ergospherical said:
For 30 I observe that ##\nabla \cdot F = 2z##, so by theorem of Gauss ##\int_A \langle F, n \rangle dS = \int 2z dV +\cancel{123\int_{\mathrm{cover}} dS} - \cancel{123\int_{\mathrm{base}} dS} = \pi R^2 h^2##.
(Directly, one could parameterise ##A## by ##\Phi(\theta, z) = (R\cos{\theta}, R\sin{\theta}, z)## to obtain a surface element ##n dS = R(\cos{\theta}, \sin{\theta}, 0)d\theta dz## and then an integral ##\int_A \langle F, n \rangle dS = R^2 \int z dz \int d\theta = \pi R^2 h^2##).
I guess it will do no harm to elaborate on these two solutions for those who do not "see" it at once.
One possible parameterization is
\begin{align*}
\phi\, : \,[0,2\pi) \times [0,h]&\longrightarrow A\\
(\varphi,z)&\longmapsto (R\cos\varphi ,R\sin\varphi ,z)
\end{align*}
The height of the cylinder is given by ##z,## and for a fixed ##z## we have a circle parallel to the plane ##z=0## described by polar coordinates. ##\phi## is a bijection because every point on ##A## has exactly one pair of parameters ##(\phi,z).##
We use this parameterization ##\phi## to calculate the surface integral.
\begin{align*}
\int_A \langle F,n \rangle\,d^2r&=\int_{0}^{2\pi}\int_{0}^{h}\langle F\circ\phi,n \rangle\,\|\partial_\varphi \phi \times \partial_z \phi\| \,dz\,d\varphi \\
&=\int_{0}^{2\pi}\int_{0}^{h}\langle F\circ\phi,\partial_\varphi \phi \times \partial_z \phi \rangle\, \,dz\,d\varphi \\
&=\int_{0}^{2\pi}\int_{0}^{h}\langle \begin{pmatrix}
zR\cos\varphi \\zR\sin\varphi \\123 \end{pmatrix},\begin{pmatrix}
R\cos\varphi \\R\sin\varphi \\0 \end{pmatrix} \rangle\, \,dz\,d\varphi \\
&=\int_{0}^{2\pi}\int_{0}^{h} zR^2\left(\cos^2\varphi +\sin^2\varphi \right)\,dz\,d\varphi \\
&=\int_{0}^{2\pi}\int_{0}^{h} zR^2\,dz\,d\varphi = 2\pi R^2\left[\dfrac{z^2}{2}\right]_{z=0}^{z=h}=h^2\pi R^2
\end{align*}
We could alternatively use Gauß's divergence theorem. This uses closed surfaces, so we have to consider base and cover. Let ##C## be the volume of the cylinder, ##D_1## its base, and ##D_2## its cover. Then ##\partial Z=A\cup D_1\cup D_2.## Note that the two integrals over ##D_1## and ##D_2## cancel each other since the normal vectors ##n## are parallel to the ##z##-axis, but pointing into opposite directions, and the third component of ##F## is constant.
\begin{align*}
\int_{D_1} \langle F,n \rangle\,d^2r + \int_{D_2} \langle F,n \rangle \, d^2r &= \int_{D_1}\langle F,\begin{pmatrix}
0\\0\\1 \end{pmatrix} \rangle \,d^2r+ \int_{D_2}\langle F,\begin{pmatrix}
0\\0\\-1 \end{pmatrix} \rangle\,d^2r\\&=
\int_{D_1}123\,d^2r + \int_{D_2}-123\,d^2r \\&=123(\pi R^2-\pi R^2)=0
\end{align*}
Therefore
$$
\int_{\partial Z}\langle F,n \rangle\,d^2r = \int_{A}\langle F,n \rangle\,d^2r+\int_{D_1}\langle F,n \rangle\,d^2r+\int_{D_2}\langle F,n \rangle\,d^2r=\int_{A}\langle F,n \rangle\,d^2r
$$
Now we apply Gauß's divergence theorem
\begin{align*}
\int_{\partial Z}\langle F,n \rangle\,d^2r&=\int_Z\operatorname{div} F\,d^3r\\&=
\int_{0}^{R}\int_{0}^{2\pi}\int_{0}^{h}(\operatorname{div}F)\circ \phi \cdot |\det J_\phi|\,dz\,d\varphi \,d\rho\\
&=\int_{0}^{R}\int_{0}^{2\pi}\int_{0}^{h} 2z\rho \,dz\,d\varphi \,d\rho\\
&=2\cdot \dfrac{R^2}{2}\cdot\dfrac{h^2}{2}\cdot 2\pi = h^2\pi R^2
\end{align*}
