Finding hypervolume lying between Gaussian function and x-y-z plane over $\mathbb{R}^3$

Question

Define the 3-variable Gaussian function by $G(x,y,z) = e^{-(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2})}$. Find the hypervolume lying between this surface and the x-y-z hyperplane, over the entire domain of $\mathbb{R}^3$; i.e., the volume in $\mathbb{R}^4$ defined by : $$ w \in [0,G(x,y,z)], \quad x\in -(\infty,+\infty), y\in -(\infty,+\infty), z\in -(\infty,+\infty) . $$ I guess we have to compute an integral but I might be wrong and if it's that case I have a trouble setting up the integral and visualize it (if that's possible).

Answer 1

Let $F(t) = \int_{-\infty}^\infty e^{-\frac{x^2}{t^2}}\, dx$. Using the fact that $\int_{-\infty}^\infty e^{-x^2}\, dx = \sqrt{\pi}$, we have $F(t) = t\sqrt{\pi}$, by the $u$-substitution $u = \frac{x}{t}$.

Hence, the hypervolume is $$\int_{\Bbb R^3} \int_0^{G(x,y,z)} 1\, dw\, dV = \int_{-\infty}^\infty \int_{-\infty}^\infty \int_{-\infty}^\infty e^{-\left(\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2}\right)}\, dz\, dy\, dx = F(a)F(b)F(c) = abc\,\pi^{3/2}.$$