Posted by Antonio on Wednesday, August 1, 2012 at 8:18am.
Flux = I = ∫∫F.n dS
for S over the surface, g(x,y,z), where
g(x,y,z) is the part of
x=2y²+2z²+3 where x≤5, which translates to
g(x,y,z)=x-2y²-2z²-3
and the region is the part of the paraboloid where y²+z²≤1 after substitution of x≤5.
This region is circular with radius =1, which facilitates integration later.
Now:
F = <0,y,z>
n=∇g/||g||
so I can be evaluated:
I = ∫∫F.n dS
The surface integral can be evaluated by projecting it to the region R on the y-z plane by multiplying:
dS over S = ||g|| dA over R
So
I = ∫∫F.n dS over S
= ∫∫F.n ||g|| dA over R
where R is the circular region r≤1 in polar coordinates.
The way I have defined g(x,y,z) results in negative flux.
Post if you could use more help.