Evaluating a Vector Field Through a Surface with the Divergence Theorem

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Discussion Overview

The discussion revolves around evaluating the flux of the vector field \(\vec{F} = \langle x, y, z \rangle\) through a surface defined by the equation \(z = 4 - x^2 - y^2\) above the xy-plane. Participants explore methods to compute this using surface integrals and the divergence theorem, addressing both theoretical and practical aspects of the problem.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant seeks step-by-step guidance on evaluating the flux and expresses confusion over obtaining the correct answer.
  • Another participant confirms the vector field and surface equation, suggesting the use of the divergence theorem and providing the formula for the unit normal vector.
  • A third participant states that the divergence of the vector field is 3 and proposes using cylindrical coordinates to compute the volume integral for flux.
  • Another participant offers an alternative approach by recognizing the surface as a level surface and suggests using the gradient to find the normal vector without separately calculating it.
  • One participant reflects on their previous mistakes regarding the limits of integration for the surface.

Areas of Agreement / Disagreement

Participants present multiple approaches to the problem, and while there is some agreement on the use of the divergence theorem, there is no consensus on the specific methods or limits of integration to be used. The discussion remains unresolved with various perspectives on how to proceed.

Contextual Notes

Some participants mention the need for careful consideration of limits and normalization of components when integrating over different surfaces, indicating potential complexities in the calculations.

Phymath
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ok this probley seems simple but i just need to see how to do it, ok well how do u evaluate this...
find the flux of
the vector field...
[tex]\vec{F}=<x,y,z>[/tex]

through this surface above the xy-plane..
[tex]z = 4-x^2-y^2[/tex]

how do u evaluate this with surface integrals method and the divergence therom

[tex]dS = \sqrt{f_{x}^2+f_{y}^2+1}dA[/tex] for surface inergrals

thanks very much to anyone who can show me this step by step, i keep getting the wrong answer and i don't know why...
 
Last edited:
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Hi phy_math,

just to make sure I understand you, is this right:

[tex]\begin{array}{l}<br /> \vec F = \vec x = x\hat x + y\hat y + z\hat z \\ <br /> f(x,y,z) = z - (4 - x^2 - y^2) = 0\\ <br /> \end{array}[/tex]

Since then ...

[tex]dS = dxdy\sqrt {(\frac{{\partial f}}{{\partial x}})^2 + (\frac{{\partial f}}{{\partial y}})^2 + 1}[/tex] makes sense

... where you can replace [itex]dA = dxdy[/itex].


To get the total flux of [itex]\vec F[/itex] through the surface, you want the integral [itex]\phi = \int\limits_S {\vec F \bullet \hat ndS}[/itex], where [itex]\hat n = n_x \hat x + n_y \hat y + n_z \hat z[/itex] is the unit normal to the surface.

So, to do this integral, you'll need to calculate the unit normal to the surface. You've already got [itex]\vec F[/itex] and can calculate [itex]dS[/itex] from the above formula.

Have you seen the equation [itex]\nabla f \bullet \hat n = 0[/itex] before?

Also, if you're calculating this flux over the whole of space remember that, by definition, you can write

[tex]\int\limits_0^\infty {\int\limits_0^\infty {f(x,y)dxdy} } = \begin{array}{*{20}c}<br /> {\lim } \\<br /> {X,Y \to \infty } \\<br /> \end{array}\int\limits_0^Y {\int\limits_0^X {f(x,y)dxdy} }[/tex]

That might help as well. See how you get on.
 
Last edited:
Correct me if I am wrong, I just learned this last week.

By the divergence theorem, [tex]\int_S \vec F \cdot \vec n \ dS = \int_V \nabla \cdot \vec F \ dV[/tex]

Since [tex]\vec F = x \vec i + y \vec j + z \vec k[/tex], then [tex]\nabla \cdot \vec F = 3[/tex].

A quick sketch of the surface shows a inverted parabolid. This volume should be easy to find via cylindrical coordinates.

[tex]flux=\int_{0}^{2\pi}\int_{0}^{2}\int_{0}^{4-r^2} 3r \ dz \ dr \ d\theta[/tex]

Side note: Does anyone know if there is a shortcut to writing the "[*tex]...[*/tex]" tags?
 
By the way, a simpler way to find [itex]\int\limits_S {\vec F \bullet \hat ndS}[/itex]
without calculating n and dS separately is this:

The surface is given by z- (4- x2- y2)= 0 or z+ x2+ y2= 4, a constant. If we take F(x,y,z)= z+ x2+ y2 then the surface is a "level surface" of F and so grad F= <2x, 2y, 1> is perpendicular to the surface and n dS is <2x, 2y, 1>dxdy.

Notice that we have "dropped down" to the xy-plane. If the z-component of grad F had not been 1, we would have had to make it 1.

For example if the surface is x2+ y2+ z2= R2 (a sphere) we would take F(x,y,z)= x2+ y2+ z2 so that grad F= <2x, 2y, 2z>. Now we have to decide in what plane we want to do the integration (with a sphere it really doesn't matter!). If we choose to integrate in the xy-plane, then we divide the entire vector by 2z to "normalize" the z-component: <x/z, y/z, 1>dx dy. Of course, we would have to divide the problem into two parts, z positive or z negative, to integrate over the entire sphere.
 
thanks all, its good to know ur all here to help! you i did similar things, turns out i was just basically using the wrong limits for the surface thanks everyone!
 

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