of a vector be negative?

Yes, the circulation of a vector can be negative. This indicates that the vector field is flowing in a counterclockwise direction around the contour. This means that the vector is pointing in the opposite direction of the contour's tangent vector at each point along the path.

5. What is the physical significance of computing circulation of a vector around a contour?

The circulation of a vector is an important concept in fluid dynamics and electromagnetism. It allows scientists to understand the flow of fluids and electromagnetic fields around closed paths, which is crucial in many real-world applications such as designing aircraft wings, analyzing weather patterns, and predicting the behavior of electric currents.

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