Let's triple the fun!!

Well, not worth Level 5!

We can write the integral as -

$\displaystyle \int_{-\infty}^{\infty}{{e}^{-{y}^{2}}} dy \int_{-\infty}^{\infty}{{e}^{-{z}^{2}}} dz \int_{-\infty}^{\infty}{{e}^{-{x}^{2}}} dx$

We know that Gaussian integral $\displaystyle \int_{-\infty}^{\infty}{{e}^{-{a}^{2}}} da = \sqrt{\pi}$

So, we can write the integral as -

$\displaystyle (\sqrt{\pi})(\sqrt{\pi})(2\text{Pi}(x))$

Hence,

$\displaystyle \text{Pi}(x)\sqrt{4}\pi$

Switching to spherical coordinates, the integral becomes: $\int_{0}^{2\pi}\int_{0}^{\pi}\int_0^{\infty}e^{-\rho^2}\rho^2\sin{\phi}d\rho d\phi d\theta$ $=4\pi\int_0^{\infty}e^{-\rho^2}\rho^2d\rho$ $=4\pi\Big[-\frac{1}{2}\rho e^{-\rho^2}+\int \frac{1}{2}e^{-\rho^2}\Big]_0^\infty$ $=2\pi\int_0^\infty e^{-\rho^2}d\rho$ $\Rightarrow \sqrt{k}=2 \Rightarrow k =4$