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Flux and divergence theorem

  • Thread starter -EquinoX-
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  • #1
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Homework Statement



http://img16.imageshack.us/img16/88/fluxm.th.jpg [Broken]

Homework Equations





The Attempt at a Solution



I've tried to find the divergence of F and I got 3x^2 + 3y^2 + 3z^2 and as this is a variable I need to set up the integral... how do I set the integral
 
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Answers and Replies

  • #2
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i think using cylindrical coordinate will be fine
 
  • #3
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how would you do it in cylindrical coordinate? is it something like this?

do you mean that I don't need to use the divF of it?
 
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  • #4
679
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how would you do it in cylindrical coordinate? is it something like this?

do you mean that I don't need to use the divF of it?
you mean how to express the integrated region?just set r from 0 to 3, z from 0 to10.
using divF is not a bad choice, indeed in this case it won't save you very much work,because it's pretty symmetric casem a you can try to calculate it directly,
 
  • #5
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to use divergence I will need tot calculate the volume right? and I then will need to convert x, y, and z to cylindrical coordinate... which is x = r cos(theta), y = r cos(theta) , so in this case it's x^3cos(theta)^3.. right?
 
  • #6
djeitnstine
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Rule of thumb...x axis and y axis are not parallel....y=r sin(theta)... unless you're dealing with some weird system...
 
  • #7
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oops.. bad copy and paste mistake.. so in the end it will be:

\int0^10 \int_0^3 27cos(\theta) + 27sin(\theta) + z^3 dr d\theta

now what is r here?
 
  • #8
djeitnstine
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This is a triple integral so there should be 3 limits. Also it seems that [tex]3x^2 + 3y^2 + 3z^2[/tex] is your field, so in cylindrical it would be [tex]\int_0^{2\pi} \int_0^3 \int_0^{10} 3r^2 cos^2(\theta) + 3r^2 sin^2(\theta) + 3z^2 r dz dr d\theta[/tex]

As you can see the field cancels out nicely in cylindrical. Also note the extra 'r' and the order of the limits.

The field seems symmetrical so you should be able to go from 0-->pi/2 and multiply by 2

latex seems to be ok again
 
  • #9
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This is a triple integral so there should be 3 limits. Also it seems that [tex]3x^2 + 3y^2 + 3z^2[/tex] is your field, so in cylindrical it would be [tex]\int_0^{2\pi} \int_0^3 \int_0^{10} 3r^2 cos^2(\theta) + 3r^2 sin^2(\theta) + 3z^2 r dz dr d\theta[/tex]

As you can see the field cancels out nicely in cylindrical. Also note the extra 'r' and the order of the limits.

The field seems symmetrical so you should be able to go from 0-->pi/2 and multiply by 2

latex seems to be ok again
so it could then be simplified as:

[tex]\int_0^{2\pi} \int_0^3 \int_0^{10} 3r^2 + 3z^2 r dz dr d\theta[/tex]
[tex]\int_0^{2\pi} \int_0^3 30r^2 + 1000r dr d\theta[/tex]
[tex]\int_0^{2\pi} 4770 d\theta[/tex]
[tex]9540\pi[/tex]

is this wrong?

my confusion is that why is it 3z^2r ?? why is there extra r at the end of z?
 
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  • #10
djeitnstine
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This is because when converting to spherical the area of the differential section is [tex]rdrd\theta[/tex] and adapted to cylindrical is [tex]rdzdrd\theta[/tex] since the differential z is linear
 
  • #11
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the answer that I got above doesn't match.. wonder why..
 
  • #12
djeitnstine
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Oh I needed brackets, I am sorry

[tex]
\int_0^{2\pi} \int_0^3 \int_0^{10} (3r^2 + 3z^2) r dz dr d\theta
[/tex]

r is ditributed
 
  • #13
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that's why I ask first place why is it 3z^2r because usually r is distributed.. thanks for clearing that up
 
  • #14
564
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Oh I needed brackets, I am sorry

[tex]
\int_0^{2\pi} \int_0^3 \int_0^{10} (3r^2 + 3z^2) r dz dr d\theta
[/tex]

r is ditributed
here's what I got:

[tex]\int_0^{2\pi} \int_0^3 \int_0^{10} (3r^3 + 3z^2r) dz dr d\theta [/tex]
[tex]\int_0^{2\pi} \int_0^3 30r^3 + 27r dr d\theta [/tex]
[tex]\int_0^{2\pi} 2916/4 d\theta [/tex]
[tex]\frac{2916\pi}{2} [/tex]


the answer still doesn't match.. am I doing something wrong here?
 
  • #15
564
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thanks I got it now :)
 

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