= Solution
The function $h$ is measurable and nonnegative. In first-quadrant polar coordinates, including the Jacobian, its integral is
$$
\left(\int_0^\infty e^{-r^2}r^3dr\right)\left(\int_0^{\pi/2}(\cos\theta)^{3/2}(\sin\theta)^{1/2}d\theta\right).
$$
The radial factor is $1/2$ by the supplied fact and the angular integrand is at most one, giving finiteness. The nonnegative change-of-variables formula is valid before finiteness is known, and this bound proves $h\in L^1$.
Fubini may now factor the Cartesian integral. Substituting $u=x^2$, $v=y^2$ in the respective factors gives
$$
\iint h\,dxdy=\frac14\left(\int_0^\infty e^{-u}u^{1/4}du\right)\left(\int_0^\infty e^{-v}v^{-1/4}dv\right).
$$
Both are finite: their powers at zero exceed $-1$, and the exponential controls infinity. Comparing with the polar expression proves
$$
\boxed{\int_0^{\pi/2}(\cos\theta)^{3/2}(\sin\theta)^{1/2}d\theta=\frac12\Gamma(5/4)\Gamma(3/4),}
$$
which is exactly the product of integrals requested.
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