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I need help evaluating the following integral

$$\frac{ \sqrt 2 \sigma}{2 \epsilon_0} \int_0^R \frac{r \,dr}{ \sqrt{(z- \frac{ rh}{R} ) ^2 + r^2} }$$

This integral pertains to the Electric potential of an upside down hollow cone with height h and base radius R at an arbitrary point along the z axis. Apparently if I let $R = h$ the integral becomes:

$$\frac{ \sqrt 2 \sigma}{2 \epsilon_0} \int_0^R \frac{r \,dr}{ \sqrt{(z- r) ^2 + r^2} } = \frac{ \sqrt 2 \sigma}{2 \epsilon_0}[ \ln(1 + \sqrt2) - 1 ]$$

I am not sure what kind of substitution I should make if one at all. any suggestions would be nice. I tried wolfram alpha but It didn't work.

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The left side in the second displayed equation seems to involve the parameter $z$, while the right side doesn't. – André Nicolas Jan 4 '13 at 19:05
Yeah I hadn't given it the thought it deserved, thanks – Cactus BAMF Jan 4 '13 at 19:08
up vote 3 down vote accepted

The integral you seek takes the general form

$$\int_0^R dr \: \frac{r}{\sqrt{r^2+a r+b}} $$

Rewrite this integral as follows:

$$\int_0^R dr \: \frac{r}{\sqrt{(r+a/2)^2+b-a^2/4}} $$

Now rewrite further as

$$\int_0^R dr \: \frac{r+a/2}{\sqrt{(r+a/2)^2+b-a^2/4}} - \int_0^R dr \: \frac{a/2}{\sqrt{(r+a/2)^2+b-a^2/4}}$$

You should be able to take it from here.

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Thank you very much – Cactus BAMF Jan 4 '13 at 19:37

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