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Evaluating Volume Integrals and Divergence Theorm

  1. Nov 30, 2013 #1
    1. The problem statement, all variables and given/known data
    Evaluate the integral as either a volume integral of a surface integral, whichever is easier.

    [itex]\iiint \nabla .F\,d\tau[/itex] over the region [itex]x^2+y^2+z^2 \leq 25[/itex], where [itex]F=(x^2+y^2+z^2)(x*i+y*j+z*k)[/itex]

    2. Relevant equations
    [itex]\iiint \nabla .F\,d\tau =\iint F.n\,d\sigma[/itex]


    3. The attempt at a solution
    I believe this would be easier to do as a volume integral though I am honestly not sure saying I am not really understand how things like line integrals and surface integrals work in the sense of things like Stokes Theorm and Div. Theorm.

    For the volume intregral I did just integrated the Div(F) from -5 to 5 for all three integrals saying that is the range each of the variables will go through. I plugged this into Wolfram Alpha and got an answer of 25000 though I am not sure if I did this question correctly.

    If I wanted to do a solve this using the right hand side of the Div. Theorem how would I approach this?
     
  2. jcsd
  3. Nov 30, 2013 #2

    LCKurtz

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    If you let all three variables go from -5 to 5 you are describing a region shaped like a rectangular block. You have a sphere.
     
  4. Nov 30, 2013 #3

    vela

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    That's not what I got from doing what you said you did, which suggests you didn't calculate div F correctly. Show us how you calculated div F.

    You may want to switch to spherical coordinates to do the integrals.
     
  5. Nov 30, 2013 #4
    Ahh, I knew I was forgetting something. For this case once I converted x,y, and z into sphrerical coordinates wouldnt the limits of integration be r=0 to r=5, θ= 0 to 2pi, and [itex]\phi[/itex]=0 to 2pi? Also just to double check. Would taking the volume integral actually easier in this case?
     
  6. Nov 30, 2013 #5
    I was just thinking about it but would actually turning it into a surface integral make it easier? Saying you would have the double integral of the force dotted into the normal, which would give out [itex](x^2+y^2+z^2)^2/5[/itex] since the normal to the sphere is just the normalized position vector? And since this is only going to be on the surface of the sphere we know that [itex](x^2+y^2+z^2)=r=5[/itex] . So would the answer then be 20pi^2? Since I would be able to draw out a consant 5, and then all that would be left would to do the integral that has no terms left inside with the limits of the two intergals both being from 0 to 2pi?
     
  7. Nov 30, 2013 #6

    vela

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    The answer is ##12500\pi##.
     
  8. Nov 30, 2013 #7

    vela

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    You're saying the integral is equal to that, or F dotted into the normal? Where did the 5 come from?

    ##r = \sqrt{x^2+y^2+z^2}##

    Those limits aren't correct. The polar angle doesn't go from 0 to ##2\pi##.
     
  9. Dec 1, 2013 #8
    Sorry, was typing that in a hurry saying I was having to leave right then.

    What I was trying to say, since the normal to the sphere is the normalized position vector, which would be [itex]{x,y,z}/5[/itex] since we know the distance to the surface is 5. If we dot this into the force that would create, [itex](x^2+y^2+z^2)*{x,y,z}.{x,y,z}/5[/itex] after doing the dot product turns into [itex](x^2+y^2+z^2)^2/5[/itex].

    After this I did mess up and say (x^2+y^2+z^2)=5 where I should of said it is equal to 25. So that means the integrad is equal to 25^2/5?

    And since the azimuthal angle is rotating completely around that means the polar angle only actually needs to go from 0 to pi correct?

    I must be making some sort of mistake somewhere saying the answer I am getting is only like 250pi. Is there anything that you can see that I am doing wrong? Or is my entire train of thought forgetting a crucial point?
     
  10. Dec 1, 2013 #9

    vela

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    What did you use for ##d\sigma##? Remember it's proportional to ##r^2##.
     
  11. Dec 1, 2013 #10
    Ahh [itex]d\sigma=r^2sin(\theta) d\theta d\phi[/itex]. Forgot that it was not just dd/theta d\phi So that means that it would be [itex]\iint r^6/5*sin(\theta)*d\theta*d\phi[/itex] if the limits are \theta=0 to pi and \phi=0 to 2pi?
     
  12. Dec 1, 2013 #11

    vela

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    Yup.
     
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