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Showing a vector field is imcompressible

  1. Aug 26, 2015 #1
    1. The problem statement, all variables and given/known data
    attachment.php?attachmentid=455997&d=1440616258.jpg

    2. Relevant equations


    3. The attempt at a solution

    As you can see, the solution is shown just below the question.

    Essentially, I don't understand how the x, y and z component of the vector field has been separated because the numerator of the vector field's fraction is: (x^2 + y^2 + z^2)^(1/2)

    It seems like they've just taken x^2 out and square rooted it to get 'x', but you can't do that, can you?
     
  2. jcsd
  3. Aug 26, 2015 #2

    DEvens

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  4. Aug 26, 2015 #3

    LCKurtz

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    That isn't what they have done. The numerator is bold faced ##\bf{r}## and the denominator is ##r##. The first is the vector and the second its magnitude. r = xi + yj + zk.
     
  5. Aug 26, 2015 #4
    Do I treat the non-bold r as a constant?

    If sub in bold r, I get:

    G = [(x^2 + y^2 + z^2)^0.5 ] / 4*pi*r^3
     
  6. Aug 26, 2015 #5

    LCKurtz

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    I just noticed in your graphic under (b) they have ##{\bf r} = \sqrt{x^2+y^2+z^2}## That should not have been a bold face r. The bold face r represents the vector and the plain r its magnitude, which is not constant.
     
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