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Evaluating a surface integral

  • Thread starter wifi
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  • #1
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Problem:

Use the fact that [tex]\int_S \vec{v} \cdot d\vec{S}=\int_S \vec{v} \cdot \frac{\nabla f}{\partial f/\partial x} dy\ dz[/tex]

to evaluate the integral for ##S=\{(x,y,z):y=x^2 ; 0 \geq x \geq 2; 0 \geq z \geq 3 \}## and ##\vec{v}=(3z^2, 6, 6xz)##.

Attempt at a Solution:

I'm having trouble setting up this integral. If I knew what ##f## was, I could easily calculate the gradient, as well as the partial wrt to x. I'd still need to figure out the limits of integration though.
 

Answers and Replies

  • #2
arildno
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f(x,y,z) ought to be the scalar function describing the surface S through the relationship:
f(x,y,z)=0
 
  • #3
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f(x,y,z) ought to be the scalar function describing the surface S through the relationship:
f(x,y,z)=0
Hmm. I'm not sure you'd get this from the info given. Any hints?
 
  • #4
arildno
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Because
1. It fits. Sort of (I admit I haven't seen closely, though).
2. Otherwise, it would be utterly meaningless, since you would have no way to evaluate the second expression due to lack of knowledge of f.
 
  • #5
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More specifically I meant, how would you find ##f## analytically from this info given?
 
  • #6
pasmith
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More specifically I meant, how would you find ##f## analytically from this info given?
From here:

##S=\{(x,y,z):y=x^2 ; 0 \geq x \geq 2; 0 \geq z \geq 3 \}##
If [itex]y = x^2[/itex] then [itex]y - x^2 = 0[/itex].

(Your inequalities are the wrong way round: you want [itex]0 \leq x \leq 2[/itex] etc.)
 
  • #7
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From here:



If [itex]y = x^2[/itex] then [itex]y - x^2 = 0[/itex].

(Your inequalities are the wrong way round: you want [itex]0 \leq x \leq 2[/itex] etc.)
So ##f(x,y,z)=y-x^2##?
 
  • #8
LCKurtz
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I would parameterize the surface as ##\vec R(x,z)## and use the formula$$
\iint \vec v\cdot d\vec S =\pm \iint_{x,z}\vec v \cdot \vec R_x\times \vec R_z~dxdz$$where the choice of signs depends on the orientation of the surface which, by the way, you need to specify.
 

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