Expert: Alon Mandes - 4/24/2010

Question
QUESTION: regarding the answer you posted for stokes' theorem.
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.

I cant get the correct answer. And there are minor errors in your curl F. should be ROt(F)=(-2y-2z)[i] + (-2x-2z)[j] + (-2x-2y)[k].

and here ∫∫[-2y-2z,-2x-2z,-2x-2y]•[1,1,1] dxdy =
-2∫∫[y+z,x+z,x+y]•[1,1,1] dxdy.

But I still cant get the correct answer after correcting the errors.

ANSWER: OK , -2∫∫[y+z,x+z,x+y]•[1,1,1] dxdy = -2∫∫y+z+x+z+x+z dxdy = -4∫∫(x+y+z) dxdy .
Note that our surface is x+y+z=3a/2 , Thus , our integral become :
-6∫∫dxdy . All you need to calculate now is the boundaries for x & y .

Alon.


---------- FOLLOW-UP ----------

QUESTION: Am I correct to say the boundary is the same ?
y=[0,3a/2-x]
x=[0,a]and apply it to -6∫∫dxdy ?

How do you get -6∫∫dxdy ?

My answer is -6a^3. But it is wrong.

Answer
Sorry, my mistake . -4∫∫(x+y+z) dxdy = -4∫∫ (3a/2)dxdy = -6a∫∫dxdy . The boundaries
are correct . This will lead to the answer : -6a^3 . The sign of - can be nuneutralized
by taking the normal vector facing downward , say [-1,-1,-1] .

Alon.