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    V = 2x^2yi - 2xy^2j - xyzk
    - Determine the divergence of this vector field and evaluate the integral of this quantity over the interior of the cube of side a in z >= 0, whose base has vertices (0, 0, 0), (a, 0, 0), (0, a, 0) and (a, a, 0).
    - Evaluate the flux of V over the surface of the cube and thereby verify the divergence theorem.

    Div(V) = 6xy - 4xy - xy = xy
    :. Triple Integral Div(V) dtau
    = Triple Integral xy dx dy dz
    = Integral (0 to a) x dx . Integral (0 to a) y dy . Integral (0 to a) 1 dz
    = (x^2/2)|(x = a) . (y^2/2)|(x = a) . (z)|(z = a)
    = a^2/2 . a^2/2 . a
    = a^5/4
    = Total flux across interior of cube

    I'm not sure how to do the last part though, how do I find n, a unit vector normal to the surface of the cube?

    Also, when evaluating Double Integral (V.n) dsigma (Flux of V over cube's surface), do I replace dsigma with dx dy?

    Help appreciated - I tried a couple of things but didn't get a^5/4, which of course I require to verify the theorem! (That: Triple Integral Div(V) dtau = Double Integral (V.n) dsigma)
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    div(v) = 4xy - 4xy - xy = -xy

    (int over cube) div(v) dV
    = a (int from 0 to a) (int from 0 to a) (-xy) dx dy
    = -a [(int from 0 to a) x dx] [(int from 0 to a) y dy]
    = -a (a^2/2)^2
    = -(1/4)a^5

    You miscalculated div(v).

    --

    If you just want a hint for the second part, consider each of the six faces separately. Read on for a solution.

    Face 1 (bottom, unit normal = -k)
    (int over Face 1) v.n dS = 0

    Face 2 (top, unit normal = k)
    (int over Face 2) v.n dS
    = (int over Face 2) -xya dS
    = -a (a^2/2)^2
    = -(1/4)a^5

    Face 3 (left, unit normal = -i)
    (int over Face 3) v.n dS = 0

    Face 4 (right, unit normal = i)
    (int over Face 4) v.n dS
    = (int over Face 4) 2a^2 y dS
    = 2a^2(a^2/2)*a
    = a^5

    Face 5 (front, unit normal = -j)
    (int over Face 5) v.n dS = 0

    Face 6 (back, unit normal = j)
    (int over Face 6) v.n dS
    = (int over Face 6) -2x a^2 dS
    = -a^5

    Adding the six integrals gives -(1/4)a^5
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    What level is this..?
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    Ahhh great, thanks Johnny I understand now completely, makes total sense.
    To the last poster, this is 1st year Imperial Physics stuff.
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    That makes me feel better
 
 
 
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