Volume of a volcano

The volume of the volcano is the sum of the volume under the surface $f(x,y)$ and the volume of the cylinder spanning its height (the throat of the volcano). The height of the throat is $h = 36$ and its radius is $r = \frac{1}{\sqrt{6}}$ so the volume of the throat is

$\pi r^2 h = \pi \big( \frac{1}{\sqrt{6}} \big)^2 \cdot 36 = \boxed{ 6 \pi }$

The volume under the surface $f(x,y)$ spanning $\mathbb{R^2}$ is:

$\iint_{x^2+y^2 \geq \big( \frac{1}{\sqrt{6}} \big)^2} \frac{1}{(x^2+y^2)^2}dxdy$

Switching to polar coordinates:

$\Longrightarrow \int_{0}^{2\pi} \int_{\frac{1}{\sqrt{6}}}^{\infty} \frac{1}{r^4} r dr d\theta = \int_{0}^{2\pi} \int_{\frac{1}{\sqrt{6}}}^{\infty} \frac{1}{r^3} dr d\theta$ $\Longrightarrow \int_{0}^{2\pi} d\theta \int_{\frac{1}{\sqrt{6}}}^{\infty} \frac{1}{r^3} dr = \big( 2\pi - 0 \big) \bigg[ \frac{r^{-2}}{-2} \bigg]_{\frac{1}{\sqrt{6}}}^{\infty} = - \pi \bigg( \frac{1}{\infty} - \frac{1}{\big(\frac{1}{\sqrt{6}}\big)^2} \bigg) = - \pi \big(0 - 6 \big) = \boxed{ 6 \pi }$

Therefore the total volume is $6 \pi + 6 \pi \Longrightarrow \boxed{ a = 12 }$