Author 
Topic: Proving Stokes' Theorem (Read 4300 times) 

Marissa
Newbie
Posts: 16


Proving Stokes' Theorem
« on: Apr 28^{th}, 2008, 10:58am » 
Quote Modify

How to prove stokes theorem (the integral of a vector over a closed trajectory = the integral of the rotational of the vector over a surface) for a particular problem where v = rXz where r is the vector of position (r*ur) and z a unit vector in the z direction (1*uz). The vector v is supposed to equal r*sin(theta)* u_phi where u_phi is the unit vector in the phi direction. I understand that the cross product is equal to the magnitudes of the two components times the sine of the angle between them but I don´t understand why the cross product of the unit vector in the radial direction (which is supposed to be in spherical coordinates) and the unit vector in the zdirection are equal to the negative unit vector in the phidirection. I was told that the integral of the closed trajectory is equal to v*dl and that dl is equal to r*d(theta)*u_phi. I don´t understand this either. I think I get that the like is equal to the radius integrated around its closed trajectory but where does the unit vector come from? Another question on the same problem is, what are the bounds I am supposed to integrate about for the closed trajectory? any thoughts, please. // modified problem title to be more descriptive  wwu

« Last Edit: May 28^{th}, 2008, 7:30pm by william wu » 
IP Logged 



Marissa
Newbie
Posts: 16


Re: help with this problem
« Reply #1 on: Apr 29^{th}, 2008, 4:03am » 
Quote Modify

The question is stated as follows: Verify Stokes theorem by calculating both members of [img:111:27]http://www.postimage.org/aVITqX9.jpg[/img] where S is a semisphere with r = c, z >0, and v = rXz (v,r,z are vectors) The supposed solution is 2*pi*c^{2} The definition of Stokes theorem I have been given is (and I apologize if it doesn't sound quite right): The flux of the rotational of a field vector through a surface is equal to the circulation of the vectoral field along the line bounding the surface. Any help would be greatly appreciated. Even just a push in the right direction


IP Logged 



Eigenray
wu::riddles Moderator Uberpuzzler
Gender:
Posts: 1948


Re: help with this problem
« Reply #2 on: May 5^{th}, 2008, 6:37pm » 
Quote Modify

Having just finished teaching a multivariable calc course, I will use the notation I am used to: _{C} Fdr = _{S} (x F)nd. Here F(r) = r x k, S is the upper hemisphere, and C is its boundary, a circle in the xyplane. Also, dr = Tds, where T is a unit tangent, and s is the arclength parameter, so that ds = dr, and n is a unit vector normal to the surface. In Cartesian coordinates, F = (y, x, 0), xF = (0, 0, 2). In terms of spherical coordinates, F = sinu_{}, where u_{} = (sin, cos, 0) is the unit vector in the direction. I am using the coordinates (x,y,z) = (sincos, sinsin, cos), so I guess we have different notation for which of , is which. So the problem becomes _{C} ydx  xdy = 2c^{2} = _{S} 2knd. For the first integral, you just need to parameterize the curve C: x = c cos, y = c sin, 02.


IP Logged 



Eigenray
wu::riddles Moderator Uberpuzzler
Gender:
Posts: 1948


Re: help with this problem
« Reply #3 on: May 5^{th}, 2008, 6:37pm » 
Quote Modify

To compute the second integral, there are a couple options. Here are two: 1) Parameterize the surface. Using spherical coordinates, r(,) = (c sincos, c sinsin, c cos), with 0, 02. Then nd= (r_{} x r_{}), and 2knd= 2c^{2} sincos, assuming we take n to be the outward (hence upwards) pointing normal. We then integrate this over the parameterization, 0, 02. 2) Think of the surface as the level surface g = 0, and use projection onto the xyplane. Then n d= g/ g k dxdy. In this case, 2knd= 2dxdy, and we integrate over the projection onto the xyplane, which is a circle of radius c.


IP Logged 



