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Homework Help: A Mysterious Calculus Operation

  1. Aug 9, 2006 #1
    I was doing a physics problem which leads me to this equation:
    [tex]
    V = \frac{\sigma}{4\pi \epsilon_0}\int_{0}^{R}\int_{0}^{2\pi} \frac{r}{\sqrt{R^2 + r^2 - 2Rr\cos{\theta}}} \,d\theta \,dr
    [/tex]
    Stumped by the math, I turned to the solution. However, without a word of explanation, the solution jumps from the above equation to the following:
    [tex]
    V = \frac{\sigma}{4\pi \epsilon_0}R\int_{0}^{2\pi} \cos{\theta} \ln{\left ( 1 + \frac{\sqrt{2}}{\sqrt{1-\cos{\alpha}} \right ) }\,dA +8R - 2\pi R
    [/tex]
    I had no idea how that happened. I would appreciate it if someone could explain what I should to do arrive at this latter equation. I don't need a full explanation, just a push in the right direction.

    EDIT: I understand that it's an integral, I just don't understand how the integral was done. Specifically, this one:
    [tex]
    \int_{0}^{R} \frac{r}{\sqrt{R^2 + r^2 - 2Rr\cos{\theta}}} \,dr
    [/tex]
    Thanks.
     
    Last edited: Aug 10, 2006
  2. jcsd
  3. Aug 10, 2006 #2
    To find the solution you first have to perform the integration with respect [tex]r[/tex].

    By the way, you have to observe that:

    [tex]\frac{r}{\sqrt{r^2 +R^2 - 2 R r \cos \theta}} = \frac{r - R \cos \theta}{\sqrt{(r-R \cos \theta)^2 + R^2 \sin^2 \theta}} + \frac{R \cos \theta}{\sqrt{(r-R \cos \theta)^2 + R^2 \sin^2 \theta}}}[/tex].
     
    Last edited: Aug 10, 2006
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