What does it mean to be divergence thorem applicable? To be able to use divergence theorem, $S$ needs to be piecewise smooth curved, and $\vec{F}(x,y,z)$ needs to be continuously differentiable.
$$\int \int_{S} \vec{F} * \vec{n} dS$$ where
$$\vec{F}(x,y,z) = <yz\tan^{-1}(y^2 + z^2) - 2x, xze^{x-z} - 2y, \sin(x^2+y)+5z>$$
and S is the sphere of radius 1 centered at the origin.
How do I check if they are piecewise smooth curved without looking at the graph? And how do I check if $\vec{F}(x,y,z)$ is continuously differentiable? 
Could someone show steps to checking?
 A: With regards to your question about why $F$ is differentiable.  
It could depend on the development in a course or textbook. This makes the question more difficult. I may make explicit and implicit assumptions as I go. A lot of details may need to be filled in, but I hope you find it a decent outline of some main ideas that could occur in some particular approach to the problem. 
I will assume that we know that functions such as
\begin{align}
\pi_1:\mathbb R^3&\to\mathbb R\\
(x,y,z)&\mapsto x
\end{align}
are continuously differentiable. For instance, consider the definition at http://en.wikipedia.org/wiki/Differentiable_function#Differentiability_in_higher_dimensions
I am going to replace the word continuously differentiable with morphism, and not worry if I am justified in this. 
I will assume that the product of two morphisms is also a morphism. 
I will assume that if $f$, $g$ and $h$ are morphisms from $X$ to $\mathbb R$, then the map 
\begin{align}
F:X&\to\mathbb R^3\\
x&\mapsto(f(x),g(x),h(x))
\end{align}
is also a morphism. 
Thus, you may note that each of 
\begin{align}
f,g,h:\mathbb R^3&\to\mathbb R\\
(x,y,z)&\overset{f}\mapsto yz\tan^{−1}(y^2+z^2)−2x,\\
(x,y,z)&\overset{g}\mapsto xze^{x−z}−2y,\\
(x,y,z)&\overset{h}\mapsto \sin(x^2+y)+5z
\end{align}
is a morphism. 
Finally, we note that $F=(f,g,h)$ is a morphism. 
