Verify the Divergence Theorem The vector field $  \ \large \vec F=(xy,yz,zx) \ $ , on the closed cylinder $ \ x^2+y^2 \leq 1 , \ \ z=0 , \ \ z \leq 1$. 
Verify the Divergence Theorem. 
Answer:
$ (i) $ we have to calculate here $ \iint_S F \cdot n \ dS$ 
Let $ z=1-x^2-y^2 \ $
Then $ \vec F=(xy,y(1-x^2-y^2) , x(1-x^2-y^2)) \ $
Now , 
$$ F_x=(y,-2xy, 1-3x^2) , \ \ F_y=(x, 1-x^2-3y^2, -2xy) $$
Then, 
$$ F_x \times F_y=(13x^2y^2+3x^4+2x^2-3y^2, \ 2xy^2+x-3x^3, \  x^2y+y-3y^3) $$
Then, 
$$ \iint_S F \cdot n \, dS= \int_{-1}^1 \int_{-1}^1 \vec F \cdot (F_x \times F_y) \, dx\,dy$$
But this becomes complicated .
I am confused . 
Am I right so far . 
Help me out
 A: The iterated integral
$$
\int_{-1}^1 \int_{-1}^1 \vec F \cdot (F_x \times F_y) \, dx\,dy
$$
doesn't make sense.  I think you're confusing the vector field be integrated with a vector function that parametrizes the solid.  If $D$ is a region in the $uv$-plane, and $\vec r \colon D \to S$ parametrizes the solid $S$, then 
$$
    \iint_S \vec F \cdot \vec n\,dS = \iint_{D} \vec F(\vec r(u,v))\cdot (\vec r_u \times \vec r_v) \,dA
$$
But your solid is both more simple and more complicated than that.  
It's more complicated because it has three faces: the flat top and bottom, and the curved sides.  So to set up $\iint_S \vec F \cdot \vec n\,dS$ by parametrizing the surface requires three separate integrals.
It's more simple because the faces are nice and have easy normals.  On the top face, $\vec n = \vec k$, on the bottom face, $\vec n = - \vec k$.  Around the curved sides of the cylinder, the normal at $\vec x$ is $\vec x$ itself.  So
\begin{align*}
    \iint_S \vec F\cdot\vec n\,dS
    &=\iint_{\text{top}}(xy,yz,zx)\cdot \vec k\,dS
    +\iint_{\text{bottom}}(xy,yz,zx)\cdot (-\vec k)\,dS
    +\iint_{\text{sides}}(xy,yz,zx)\cdot (x,y,z)\,dS
\end{align*}
On the top face, $z=1$, and on the bottom face, $z=0$.  So
$$
\iint_S \vec F\cdot\vec n\,dS
    = \iint_{\text{top}}x\,dS + 0 + \iint_{\text{sides}}(x^2y + y^2 z +
 z^2 x)\,dS
$$
By symmetry, everything vanishes except
$$
\iint_S \vec F\cdot\vec n\,dS
    = \iint_{\text{sides}}y^2 z\,dS 
$$
Edit to address the concern OP brought up in the first comment.  The first integral $\iint_{\text{top}}x\,dS$ is essentially the integral of $x$ over the unit disk.  For every point in the disk with a positive $x$ coordinate, the point's reflection in the $y$-axis is also in the disk and has the opposite $x$ coordinates.  So you can cancel all the points on one side with their reflections on the other side.  
The same thing works with $\iint_{\text{sides}}x^2y\,dS$ and $\iint_{\text{sides}}z^2 x\,dS$, using either the $y$ or $x$ coordinate to cancel.  But the remaining term $\iint_{\text{sides}}y^2z\,dS$ does not have that symmetry.
On the sides, $dS = d\theta \,dz$.  So
$$
    \iint_{\text{sides}}y^2 z\,dS 
    = \int_0^{2\pi} \int_0^1 \sin^2\theta\,d\theta\,dz = \frac{\pi}{2}
$$
On the other hand $\nabla\cdot(xy,yz,zx) = y + z + x$.
Let $E$ be the solid cylinder enclosed by $S$.  Then
$$
    \iiint_E \nabla\cdot\vec F\,dV = \iiint_E (x+y+z)\,dV
    = \iiint_E z \,dV
$$
Again by symmetry, the $x$ and $y$ integrals evaluate to zero.
Within the cylinder, $dV = r \,dr\,d\theta\,dz$.  So
$$
    \iiint_E z \,dV = \int_0^{2\pi} \int_0^1 \int_0^1 z r \,dr\,d\theta\,dz = \frac{\pi}{2}
$$
A: You should have started with choosing the paramaterization of your surface. We have three parts. We have the upper part the side part and the bottom part. I only do the side part. A paramterization for the side part is:
\begin{align}
\phi(t,\theta) =
\begin{pmatrix}
\cos \theta\\
\sin\theta\\
t
\end{pmatrix}
\end{align}
$t\in [0,1]$ and $\theta \in[0,2\pi]$. Then the normal vector is:
\begin{align}
\partial_\theta\phi(t,\theta) \times \partial_t\phi(t,\theta) = 
\begin{pmatrix}
-\sin\theta \\
\cos\theta\\
0\\
\end{pmatrix} \times
\begin{pmatrix}
0\\
0\\
1\\
\end{pmatrix}
=
\begin{pmatrix}
\cos\theta \\
\sin\theta \\
0
\end{pmatrix}
\end{align}
So:
\begin{align}
\iint_S \textbf{F}\cdot \textbf{n} dS &= \int^{2\pi}_0 \int^1_0 \textbf{F}(\phi(t,\theta))\cdot( \partial_\theta\phi(t,\theta) \times \partial_t\phi(t,\theta) ) dt d\theta\\
&=  \int^{2\pi}_0 \int^1_0 \cos^2\theta \sin\theta + t\sin^2\theta dtd\theta\\
&= \frac{\pi}{2}
\end{align}
For the last integral you need to use the trig identities as usual. As I already said I'll leave the top part and the bottom part for you. 
The divergence:
\begin{align}
\iiint_K \text{div} \textbf{F} dV 
\end{align}
Where $K=\{ (x,y,z) | x^2+y^2\leq 1 , 0\leq z\leq 1\} $.  Using polar coordinates:
\begin{align}
\iiint_K \text{div}\textbf{F} dV &=\iiint_K y+z+x dV\\
&=\int^1_0\int^1_0 \int^{2\pi}_0 (r\cos\theta+r\sin\theta + z)r d\theta dr dz\\
&= \frac{\pi}{2}
\end{align}
I assume integrating that is not a problem. Now do the bottom part and the top part of the surface integral to verify the Divergence Theorem. Let me know if there is something unclear.
