= Solution
The <function> is nonnegative and <continuous> on the positive quadrant. In <polar coordinates> its product integral is, by the nonnegative <change of variables formula> and <Tonelli's theorem>,
$$
\int_0^{\pi/2}\int_0^\infty e^{-r^2}r^3(\cos\theta)^{3/2}(\sin\theta)^{1/2}\,dr\,d\theta.
$$
The angular factor is at most one, while the supplied radial integral is $1/2$, so the product integral is at most $\pi/4$ and is finite. This establishes integrability before any signed interchange is used.
Let $A$ denote the angular integral. The polar evaluation is $A/2$. In Cartesian coordinates <Fubini's theorem> instead factors the integral as
$$
\left(\int_0^\infty e^{-x^2}x^{3/2}\,dx\right)
\left(\int_0^\infty e^{-y^2}y^{1/2}\,dy\right)
=\frac14\left(\int_0^\infty e^{-u}u^{1/4}\,du\right)
\left(\int_0^\infty e^{-v}v^{-1/4}\,dv\right),
$$
using $u=x^2$, $v=y^2$. All factors converge: their powers exceed $-1$ at zero and the exponential controls infinity. Equating the evaluations gives
$$
\boxed{A=\frac12\left(\int_0^\infty e^{-u}u^{1/4}\,du\right)\left(\int_0^\infty e^{-v}v^{-1/4}\,dv\right)}.
$$
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