Surface Integral With Divergence Thm

  • #1
133
1

Homework Statement


Let ##\mathit{F}(x,y,z) = (e^y\cos z, \sqrt{x^3 + 1}\sin z, x^2 + y^2 + 3)## and let ##S## be the graph of ##z = (1-x^2-y^2)e^{(1-x^2-3y^2)}## for ##z \ge 0##, oriented by the upward unit normal. Evaluate ##\int_{S} \mathit{F} \ dS##. (Hint: Close up this surface and use the Divergence Theorem)


Homework Equations





The Attempt at a Solution


It's clear that ##div \ F = 0##, so if I was working with a closed surface, ##\int_{S} \mathit{F} \ dS## would equal 0. For the graph of ##S##, I am unsure as to what "closing up" a surface means. How would one close up a surface?
 

Answers and Replies

  • #2
30
1
Add the surface z=0 to creates closed surface. Then, the divergence theorem gives the surface integral of the sum of the surfaces; subtract the integral over what you added to get the surface integral of the desired surface.
 
  • #3
133
1
Ok, so what I think you're saying is I have to do:

##0 - \int_{plane \ z = 0} F \ dS##

which will get me only the surface integral of the graph of z?
 
  • #4
Dick
Science Advisor
Homework Helper
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Ok, so what I think you're saying is I have to do:

##0 - \int_{plane \ z = 0} F \ dS##

which will get me only the surface integral of the graph of z?

Yes. And you'll need to pick the correct direction for the normal vector for the surface z=0.
 
  • #5
133
1
For it to be outward pointing, would the z-component have to be negative, ie, (0,0,-1)?
 
  • #6
Dick
Science Advisor
Homework Helper
26,263
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For it to be outward pointing, would the z-component have to be negative, ie, (0,0,-1)?

Exactly.
 

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