# Compute circulation of vector around the contour

• princessme
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
princessme

## 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 $\oint$F.dr = $\int$∇ X F . n dS

## The Attempt at a Solution

For (a) which is solving it directly, i used the left side of the equation $\oint$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?

No, it isn't. Don't you know how $nabla\times \vec{F}$ is defined? If $\vec{F}= f(x,y,z)\vec{i}+ g(x,y,z)\vec{j}+ h(x,y,z)\vec{k}$ then
$$\nabla\times \vec{F}= \left(\frac{\partial h}{\partial y}- \frac{\partial g}{\partial z}\right)\vec{i}-\left(\frac{\partial h}{\partial x}- \frac{\partial f}{\partial z}\right)\vec{j}+ \left(\frac{\partial g}{\partial x}- \frac{\partial f}{\partial y}\right)\vec{k}$$

HallsofIvy said:
No, it isn't. Don't you know how $nabla\times \vec{F}$ is defined? If $\vec{F}= f(x,y,z)\vec{i}+ g(x,y,z)\vec{j}+ h(x,y,z)\vec{k}$ then
$$\nabla\times \vec{F}= \left(\frac{\partial h}{\partial y}- \frac{\partial g}{\partial z}\right)\vec{i}-\left(\frac{\partial h}{\partial x}- \frac{\partial f}{\partial z}\right)\vec{j}+ \left(\frac{\partial g}{\partial x}- \frac{\partial f}{\partial y}\right)\vec{k}$$

Yes I know that. But to obtain a unit normal vector n, i need to first find $\frac{∇f}{|∇f|}$ , 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?

How did you get -4pi? Also, did you manage to finish part b?

princessme said:
Yes I know that. But to obtain a unit normal vector n, i need to first find $\frac{∇f}{|∇f|}$ , 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##.

How do you know it's "k" and not negative "k"

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.