Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Change of variable (involving partial diff.)

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


    2. Relevant equations


    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.


    Attached Files:

    • 3.jpg
      File size:
      16.4 KB
    • 5.jpg
      File size:
      14.3 KB
    • 4.jpg
      File size:
      25.5 KB
  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?


    Attached Files:

    • 1.jpg
      File size:
      18.6 KB
  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?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook