So during lecture there was this problem that was worked through and i understand it right up until the very end, and i was wondering if someone could explain how he simplified it down.
So there was a double integration, and once it had been integrated twice it came out as
$$R^2(2\pi)(-1/a)e^{a\cos\theta} \text{( this is already integrated)} $$
it had been integrated with respect to $\theta$ and it now all i have to do was sub in $\pi$ as the upper limit and $0$ as the lower limit
it came out as $2\pi a(e^{-a}-e^a)$ and im not quite sure how
The result as you substitute in $0$ and $\pi$ for $\theta$ is as follows:
\begin{align} R^2 (2 \pi) \left( - \frac1a\right)\left[ \exp(a \cos\theta) \right]_0^\pi &= R^2 (2 \pi) \left( - \frac1a\right) \left[\exp(a \cos\pi) - \exp(a \cos 0) \right]\\ &= R^2 (2 \pi) \left( - \frac1a\right) \left[\exp(a (-1)) - \exp(a (1)) \right]\\ &= R^2 (2 \pi) \left( - \frac1a\right) \left[\exp(-a) - \exp(a) \right]\\ \end{align}
You mentioned that it is a double integration, but it seems that you have not integrated with $R$ or is $R$ a constant. With the information that is given, there is no way to explain away the $R$ term unless $R$ is somehow equal to $a$ and the negative sign is just a typo.