The ends of a circular cylinder of radius $a$ and axis $OZ$ are in the planes $z = 0$, $z = 2a$. If $S$ denotes the complete surface of the cylinder (including the plane ends), then how to evaluate the surface integral $$ \int\int_S z \, dx \, dy? $$
My Attempt:
On the surface $S_1$ of the cylinder, we have $$ \mathbf{r} = \big( a \cos \phi, a \sin \phi, z \big), $$ where $$ 0 \leq \phi \leq 2 \pi, \ 0 \leq z \leq 2a. $$ Then we have $$ \begin{align} \mathbf{r}_\phi &= \big( -a \sin \phi, \, a \cos \phi, \, 0 \big), \\ \mathbf{r}_z &= \big( 0, \, 0, \, 1 \big), \end{align} $$ and so $$ \mathbf{r}_\phi \times \mathbf{r}_z = \big( a \cos \phi, \, a \sin \phi, \, 0 \big). $$ Therefore the element $dS_1$ of the surface area is given by $$ dS_1 = \left| \mathbf{r}_\phi \times \mathbf{r}_z \right| = a, $$ and so \begin{align} \int\int_{S_1} z \, dS_1 &= \int_{z=0}^{z=2a} \int_{\phi = 0}^{\phi=2\pi} az \, d\phi \, dz \\ &= \int_{z=0}^{z=2a} 2\pi a z \, dz \\ &= \pi a \left[z^2 \right]_{z=0}^{z=2a} \\ &= 4 \pi a^3. \end{align}
On the surface $S_2$ of the end $z = 0$, we have $$ \mathbf{r} = \big( r \cos \phi, \, r \sin \phi, 0 \big), $$ where $$ 0 \leq r \leq a, \ 0 \leq \phi \leq 2 \pi. $$ So we have \begin{align} \mathbf{r}_r &= \big( \cos \phi, \, \sin \phi, \, 0 \big), \\ \mathbf{r}_{\phi} &= \big( -r \sin \phi, \, r \cos \phi, \, 0 \big), \end{align} and thus $$ \mathbf{r}_r \times \mathbf{r}_{\phi} = \big( 0, \, 0, \, r \big). $$ Thus the element $dS_2$ of the surface area is given by $$ dS_2 = \left| \mathbf{r}_r \times \mathbf{r}_{\phi} \right| = r, $$ and \begin{align} \int \int_{S_2} z \, dS_2 &= \int_{r=0}^{r=a} \int_{\phi=0}^{\phi = 2\pi} 0 r \, d\phi \, dz \\ &= 0. \end{align}
Finally, on the surface $S_3$ of the end $z = 2a$, we have $$ \mathbf{r} = \big( r \cos \phi, \, r \sin \phi, 2a \big), $$ where $$ 0 \leq r \leq a, \ 0 \leq \phi \leq 2 \pi. $$ So we have \begin{align} \mathbf{r}_r &= \big( \cos \phi, \, \sin \phi, \, 0 \big), \\ \mathbf{r}_{\phi} &= \big( -r \sin \phi, \, r \cos \phi, \, 0 \big), \end{align} and thus $$ \mathbf{r}_r \times \mathbf{r}_{\phi} = \big( 0, \, 0, \, r \big). $$ Thus the element $dS_3$ of the surface area is given by $$ dS_3 = \left| \mathbf{r}_r \times \mathbf{r}_{\phi} \right| = r, $$ and \begin{align} \int \int_{S_3} z \, dS_3 &= \int_{r=0}^{r=a} \int_{\phi=0}^{\phi = 2\pi} 2ar \, d\phi \, dz \\ &= \int_{r=0}^{r=a} 4\pi a r \, dr \\ &= 2 \pi a \left[ r^2 \right]_{r=0}^{r=a} \\ &= 2 \pi a^3. \end{align}
Hence the total surface integral is given by $$ \int\int_S z \, dS = \int\int_{S_1} z \, dS_1 + \int\int_{S_2} z \, dS_2 + \int\int_{S_3} z \, dS_3 = 4\pi a^3 + 0 + 2\pi a^3 = 6\pi a^3. $$
Is this solution correct and clear enough in each and every detail? Or, are there any problems?
It's quite clear and systematic to me. Interestingly we can evaluate as the following for this particular case; define $z'=z-a$ $$ \int\int_{S}z\phantom{x}dx\phantom{.}dy\phantom{x}=\phantom{x}\int\int_{S}z'\phantom{x}dx\phantom{.}dy\phantom{x}+a\int\int_{S}\phantom{x}dx\phantom{.}dy $$
The first term on the right evaluates to zero because of symmetry. The second term is simply product between $a$ and the cylinder's surface area.