Vector Calculus

  • Thread starter Hoofbeat
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  • #1
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Could someone help me with the following question:

====
Q. A = (y,-x,0). Find Integral A.dl for a closed loop on the surface of the cylinder (x-3)^2 + y^2 = 2.
====

I'm managed to ascertain the following:
*I can use Stokes' theorem to solve the problem
*The curl (A) = (0,0,-2)
*Radius = sqrt2
*The normal to the cylinder, n, is equal to k (as cylinder lies in xy plane).

However, I have no clue how to do the actual surface integral! I'm fairly sure I need to transform into cylindrical coordinates, but I honestly have no idea how to use the actual 'cylinder' in the problem (is it to provide the limits?!) not how I form the integral? Please could someone offer some help. Thanks :cool:
 

Answers and Replies

  • #2
ehild
Homework Helper
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Hoofbeat said:
Could someone help me with the following question:

====
Q. A = (y,-x,0). Find Integral A.dl for a closed loop on the surface of the cylinder (x-3)^2 + y^2 = 2.
====

I'm managed to ascertain the following:
*I can use Stokes' theorem to solve the problem
*The curl (A) = (0,0,-2)
*Radius = sqrt2
*The normal to the cylinder, n, is equal to k (as cylinder lies in xy plane).

However, I have no clue how to do the actual surface integral! :cool:

It is all right up to now. The surface, for you have to integrate curl A, lies in the xy plane, its boundary is a circle with radius sqrt(2), and you are very lucky as curlA is constant on this surface and is perpendicular to it. You know that to get the surface integral of a vector-vector function A(r) you have to multiply the normal component of A with the surface element and integrate - you have to integrate a constant function now....

ehild
 
  • #3
48
0
Thanks. So now do I write ds in terms of cylindrical coordinates then integrate cur(lA).n ds? ie.

curl(A).n = (0,0,-2)

and ds = r.d(theta).dz
thus integral is -2 dz?
 
Last edited:
  • #4
ehild
Homework Helper
15,543
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Hoofbeat said:
Thanks. So now do I write ds in terms of cylindrical coordinates then integrate cur(lA).n ds? ie.

curl(A).n = (0,0,-2)

and ds = r.d(theta).dz
thus integral is -2 dz?

No, you have to integral the curl for the area the line encloses. If dS is the surface element and n is its normal unit vector, you have to calculate

[tex] \int { curl \vec{A}\vec n dS} [/tex].

The curl is constant, so you can write it in front of the integral, and it is multiplied by the projection of the area that is normal to the curl. If the line encloses the axis of the cylinder this area is r^2pi, and it is zero otherwise.


ehild
 
Last edited:

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