• Support PF! Buy your school textbooks, materials and every day products Here!

Problem with a simple Stoke's Theorem question

  • Thread starter Moham1287
  • Start date
  • #1
7
0

Homework Statement


Using Stoke's Theorem, evaluate the contour integral:
[tex]\oint F.dr[/tex]
as an integral over an appropriately chosen 2 dimensional surface.

Use F = [tex](e^{x}y+cos\siny,e^{x}+sinx\cosy,ycosz)[/tex] and take the contour C to be the boundary of the rectangle with the vertices (0,0,0), (A,0,0), (A,B,0), (0,B,0) oriented anticlockwise.

Then evaluate the same integral directly as a contour integral.

Homework Equations


Stoke's Theorem,

[tex]\int\int_{S}(curl F).n\;d^{2}A=\oint F.dr[/tex]
where n is unit normal vector.

The Attempt at a Solution



I got Curl F to be (cos z)i by the standard method using a matrix. I set n as -k as the surface is in the xy plane so the normal vector is along the z direction and I got the negative by the right hand rule.

This gives:

[tex]\int^{0}_{A}dx\int^{0}_{B}(-1)k.(cosz)i\dy[/tex]

(-k).(cos z)i is 0

which makes the double integral nothing

However when I solve the same thing directly as a contour integral I get an answer of -B.

Going anti clockwise from the origin, (so from (0,0,0) to (A,0,0) is I, (A,0,0) to (A,B,0) is II, (A,B,0) to (0,B,0) is III and (0,B,0) to (0,0,0) is IV) I get:
I=0
II=[tex]e^{A}B + sinA\;sinB[/tex]
III=[tex]-(e^{A}B+sinA\;sinB)[/tex]
IV=[tex]-B[/tex]

Which, when added together, gives -B.... I think there must be a glaring error somewhere. I can write up my calculations for the contour integral if that would help solve it... Any help with this would be much appreciated.
 

Answers and Replies

  • #2
gabbagabbahey
Homework Helper
Gold Member
5,002
6
This gives:

[tex]\int^{0}_{A}dx\int^{0}_{B}(-1)k.(cosz)i\dy[/tex]

(-k).(cos z)i is 0

which makes the double integral nothing
Technically, the normal to the surface should be +k as given by the right-hand rule when your path is anticlockwise. Luckily, it didn't affect your answer in this case as -1*0 is still zero..

However when I solve the same thing directly as a contour integral I get an answer of -B.

Going anti clockwise from the origin, (so from (0,0,0) to (A,0,0) is I, (A,0,0) to (A,B,0) is II, (A,B,0) to (0,B,0) is III and (0,B,0) to (0,0,0) is IV) I get:
I=0
II=[tex]e^{A}B + sinA\;sinB[/tex]
III=[tex]-(e^{A}B+sinA\;sinB)[/tex]
IV=[tex]-B[/tex]
Your error is in integral III; recheck that calculation and post your work for it if you can't find the error.
 
  • #3
7
0
I got it! Thanks a lot. Ha, I somehow messed up with the right hand rule, which is pretty basic, but that wasn't the problem. I just made a copying error and forgot about one of the [tex]e^{0}B[/tex] which should have become B, not nothing. Thanks a lot!
 

Related Threads for: Problem with a simple Stoke's Theorem question

  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
3
Views
5K
  • Last Post
Replies
7
Views
896
Replies
3
Views
2K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
3
Views
464
Top