Verifying Stokes' Theorem on F, S in First Octant

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SUMMARY

This discussion focuses on verifying Stokes' Theorem for the vector field F = (y²)i + (z²)j + (x²)k over the surface S defined by the first octant portion of the plane x + y + z = 1. The user attempts to compute the surface integral and finds a discrepancy, obtaining 1/4 instead of the expected result of -1. The error is identified as neglecting to compute the curl of F before performing the surface integral, which is essential for applying Stokes' Theorem correctly.

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Homework Statement


Verify Stokes' Theorem for F and S
F=(y^2)i+(z^2)j+(x^2)k
S is the first octant portion of x+y+z=1

Homework Equations





The Attempt at a Solution


I know that it should be equal to -1 from Stoke's theorem, but I keep getting 1/4 when I use the normal surface integral way. (I have to do it both ways)

The integral I am using is
[tex]\iint x^2 + y^2 + (1-x-y)^2[/tex]

What am I doing wrong?? :(
 
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I don't think you took the curl of F before you dotted with dS.
 
“The line integral of [tex]\v{F}[/tex] around any closed contour C equals the surface integral (flux) of [tex]curl \v{F}[/tex] over any surface bounded by C

[tex]\oint_{C}\v{F}\bullet\v{dl}=\int_{S}\left(Curl\v{F}\right)\bullet\v{dS}[/tex]

Found these in some of my old undergrad notes...
 

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