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May 30, 2016
Posted by **Antonio** on Wednesday, August 1, 2012 at 8:18am.

- Calculus -
**MathMate**, Wednesday, August 1, 2012 at 9:29amFlux = 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.