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Can't Get Integral To Come Out Right

  1. Nov 4, 2012 #1
    1. The problem statement, all variables and given/known data

    What is ∫r^3/(r^2 + h^2) ^ (3/2)

    2. Relevant equations

    uv - ∫v du

    3. The attempt at a solution

    Using Maxima gives me (r^2 + 2h^2) / sqrt(r^2 + h^2), but when I solve it using integration by parts, I get (3r^2 + 4h^2) / sqrt (r^2 + h^2).

    I do:

    dv = r(r^2 + h^2) ^ (-3/2)
    v = -1/sqrt(r^2 + h^2)

    u = r^2
    du = 2r dr

    Then I plug it into the int. by parts formula.

    uv - ∫v du = -r^2/ sqrt (r^2 + h^2) - ∫ - 2r dr / (r^2 + h^2)

    And then I integrate once more and I'm left with:

    -r^2 / sqrt (r^2 + h^2) + 4*sqrt (r^2 + h^2)

    I know that this should work out almost exactly like example 7 from this page, but it just won't. What's going on??
     
  2. jcsd
  3. Nov 4, 2012 #2

    LCKurtz

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    Try a straight u substitution instead of parts. Write your numerator as ##r^2\cdot r\, dr## and let ##u=r^2+h^2##.
     
  4. Nov 4, 2012 #3
    Doesn't seem to work.

    Letting u = r ^2 + h^2 takes care of r dr but not r^2.

    I even tried setting r^2 = u - h^2 , but that doesn't get me anywhere either.
     
  5. Nov 4, 2012 #4

    SammyS

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    That should (and does) work !

    Break the integrand into the sum/difference of two fractions.
     
  6. Nov 4, 2012 #5

    LCKurtz

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    Sure it does. Put ##u - h^2## in for that ##r^2##.
     
  7. Nov 5, 2012 #6
    Okay, thanks guys. I split the fraction up into two.

    ∫u/(2*u^(3/2)) - ∫h^2/(2*u^(3/2)

    I wasn't able to get it to come out right until I considered the h^2 as a constant and factored it out along with the 1/2.

    1/2∫u/(u^(3/2)) - h^2/2 ∫1/(2*u^(3/2)

    Why can I consider h^2 as a constant here? Also, why did integration by parts provide the wrong answer for me??
     
  8. Nov 5, 2012 #7

    SammyS

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    Well, h is not a function of r, is it ?
     
  9. Nov 5, 2012 #8
    I mean, h^2 = u - r^2 which means to me that the value of h^2 is dependent on the value of r. How could it not be?
     
  10. Nov 5, 2012 #9

    uart

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    No you're confusing yourself. u and r^2 are both functions of r, but their difference is not.

    Consider this trivial example. Let [itex]y= 1 - x^2[/itex]. Now y is a function of x, obviously. But x^2 is also a function of x.

    So if we rearrange this to get [itex]1 = y - x^2[/itex], does this mean that "1" is a function of x? No, certainly not.
     
    Last edited: Nov 5, 2012
  11. Nov 5, 2012 #10
    I might be reaching here but let's see.

    I have:

    u = h^2 + r^2

    and

    r^2

    u - r^2 = (h^2 + r^2) - r^2 = h^2

    Thus, r^2 vanishes and h^2 is no longer of a function of r. Thus, I can factor it out of my integral.

    Am I seeing this right?
     
  12. Nov 5, 2012 #11

    dextercioby

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    Hmm, let [itex] r = h \sinh t [/itex] as a change of variable.
     
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