- #1

Moham1287

- 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.