Multivariable Calc help

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  • #1
cronxeh
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Hey can you guys check my answer.

Question: Use the Divergence Theorem to calculate the Flux of the vector field F(x,y,z)=xi + y^2j - zk through the unit sphere centered at the origin with the outward orientation

Solution: div(F) = 1 + 2y - 1 = 2y
Flux = [tex]\int_{W} div(F) dV = \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} 2y \ dxdydz = \int_{0}^{1} \int_{0}^{1} 2y \ dydz = \int_{0}^{1} 1 \ dz = 1 [/tex]

So for another method would I have to use [tex]r=x^2 + y^2 + z^2[/tex] obtain the [tex]dr=2x + 2y + 2z[/tex] and do [tex] \int_{R} F dr = \int_{S} (xi + y^2j -zk)(2x+2y+2z)dA=\int_{S} (2x^2 + 2y^3 - 2z^2)dA = \int_{0}^{1} \int_{0}^{1} 2/3 + 2y^3 - 2z^2 \ dydz = \int_{0}^{1} 2/3 + 1/2 - 2z^2 dz = 2/3 + 1/2 - 2/3 = \frac{1}{2} [/tex]

Is this method incorrect or was the first one incorrect? Are they both wrong? :confused:
 

Answers and Replies

  • #2
Brad Barker
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ermm... wouldn't it be easiest to use spherical coordinates?


edit: and in that second method, why are you calculating a line integral?

gauss' theorem doesn't have line integrals in there at all.

look up what gauss' theorem tell you...

i imagine you wanted to solve the problem the easy (integrating the divergence) way and the harder (what integrating the divergence is equal to :wink: ) way.


so to answer your question... both methods are incorrect! :eek:
 
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  • #3
Brad Barker
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yeah, the way you have written the limits of integration, you are taking the divergence of that vector field through the unit cube in your first method. (haven't bothered with your second method--get your first one fixed!)
 
  • #4
Hurkyl
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Do you have any idea what regions you're supposed to be integrating over, for either calculation?
 
  • #5
Brad Barker
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Hurkyl said:
Do you have any idea what regions you're supposed to be integrating over, for either calculation?

and is it just me or is he trying to use stokes' theorem in that second attempt?

:eek:

edit: no...closer look tells me that... it's not even that.

your second method is way off base.


look back at your textbook and see the two ways of calculating flux.
 
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  • #6
Hurkyl
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The second method is surely supposed to be the definition of flux: [itex]\iint_{\delta R} \vec{F} \cdot \hat{n} \, dA[/itex].
 
  • #7
Brad Barker
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Hurkyl said:
The second method is surely supposed to be the definition of flux: [itex]\iint_{\delta R} \vec{F} \cdot \hat{n} \, dA[/itex].

yeah, that's the other--probably harder--way to do the problem. except... it looks like he was trying to use a line integral or...something. :confused:
 
  • #8
cronxeh
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Ok I just realized my mistake for divergence method ( I hope )

div F = 2y

Should the flux be: [tex]\int_{W} div(F) \ dV =[/tex]

[tex]\int_{-1}^{1} \int_{-sqrt(1-x^2)}^{sqrt(1-x^2)} \int_{-sqrt(1-x^2-y^2)}^{sqrt(1-x^2-y^2)} 2y \ dzdydx[/tex]
 
  • #9
cronxeh
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And in spherical coordinates:

Flux = [tex]\int_{0}^{2pi} \int_{0}^{pi} \int_{0}^{1} 2y \ p^2 \ sin(phi) \ dp \ dphi \ dtheta[/tex]
 
  • #10
Hurkyl
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Yah, but don't forget you should convert y to spherical coordinates too.
 
  • #11
cronxeh
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that would be 2y ->> 2(p)sin(theta)sin(phi) ?
 
  • #12
HallsofIvy
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Because in spherical coordinates

x= ρcosθsinφ
y= ρsinθsinφ
z= ρcosφ
 
  • #13
Brad Barker
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yeah, good going.
 

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