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How to solve this double integration?

  1. Apr 25, 2014 #1

    Red

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    1. The problem statement, all variables and given/known data
    Hi Guys, I need help to solve this double integration. This integration is over r and theta. The rest are constant.

    2. Relevant equations
    ∫[itex]^{R_{2}}_{r=0}[/itex] ∫[itex]^{\pi}_{\theta=0}[/itex] r[itex]^{2}[/itex] sin([itex]\theta[/itex]) dr d[itex]\theta[/itex] / ((D[itex]^{2}[/itex]+r[itex]^{2}[/itex] - 2rd cos([itex]\theta[/itex]))[itex]^{2}[/itex] - R[itex]_{1}[/itex]^2)[itex]^{3}[/itex], r from 0 to R[itex]_{2}[/itex], theta from 0 to pi.

    The slash stand for division, I am not sure how to create a proper fraction here so I used slash.

    3. The attempt at a solution
    I have attempted to solve this problem by reducing it to s hollow shell, ie r is a constant, but the integratino seem to be the same for theta.

    Thank you for your help!
     
    Last edited: Apr 25, 2014
  2. jcsd
  3. Apr 25, 2014 #2

    haruspex

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    I assume D is the same as d.
    Are you sure about the integrand? It looks dimensionally wrong. R12 is a quadratic term being subtracted from a quartic term.
    To get fractions in LaTex use \frac{}{}.
     
  4. Apr 25, 2014 #3

    Simon Bridge

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    Oh messy - you used the equation editor right?
    Here, let me help...

    $$\int_{r=0}^{R_2} \int_{\theta=0}^{\pi}
    \frac{ r^2 \sin\theta \;\text{d}r\text{d}\theta }{ \big( (D^2+r^2-2rD\cos\theta)^2-R_1^2\big)^3 }$$ ... this what you meant?

    To see how I did that, use the "quote" button at the bottom of this post.
    I have the same reservations about the denominator as haruspex... suspect the square about the cosine rule term is misplaced.

    It would probably help to know what problem you are actually trying to solve here.
     
  5. Apr 26, 2014 #4

    Red

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    Thank you Simon Bridge! It look much nicer now. I realized I made a mistake that you and haruspex pointed out, I have corrected the integration and it looks like this:

    $$\int_{r=0}^{R_2} \int_{\theta=0}^{\pi}
    \frac{ r^2 \sin\theta \;\text{d}r\text{d}\theta }{ \big( D^2+r^2-2rD\cos\theta-R_1^2\big)^3 }$$

    I realized I can solve the theta part of this integration, however I do not know how to proceed on with the r part. This is what I get after solving the theta part:

    $$\frac{ 1 }{ 4 D }\int_{r=0}^{R_2} \frac{ r }{ \big( (D-r)^2-R_1^2\big)^2 } - \frac{ r }{ \big( (D+r)^2-R_1^2\big)^2 } \text{d}r $$

    Can you help me on this integration? Thank you!

    I am using the Hamaker approach to derive the van der Waal potential between two sphere of radius R[itex]_{1}[/itex] and R[itex]_{2}[/itex], the distance between the center of the two sphere is D.
     
  6. Apr 26, 2014 #5

    Simon Bridge

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    I think you want a substitution like ##R_1\sec u = D-r## for the first one and ##R_1\sec u = D+r## for the second one.
    But you may be able to just do it by partial fractions.
     
  7. May 1, 2014 #6

    Red

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    Thank you for your help!
     
  8. May 1, 2014 #7

    adjacent

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    You can also right click > Show math as > TeX commands.
     
  9. May 4, 2014 #8

    Red

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    Thanks adjacent! I have just learn TeXstudio so I am kind of more familiar with this kind of text input now.
     
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