\[S\]
is \[\int_{S} \mathbf{F} \cdot \mathbf{n} dS\]
.
If \[S\]
is the surface \[x^2 +y^2 +z^2 =9\]
then \[\mathbf{n} = \frac{2x \mathbf{i} + 2y \mathbf{j} + 2z \mathbf{k}}{\sqrt{4x^2 + 4y^2 = 4z^2}} = \frac{x \mathbf{i} + y \mathbf{j} + z \mathbf{k}}{r}\]
and \[\mathbf{F} = \frac{x}{\sqrt{2}} \mathbf{i} + \frac{y}{\sqrt{2}} \mathbf{j} + \frac{z}{\sqrt{2}} \mathbf{k}\]
The integral is
\[\begin{equation} \begin{aligned} \int_S \mathbf{F} \cdot \mathbf{n} dS &=
(\frac{x}{\sqrt{2}} \mathbf{i} + \frac{y}{\sqrt{2}} \mathbf{j} + \frac{z}{\sqrt{2}} \mathbf{k}) \cdot (\frac{x}{r} \mathbf{i} + \frac{y}{r} \mathbf{j} + \frac{z}{r} \mathbf{k}) \\ &=
\int_S \frac{x^2 +y^2 + z^2}{r \sqrt{2}} dS \end{aligned} \end{equation}\]
In spherical polar coordinates the integral becomes
\[\begin{equation} \begin{aligned} A &=
\int^{\pi}_0 \int^{2 \pi}_0 \frac{r^2}{r \sqrt{2}} r^2 sin \theta d \theta d \phi \\ &=
\frac{2 \pi r^3}{\sqrt{2}} \int^{\pi}_0 sin \theta d \theta \\ &=
\frac{2 \pi r^3}{\sqrt{2}} [- cos \theta]^{\pi}_0 \\ &=
\frac{2 \pi r^3}{\sqrt{2}}(-cos \pi -(-cos 0)) \\ &=
\frac{108 \pi r^3}{\sqrt{2}} \end{aligned} \end{equation}\]