Let $t \in A:= \Bbb{R}\setminus\{2k\pi,2k\pi +\pi/2:k \in Z\}$
$x\log(xy)=x\log{x} +x\log{y}$ thus for $x=r\cos{t}$ and $y=r \sin{t}$
we have that $f(r,t) \to 0$ as $r \to 0$ using the fact that $r\log{r} \to 0$ as $r \to 0$
Since $t$ is arbitrary in $A$ we conclude that the limit exists.
Applying divergence theorem, you get flux over closed surface $S_2$.
In spherical coordinates, $x^2 + y^2 + z^2 = \rho^2$
where $x = \rho \cos\theta \sin\phi, y = \rho \sin\theta \sin\phi. z = \rho \cos\phi$
As it is part of the sphere above $z = 0, \phi \leq \pi/2$.
So the volume integral is,
$ \displaystyle \int_0^{2\pi} \int_0^{\pi/2} \int_0^1 \rho^4 \sin\phi ~d\rho ~d\phi ~d\theta = \frac{2\pi}{5}$
Now the unit normal vector to the disk surface $S1: x^2 + y^2 \leq 1, z = 0 ~$ is $~\hat n = (0, 0, -1)$
$\vec F \cdot \hat n = - y^2$
Using polar coordinates, the surface integral over disk surface is,
$ \displaystyle \int_0^{2\pi} \int_0^1 - r^3\sin^2\theta ~ dr ~ d\theta = - \frac{\pi}{4}$
So flux over spherical surface $S$ is,
$ \displaystyle \frac{2\pi}{5} - \left(- \frac{\pi}{4}\right) = \frac{13\pi}{20}$
Best Answer
$\sqrt{2}$ comes from the equation: \begin{align} &\cos(\theta)+\sin(\theta)\\ =&\sqrt{2}[\frac{\sqrt{2}}{2}\cos(\theta)+\frac{\sqrt{2}}{2}\sin(\theta)]\\ =&\sqrt{2}[\cos(\theta)\sin(\frac{\pi}{4})+\sin(\theta)\cos(\frac{\pi}{4})] \end{align}