-edit, i really need to know which part i got wrong as i'm having trouble proceeding, please help me, thanks-
Applying Stokes' Theorem, evaluate the integral
Curve C is the intersection of the boundary surface of the cube $0≤x,y,z≤a$, with the plane $x+y+z=3a/2$ where a = 1.17. The contour C is oriented counterclockwise if seen from the positive direction of the Ox-axes.
the problem http://dl.dropbox.com/u/5681270/Screen%20Shot%202011-11-16%20at%2010.32.40%20PM.png
my workings:
Let S be the hexagon that lies in the cube 0<=x,y,z <=a and on the plane x + y + z = 3a/2
then we can parameterize it using x,y obtaining z = 3a/2 - x - y.
normal vector to plane = | {i j k}, {1 0 -1}, {0 1 -1} | = i + j + k
curl F = |{i j k}, {dx dy dz}, {$y^{2}-z^{2}$ $z^{2}-x^{2}$ $x^{2}-y^{2} $}| = ($-2y-2z$)i + ($-2z-2x$)j + ($-2x-2y$)k
using stroke's theorem
the integral = $\int \int_{S}$ curl F.N dS
= $\int \int_{0<=x,y<=a} -4y - 4x - 4(\frac{3a}{2} - x - y) dx dy$
=$\int \int_{0<=x,y<=a} 6a$ dx dy = 6a $\int \int_{0<=x,y<=a}1 dx dy$ = 6a * area = 6a^3.
which part of my workings is wrong? thanks for looking through