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Change of variable (involving partial diff.)

  1. Jul 10, 2011 #1
    1. The problem statement, all variables and given/known data

    attachment.php?attachmentid=37092&stc=1&d=1310359222.jpg

    2. Relevant equations

    attachment.php?attachmentid=37094&stc=1&d=1310359249.jpg

    3. The attempt at a solution
    I am not sure part b and c.

    Also, I don't know how to answer the last question.

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  2. jcsd
  3. Jul 11, 2011 #2
    In your solution for part 3, you have a factor of r3 in the denominator outside the integral (in each of the three integrals). Remember r = x + y + z, so you cannot take it outside the integral.

    Your approach for part c looks confusing. We'll talk about part c, let's get through part b first.
     
  4. Jul 11, 2011 #3
    It seems [itex]r=\sqrt{x^{2}+y^{2}+z^{2}}[/itex], isn't it?
     
  5. Jul 11, 2011 #4
    Ah, ofcourse! My bad.
     
  6. Jul 11, 2011 #5
    Well if I intergrate r at the same time, it becomes [itex]\int\frac{x}{({x^{2}+y^{2}+z^{2}})^\frac{3}{2}}[/itex]

    Which is difficult to integrate...
     
  7. Jul 11, 2011 #6
    Not really. For example, for x integral take x2 + y2 + z2 = t, then 2xdx = dt and your integral is simply dt/2*(t^1.5)
     
  8. Jul 11, 2011 #7
    Oh. I forget that I can use substitution.
    But how about the new interval?
    From 0 to x^2+y^2+z^2?
     
  9. Jul 11, 2011 #8
    Common convention is to express potential relative to potential at infinity, so limits would be[itex]\infty[/itex] to x2 + y2 + z2.
     
  10. Jul 11, 2011 #9
    Now, I find that the integral for x, y and z are the same.
    Is that correct now?


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  11. Jul 11, 2011 #10

    2 things. First, although you wrote 0 as lower limit, you realize it's actually infinity (as I mentioned in previous post). Having the correct upper and lower limit will help you fix the sign of final value for x integral you get.
    Second, note it's the x integral; it is accompanied with a [itex]\hat{x}[/itex] and similarly y and z integral are accompanied with respective unit vectors. Together they form [itex]\vec{E}[/itex] which has to look same as 8.11 (this is part c!)
     
  12. Jul 11, 2011 #11
    Do you mean I should put infinty as the upper limit and x^+y^+z^2 as lower limit?
    What is the physical meaning if I put infinty as the lower limit and x^+y^+z^2 as the upper limit?


    I dont understand this point.
    Do you mean that I miss out the vector?

    However, inside the integral is a dot product. i.e.[itex]F\bullet dr[/itex]
    Since dot product produce scalar only, there should be not any vector left.
     
  13. Jul 11, 2011 #12
    Look at it this way - x2 + y2 + z2 = t (where y and z and not constant anymore). So, 2xdx + 2ydy + 2zdz = dt. This substitution makes more sense than taking x2 + y2 + z2 = t with y and z constant as I had suggested previously. Because now the limit of integral is infinity and x2 + y2 + z2. I am not sure about limits of the integral if y and z are kept constant.

    On the meaning of upper lower limits - Griffith defines potential like potential.jpg I cannot possibly give a better explanation. I highly recommend you have a look (Sec 2.3 in Griffith - Intro to electrodynamics)

    Ignore this. I jumped on part c in my head.
     
    Last edited: Jul 11, 2011
  14. Jul 11, 2011 #13
    So [itex]\phi=\frac{q}{r}[/itex]?
    If it is correct, let's move to part c.

    I would like to know if I get the correct cylindrical form.
     
  15. Jul 11, 2011 #14
    Yes that is correct. You can check that by using [itex]\vec{E}[/itex] = -[itex]\nabla[/itex] [itex]\phi[/itex] and comparing it with 8.11
    To get [itex]\phi[/itex] in cylindrical form, you have to write [itex]\vec{E}[/itex] in cylindrical coordinates and follow the procedure as above. (To know how to go to cylindrical from spherical, look at http://is.gd/JFitBh)
     
  16. Jul 11, 2011 #15
    Do you mean that my approach above can't get the answer?
    If so, why?
     
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