$\int_{S} v.n dS = ?$ where $v = xz \hat{i} + 2yz\hat{j} + 3xy\hat{k}$ Let $n(x,y,z)$ be the unit outward normal of the surface $S$ of the cylinder $x^2 + y^2 \leq 1 , 0 \leq z \leq 3$.
We have to compute $\int_{S} v.n dS = ?$ where $v = xz \hat{i} + 2yz\hat{j} + 3xy\hat{k}$.
the unit normal vector is given by the gradient of $S$, But how to find the gradient I think I would take gradient of $x^2 + y^2 -4$ that is $(2x,2y,0)$.
Now how do we evaluate $\int_{S} v.n dS = ?$ that is $\int_{x}\int_{y} (2x^2z + 4y^2z dx dy$ where $x,y$ vary such that $x^2 + y^2 \leq 4$ ?
 A: The integral breaks up into a top and bottom caps and the side of the cylinder.  For the side the normal vector is clearly $\hat \rho$, while it is $\pm \hat z$ for the caps.  You might want to convert $\hat \rho$ to Cartesian.  
Note that the points on the side of the cylinder are of the form $2(\cos\phi,\sin\phi,z_0)$ so a unit tangent vector is $(-\sin\phi,\cos\phi,0)$ and a normal would be $(\cos\phi,\sin\phi,0)$ which you may convert to $x,y$.
However, it is simplest here to compute the dot product in Cartesian and the convert to cylindrical since your surface element (for the side) is then simply $2d\phi dz$.
A: This answer shall be more explicit than my predecessors but I agree with the calculations of ZeroTheHero. There are two clear alternatives: primarily I suggest using the divergence theorem which is made advantageous by the form of $\vec{v}$ and the closed nature of the surface, alternatively the surface integral may be evaluated explicitly. Since the previous answer detailed the latter alternative I shall explicitly proceed along this path before presenting the former alternative to check my computation.
The surface integral over the closed cylinder with radius 2 ($x^2+y^2\le4$) and height 3 ($0\ge z \le 3$) oriented with an outward normal may be split in three parts
$$
\oint_S \vec{v}.\hat{n} \ dS=\bigg[\underbrace{\int_{Top}}_{=T} + \underbrace{\int_{Cylinder}}_{=C} + \underbrace{\int_{Bottom}}_{=B} \bigg] \vec{v}.\hat{n} \ dS,
$$
where $\vec{v}=xz\hat{i}+2yz\hat{j}+3xy\hat{k}$. The unit normal on the cylinder is $\hat{\rho}$ by definition while on the top and bottom of the cylinder the normal is $\pm\hat{z}$. The cylindrical symmetry of the surface suggests a cylindrical parametrisation of the region of integration. Under such a parametrisation the Jacobian implies $dS|_{Top}=\rho d\rho d\phi$ while $dS|_{Bottom}=\rho d\rho d\phi$ and $dS|_{Cylinder}=2 d\phi dz$. I shall now address each integral in turn firstly
$$
T=\int_{Top}  \vec{v}.\hat{z} \ dS,
\\=\int_{0}^{2\pi}\int_{0}^2  \vec{v}(\rho,\phi,z).\hat{z} \ \rho d\rho d\phi,
\\=3\underbrace{\int_{0}^{2\pi}\cos{(\phi)}\sin{(\phi)}\ d\phi}_{=0}\int_{0}^2  \rho^3\ d\rho,
\\=0,
$$
secondly
$$
B=-\int_{Bottom}  \vec{v}.\hat{z} \ dS,
\\=-\int_{0}^{2\pi}\int_{0}^2  \vec{v}(\rho,\phi,z).\hat{z} \ \rho d\rho d\phi,
\\=-3\underbrace{\int_{0}^{2\pi}\cos{(\phi)}\sin{(\phi)}\ d\phi}_{=0}\int_{0}^2  \rho^3\ d\rho,
\\=0,
$$
thirdly (the surface integral which cannot be reduced to triviality by considering its integrand as the product of odd and even functions)
$$
C=\int_{Cylinder}  \vec{v}.\hat{\rho} \ dS,
\\=2\int_{0}^3 \int_{0}^{2\pi} (2z\cos{(\phi)}, 4z\sin{(\phi)}, 12\cos{(\phi)}\sin{(\phi)}).(\cos{(\phi)},\sin{(\phi)},0) \ d\phi dz,
\\=2\int_{0}^3 \int_{0}^{2\pi} (2z\cos^2{(\phi)}+4z\sin^2{(\phi)}) \ d\phi dz,
\\=4\int_{0}^3 \int_{0}^{2\pi} (z\underbrace{(\cos^2{(\phi)}+\sin^2{(\phi)})}_{=1}+z\sin^2{(\phi)}) \ d\phi dz,
\\=4\int_{0}^3 z dz \int_{0}^{2\pi} (1+\sin^2{(\phi)}) \ d\phi,
\\=4\times\frac{9}{2}\times\int_{0}^{2\pi} \bigg[\frac{3}{2}-\frac{1}{2}\bigg(1-2\sin^2{(\phi)}\bigg)\bigg] \ d\phi,
\\=4\times\frac{9}{2}\times\bigg[\int_{0}^{2\pi} \frac{3}{2} \ d\phi-\underbrace{\int_{0}^{2\pi} \frac{1}{2}\cos{(2\phi)} \ d\phi}_{=0}\bigg],
\\=9\times3\times2\pi,
\\=54\pi.
$$ 
Hence 
$$
\oint_S \vec{v}.\hat{n} \ dS=54\pi.
$$
To check this answer or the surface integral may be evaluated using the divergence theorem
$$
\oint_S \vec{v}.\hat{n} \ dS=\int_V \vec{\nabla}.\vec{v} \ dV,
\\=\int_V 3z \ dV,
\\=\int_0^{3} \int_0^{2\pi} \int_0^2 3z \ \rho d\rho d\phi dz,
\\=3\int_0^{3} z \ dz\int_0^{2\pi} d\phi\int_0^2 \rho \  d\rho,
\\=3\times\frac{9}{2} \times2\pi\times\frac{2^2}{2},
\\=3\times9\times2\pi,
\\=54\pi,
$$
as shown by the previous method. Clearly the latter alternative is much quicker and for that reason I recommend you use the divergence theorem.
