Divergence Theorem Stuff Watch
- 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
= 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)
(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
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
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
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
Adding the six integrals gives -(1/4)a^5
To the last poster, this is 1st year Imperial Physics stuff.