Answer
$$ 2\pi $$
Work Step by Step
Apply the spherical coordinates system as:
$x=\rho \sin \phi \cos \theta \\ y=\rho \sin \phi \sin \theta \\z=\rho \cos \phi$;
So, $\rho=\sqrt {x^2+y^2+z^2} \implies \rho^2=x^2+y^2+z^2$
The jacobian for spherical coordinates is
$\rho^2 \sin \phi$
Therefore,
$Volume= \iiint_{V} dV=\int_0^{\pi} \int_0^{\pi} \int_{0}^{\infty} \rho^3 \sin \phi e^{-\rho^2}d\rho d \phi \ d\theta \\=\int_0^{2 \pi} \int_0^{\pi} \sin \phi \times \dfrac{-(\rho^2+1)e^{-\rho^2}{2}}|_0^{\infty} d \phi \ d\theta \\=\int_0^{2 \pi} 1 d\theta \\= 2\pi $
