When I am verifying divergence theorem for $\vec{F}=(x+y^2)i-(2x)j+(2yz)k$ , surrounded by the co-ordinate planes and the plane $2x+y+2z=6$ , I am getting some errors.
While calculating the using the double integral, i.e., surface integral, I am getting $81$ as the answer.
-> In this method I am using the projection of the plane on the XY plane, and further solving.
But, when I am using the divergence theorem, and calculating the volume integral, I am getting $36$ as the answer.
I also tried various triple integral calculators online, and they were giving the same answer.
-> In this method I put the limits as :-
$Del(\vec{F})=2y+1$
z - $0\rightarrow (6-2x-y)/2 $
y - $0 \rightarrow (6-2x)$
x - $0 \rightarrow 3$
Maybe I am putting the limits wrong in the second case, or doing some procedural mistake.
The surface is tilted in the positive co-ordinate planes, or the positive octant.
When we solved using the surface integral (by the above solved projection method), we only solved for the given tilted plane, i.e., base of the tetrahedron, and the answer was $81$.
But the divergence theorem works for closed volumes, and gives the flux through the entire planes of the tetrahedron, i.e., $36$.
Thus we need to find the flux through 3 additional planes, i.e.,
$XY$ plane (comes out to be $0$)
$YZ$ plane (comes out to be $-54$)
$ZX$ plane (comes out to be $+9$)
And finally summing up all the values :- $81+0-54+9=36$ , i.e., the same answer as given by the divergence theorem.