A Gaussian Revolution

The given function is positive definite, so the volume is just $V=\int_{xy-plane}{e^{-r^2}}dS$ .

$V=\int_0^{2\pi}{\int_0^{\infty}{r e^{-r^2}dr}d\theta}$

For those wondering where the factor r comes from, it always appears when we swich to polar coordinates, it is called the Jacobian determinant for polar coordinates. If we have the coordinate mapping

$\left\{ \begin{array}{ll} x=r\cos \theta \\ y=r\sin \theta \end{array} \right.$ , then $J=\begin{bmatrix}  \frac {\partial x}{\partial r} & \frac {\partial x}{\partial \theta} \\ \frac {\partial y}{\partial r} & \frac {\partial y}{\partial \theta}  \end{bmatrix}$ $=\begin{bmatrix}  \cos \theta & -r \sin \theta \\ \sin \theta & r \cos \theta \end{bmatrix}$ and $\det J = r\cos^2 \theta+r\sin^2 \theta=r$ .

$V=\int_0^{2\pi}{d\theta} \cdot \int_0^{\infty}{r e^{-r^2}dr}$

$=2\pi \cdot \int_0^{\infty}{d -\frac{1}{2}e^{-r^2}}$

$= 2\pi \cdot (-\frac{1}{2}e^{-\infty}-(-\frac{1}{2}e^{0}))$

$=2\pi \cdot(0+\frac{1}{2})$

$=\pi$

The volume asked for is very straightforward using cylindrical coordinates and has a value of $\pi$ (see @Tom Engelsman 's solution.)

The result can be used to compute $\displaystyle \int_{-\infty}^{\infty}e^{-x^2}\,dx$ . We need to express the volume integral using cartesian coordinates:

$\displaystyle \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(x^2+y^2)}\,dx \,dy= \int_{-\infty}^{\infty}e^{-x^2}\,dx \cdot \int_{-\infty}^{\infty}e^{-y^2}\,dy=\left( \int_{-\infty}^{\infty}e^{-x^2}\,dx \right)^2=\pi \implies \displaystyle \int_{-\infty}^{\infty}e^{-x^2}\,dx=\sqrt{\pi}$

The volume in question can be computed via the double integral in cylindrical coordinates:

$V = \int_{0}^{2\pi} \int_{0}^{\infty} e^{-r^2} r dr d\theta = \boxed{\pi}.$