- #1
Moham1287
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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.