Physics Help Forum How do I find the equation for the binding surface of a vector field?

 Mar 30th 2010, 09:00 AM #1 Junior Member   Join Date: Mar 2010 Posts: 10 How do I find the equation for the binding surface of a vector field? Hi, I am working on finding stokes' theorem for different three dimensional surfaces. As we have to perform the surface integral along the binding surface, we must first find what that binding surface is. For a unit half sphere, a cylinder and a cone, the bounding surface is the same: a unit circle in the x-y plane. Why is this? Why is that the binding surface and not the rest? Also, why is it that when converting to spherical polar coordinates, the limits of integration of the cita variable are 0 to pi/2 and not pi? I don't know if i expressed myself clearly enough, thanks a million!
Mar 30th 2010, 10:09 AM   #2
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The stokes theorem allows you to find the circulation of the field along a closed loop. It says that the circulation of the field along the loop equals the integral flux of the curl taken over *any* surface that is bound by that loop. That is, you can choose whatever surface you want to integrate the curl over, the result won't change, as long as the surface is delimited by the loop around which you want to find the circulation.

 Originally Posted by maple_tree Hi, For a unit half sphere, a cylinder and a cone, the bounding surface is the same: a unit circle
I think you mistakenly inverted the two things in the sentence. The circle would be the loop, and the half sphere, etc. would be the surface.

If you want to calculate the circulation of the field around the circle, the stokes theorem tells you that it equals the flux of the curl of the field over any surface that is bound by that circle. That means that you can calculate the flux of the rotor of E over the surface of the half sphere lying on that circle, of a cone, of a ctylinder, etc. as long as they have their basis on the circle you are interested in. The number you get will always be the same.

As for the coordinates question, I will leave it to someone else because I'm having coordinates-related problems myself and I guess I would end up confusing you too.

 Apr 2nd 2010, 03:53 PM #3 Senior Member   Join Date: Mar 2010 Location: Lithuania Posts: 105 You wrote "... converting to spherical polar coordinates". Can you write this expression, this integral.
Apr 2nd 2010, 04:26 PM   #4
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 Originally Posted by maple_tree Also, why is it that when converting to spherical polar coordinates, the limits of integration of the cita variable are 0 to pi/2 and not pi?
 Originally Posted by zzzoak You wrote "... converting to spherical polar coordinates". Can you write this expression, this integral.
I'm guessing that maple_tree is confused. In some exercises where one has to calculate a volume, it is often convenient to integrate over say one quadrant and then multiply the result by 4 (if the symmetry allows it). In one quadrant, theta (see this picture to see what I call theta: File:Coord system SZ 0.svg - Wikipedia, the free encyclopedia) goes from 0 to pi/2. If you had integrated over the 4 quadrants, theta would have went from 0 to pi. So it depends where you integrate.
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