# Gauss Divergence Theorem

• May 10th 2010, 11:51 PM
HeheZz
Gauss Divergence Theorem
Verify Gauss Divergence Theorem ∭∇.F dxdydz=∬F. NdA
Where the closed surface S is the sphere x^2+y^2+z^2=9 and the vector field F = xz^2i+x^2yj+y^2zk

I have tried to solve the left hand side which appear to be (972*pi)/5
However, I cant seems to solve the right hand side to get the same answer.
I substitute x = 3sin(theta)cos(phi), y=3sin(theta)sin(phi), z=3cos(theta)
Therefore N=9sin^2(theta)cos(phi)i+9sin^2(theta)cos(phi)j+9cos(theta)sin(theta)k
and
F=27sin(θ)cos^2(θ)cos(φ)i+27sin^3(θ)cos^2(φ)sin(φ)j+27sin^2(θ)cos(θ)sin^2(φ)k
then I used ∫(0-2pi)∫(0-pi) F. N dθdφ
I got the final answer as (324*pi)/5 which does not match with left hand side.
Hope anyone can help here plz. Thanks!
• May 11th 2010, 06:29 AM
Jester
Quote:

Originally Posted by HeheZz
Verify Gauss Divergence Theorem ∭∇.F dxdydz=∬F. NdA
Where the closed surface S is the sphere x^2+y^2+z^2=9 and the vector field F = xz^2i+x^2yj+y^2zk

I have tried to solve the left hand side which appear to be (972*pi)/5
However, I cant seems to solve the right hand side to get the same answer.
I substitute x = 3sin(theta)cos(phi), y=3sin(theta)sin(phi), z=3cos(theta)
Therefore N=9sin^2(theta)cos(phi)i+9sin^2(theta)cos(phi)j+9cos(theta)sin(theta)k
and
F=27sin(θ)cos^2(θ)cos(φ)i+27sin^3(θ)cos^2(φ)sin(φ)j+27sin^2(θ)cos(θ)sin^2(φ)k
then I used ∫(0-2pi)∫(0-pi) F. N dθdφ
I got the final answer as (324*pi)/5 which does not match with left hand side.
Hope anyone can help here plz. Thanks!

Check the second component of your normal. I got $\displaystyle {\bf N}_2 = 9\sin^2(\theta)\sin(\phi)$.