I was solving a physics problem and I run in to an integral that I could not solve. Could you give me some tips to solve this kind of integral?

For context I'm letting the original vectors at the end of the post, and the link to the physics problem.

\begin{align} V &= D \int_{\alpha_0}^{\alpha}\frac{\sec^2{\alpha} \ d\alpha}{\Vert\vec{r}-\vec{r_Q} \Vert} \tag{1} \\ \Vert\vec{r}-\vec{r_Q} \Vert &= \sqrt{r^2 + D^2 \sec^2{\alpha} - 2rD\ [\sin{\theta}\sin{\phi}\tan{\alpha} + \cos{\theta}]} \tag{2} \\ \alpha_0 &= \arctan(-\frac{L}{2D}) \qquad \alpha = \arctan(\frac{L}{2D}) \end{align}

$$ \vec{r} = r\sin\theta\cos\phi\ \hat{i}\ +\ r\sin\theta\sin\phi\ \hat{j}\ +\ r\cos\theta\ \hat{k} $$ $$ \vec{r_Q} = D\tan\alpha\ \hat{j}\ +\ D\ \hat{k} $$

Physics Post

  • 2
    $\begingroup$ How can $\alpha$ be a bound of the integral and the variable of integration? $\endgroup$
    – Clayton
    Jul 31 '18 at 23:51
  • $\begingroup$ I've put the values for the boundaries in the equation below the vectors modules $\endgroup$
    – liuzp
    Jul 31 '18 at 23:55
  • $\begingroup$ then what is the variable of integration? $\endgroup$
    – Clayton
    Aug 1 '18 at 1:12

This integral as stated can be evaluated using the substitution $u=\tan\alpha$, which brings it to the form


where $\delta=\frac{L}{2D}$, $a=D^2$ , $b=-2rD\sin\theta\sin\phi$ , $c=r^2+D^2-2rD\cos\theta$.

I took a look at the physics however and it seems that you have not considered the fact that the image of the charged rod through the conducting sphere is not going to be a rod itself, but a circle because every point of the rod is mapped a different distance away from the center of the sphere. Taking this into account, I don't know how useful this calculation is going to be for you but I submit it anyway.

  • $\begingroup$ Well, thanks, for the feedback, I'm going to revise the calculations, and, just one more thing since you stated it, the radius of the image would be constant? Learning a new integral is allways good, but eventualy I'm going to solve the problem. $\endgroup$
    – liuzp
    Aug 1 '18 at 2:41
  • $\begingroup$ [Here](en.wikipedia.org/wiki/… ) you can get started with a single charge, and then use superpositions to construct the whole rod out of infinitesimal charges at the appropriate distances from the origin. Every charge is mapped to a different distance than the charge next to it, and using the rule that the link provides,I found myself that the image is going to be bent into a circular arc of constant radius. $\endgroup$ Aug 1 '18 at 2:55
  • $\begingroup$ You should fill in the details and be advised that the electric field of a uniformly charged circular arc doesn't have a nice solution in terms of simple elementary functions, so the integral you will do will be a tad more advanced. $\endgroup$ Aug 1 '18 at 2:55
  • $\begingroup$ And my professor saying that he solved in a single A4 paper. But, again thanks, you've helped more than you think. $\endgroup$
    – liuzp
    Aug 1 '18 at 3:00
  • $\begingroup$ Let us not doubt your professor's awesomeness :P Good luck! $\endgroup$ Aug 1 '18 at 4:02

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