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Integration - slightly confused

  1. Jun 7, 2014 #1
    Hi I'm trying to integrate the following [itex]q_m = -D A \frac{dc}{dx}[/itex]
    where [itex]A = 4 \pi r^2[/itex] Yes, a sphere.


    My supplied literature simplifies to [itex]q_m = -D 2 \pi r L \frac{dc}{dr}[/itex] when [itex]A = 2 \pi r L [/itex]

    Integrating to [itex]\int_{r1}^{r2} q_m \frac{dr}{r} = - \int_{c1}^{c2} 2 \pi L D dc[/itex]

    Integrated to [itex]q_m ln \frac{r_2}{r_1} = 2 \pi L D (c_1 - c_2) [/itex]

    I've had a go but unsure what to do with the [itex]r^2[/itex]

    I thought this might work but it gives a negative value for [itex]q_m[/itex]

    [itex]\int_{r1}^{r2} q_m \frac{dr}{r^2} = - \int_{c1}^{c2} 4 \pi D dc[/itex]

    Integrated to [itex]q_m ln \frac{r_2}{r_1^2} = 4 \pi D (c_1 - c_2) [/itex]


    Any ideas? Thanks.
     
    Last edited: Jun 7, 2014
  2. jcsd
  3. Jun 7, 2014 #2
    I don't follow... how does x relate to r?
     
  4. Jun 7, 2014 #3
    I've edited it whilst you were reading. Do you follow now? They replace x with r.
     
  5. Jun 7, 2014 #4
    Not really. Can you post the problem statement? It isn't even clear that r is a variable, or that there is correlation between x and r.
     
  6. Jun 7, 2014 #5
    I've attached the literature. Basically the formula is for a cylinder, I need to convert it for a sphere.
     

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  7. Jun 7, 2014 #6

    HallsofIvy

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    What you post in your last response integrates over a cylinder of radius r, length L. But you are asking about a sphere. Are you asking how to change from a cylinder to a sphere? If so then, yes, the surface area of a sphere of radius r is [tex]4\pi r^2[/tex] so that [itex]q_m= -DA\frac{dc}{dr}[/itex] (NOT [itex]\frac{dc}{dx}[/itex]) becomes [itex]q_m= -D(4\pi r^2)\frac{dc}{dr}[/itex] which can be separated as
    [tex]\frac{q_m}{r^2}dr= -4D\pi dc[/tex]
    and integrating,
    [tex]q_m\int_{r_1}^{r_2} \frac{1}{r^2}dr= -4D\pi\int_{c_1}^{c_2} dc[/tex]

    Integrate that.
     
  8. Jun 7, 2014 #7
    This?

    [itex]q_m ln (\frac{1}{r_1^2}r_2) = 4 \pi D (c_1 - c_2) [/itex]
     
  9. Jun 8, 2014 #8

    HallsofIvy

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    Then you are not "slightly confused" about integration- you seem to be saying you do not know how to integrate at all. No [itex]\int dr/r^2[/itex] is not "ln(r^2)". Using the very basic rule [itex]\int r^n dr= r^{n+1}/(n+ 1)[/itex], [itex]\int dr/r^2= \int r^{-2}dr= -r^{-1}+ C[/itex].
     
  10. Jun 8, 2014 #9
    How do I fit in [itex] r_1 [/itex] and [itex] r_2 [/itex]
     
  11. Jun 9, 2014 #10

    HallsofIvy

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    If [itex]\int f(x)dx= F(x)+ C[/itex] then
    [tex]\int_{x_1}^{x_2} f(x)dx= F(x_2)- F(x_1)[/tex].
     
  12. Jun 9, 2014 #11

    Thanks, I thought as much but seeing as I made a glaring error previously I thought I should double check.
     
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