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Vector Calculus - Divergence theorem

  • Thread starter scarlets99
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Homework Statement


1. Consider a cube with vertices at A=(0,0,0) B=(2,0,0) C=(2,2,0) D=(0,2,0) E=(0,0,2) F=(2,0,2) G=(2,2,2) H=(0,2,2)
A)Calculate the flux of the vector fieldF=xi through each face of the cube by taking the normal vectors pointing outwards.
B)Verify Gauss's divergence theorem for the cube and the vector field F by computing each side of the formula.
C)Using Gauss's divergence theorem evaluate the flux \(\displaystyle \int \int F.dS \) of the vector field F=xi+yj+\(\displaystyle z^2\)k where S is a closed surface consisting of the cylinder\(\displaystyle x^2 + y^2 = a^2\), 0<z<b and the circular disks \(\displaystyle x^2 + y^2 , a^2\) at z=0 and \(\displaystyle x^2 + y^2 = a^2\) at z=b.

I have the basics but dont know how to get an answer. My workings are attatched.

Homework Equations



Gauss's divergence theorem


The Attempt at a Solution



My workings are attatched.
 

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Answers and Replies

  • #2
gabbagabbahey
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I don't understand what you are doing in part (i)...there are six different faces and each of the will have a different normal. For example, the face at [itex]z=2[/itex] (the top face if the vertical axis is the z-axis) will have an outward unit normal of [itex]\textbf{k}[/itex], and hence will be perpendicular to [itex]\textbf{F}=x\textbf{i}[/itex] and the flux through that face will be zero. What about the other 5 faces?

For part (ii), you apparently forgot how to differentiate:

[tex]\frac{\partial}{\partial x}x=1\neq 0[/tex]
 
  • #3
vela
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To calculate the surface integral, you want to treat each face separately and then sum the 6 results. I'm not sure if your N and n vectors are supposed to correspond to the same thing. For the top face, the outward normal vector is j according the set of axes of you drew. So the flux through that face is

[tex]\int \vec{F}\cdot d\vec{S} = \int \vec{F}\cdot\hat{n}\,dxdz = \int (x\,\hat{i}\cdot\hat{j})\,dxdz=0[/tex]

For each face, you want to determine n, evaluate the dot product, and then integrate.

In part ii, your divergence calculation is wrong.
 
  • #4
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If using the divergence theorem, from my understanding, you don't need the outward unit normals. You just take the divergence of the function, make sure the surface is closed, continuously differentiable, and integrate over a cube.
 
  • #5
gabbagabbahey
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If using the divergence theorem, from my understanding, you don't need the outward unit normals. You just take the divergence of the function, make sure the surface is closed, continuously differentiable, and integrate over a cube.
He's supposed to verify that the divergence theorem holds for this function by computing the surface integral directly---and you do need the outward unit normals for that part--- and comparing it to the volume integral of the divergence.
 
  • #6
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He's supposed to verify that the divergence theorem holds for this function by computing the surface integral directly---and you do need the outward unit normals for that part--- and comparing it to the volume integral of the divergence.
Oh yea. XD I didn't read that part.

Sigh, vector calculus was my favorite part of calculus 3... I'm gonna miss it.
 

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