Expert: Alon Mandes - 4/11/2009

Question
Applying Stokes' Theorem, evaluate the integral

∫((y^2-z^2 )dx+(z^2-x^2 )dy+(x^2-y^2)dz)

Here, C is the section of the surface of the cube 0≤x,y,z≤a by the plane x+y+z=3a/2.

PLease help.

Answer
ok, 1st of all , lets write the technique to Stokes' Theorem :
∫∫ROt(F)ds = ∫Fdr
s            c
So, 1st step will be to calculate the rotor of the field function F:
In our case F=(y²-z²)[i] + (z²-x²)[j] + (x²-y²)[k] .
Thus ROt(F)=(-2y)[i] + (-2x-2z)[j] + (-2x-2y)[k]. I'll leave it to
you as an exercise to verify it !
Now, next step will be to determine our curve & the region we are
willing to integrate on :
We have been given the plane x+y+z=3a/2 -> z=(3a/2)-x-y. This plane
will intersect with the cube (x,y,z)Є[a,a,a] & the region will be :
when z=0 -> (3a/2)-x-y=0 -> y=(3a/2)-x, hence :
Boundaries of x : [0:a]
Boundaries of y : [0,(3a/2)-x]
Now all we need to do is to calculate the double integral :
∫∫[-2y,-2x-2z,-2x-2y]•[1,1,1] dxdy =
-2∫∫[y,x-z,x-y]•[1,1,1] dxdy= -2∫∫(y+x-z+x-y)dxdy=-2∫∫(2x-z)dxdy=
-2∫∫(2x-(3a/2)+x+y)dxdy=-2∫∫(3x+y)-(3a/2) dxdy =

  a  (3a/2)-x
-2∫dx∫3x+y-(3a/2) dy
0   0

& I will leave this integral to you to proceed..

Alon.