Problem with a simple Stoke's Theorem question

[Moham1287]

Homework Statement

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

Use F = (e^{x}y+cos\siny,e^{x}+sinx\cosy,ycosz) 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,

\int\int_{S}(curl F).n\;d^{2}A=\oint F.dr
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:

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

(-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=e^{A}B + sinA\;sinB
III=-(e^{A}B+sinA\;sinB)
IV=-B

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.

 

[gabbagabbahey]

This gives:

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

(-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=e^{A}B + sinA\;sinB
III=-(e^{A}B+sinA\;sinB)
IV=-B

Your error is in integral III; recheck that calculation and post your work for it if you can't find the error.

 

[Moham1287]

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 e^{0}B which should have become B, not nothing. Thanks a lot!