= Solution
Choose the <plane> normal to be $+\widehat y$. Only the second field component contributes, and on $y=0$ it is $e^{-(x^2+z^2)}$. Thus the <surface integral> factors into two <Gaussian integrals>:
$$
\boxed{\int_SF\cdot dS
=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-x^2-z^2}\,dx\,dz
=(\sqrt\pi)^2=\pi.}
$$
The original question does not specify an orientation; choosing $-\widehat y$ instead gives \b[$-\pi$].
The printed first component $\sin y/y$ is undefined on the entire <integration> <plane>. Its removable continuous extension there is $1$, since $\lim_{y\to0}\sin y/y=1$. The stated <flux> uses this natural extension; in fact the normal component alone is already well-defined and integrable on the <plane>, so the first component does not affect either oriented value.
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