Sketch the vector function and determine the divergence $ \overrightarrow v=\frac{\hat r }{r^2}$
I computed the divergence of this to be zero, but I have 2 questions:


*

*I was supposed to sketch this but my picture was wrong. I drew a point and then drew equally sized arrows outward from the point, but apparently this is wrong. The picture should have arrows that decrease as you move away from the point. Why is this?

*What is supposed to be ''surprising'' about the divergence being zero? The question specifically indicates that something about this should surprise me. 
 A: To answer the first question in the OP, note that the magnitude of the vector $\vec v=\frac{\hat r}{r^2}$ is given by 
$$|\vec v|=\frac{1}{r^2}$$
Inasmuch as $|\vec v|$ decays with increasing $r$, the field lines decay commensurately.

To answer the second question, note that for any closed and smooth surface $S$ that encompasses the origin, the outward flux $\Psi$ is
$$\Psi =\oint_S \vec v\cdot \hat n\,dS=4\pi$$
This is surprising since $\nabla \cdot \vec v=0$ for all $r\ne 0$.  Therefore, in terms of both Lebesgue and Riemann integrals, $\int_V \nabla \cdot \vec v\,dV=0$.  But naïve application of the divergence theorem would show that
$$0=\int_V \nabla \cdot \vec v\,dV=\oint_S \vec v\cdot \hat n\,dS=4\pi \tag 1$$
The reason that the logic used in $(1)$ is flawed is that the divergence theorem is inapplicable since $\nabla \cdot \vec v$ does not exist at the origin.


NOTE:  Broadening the Interpretation of $\nabla \cdot \left(\frac{\hat r}{r^2}\right)$ 

In THIS ANSWER, I provided a rigorous development to show that the Dirac Delta can be regularized as
$$\bbox[5px,border:2px solid #C0A000]{\lim_{a\to 0} \nabla \cdot \vec \psi(\vec r;a)\sim 4\pi \delta(\vec r)} \tag 2$$
where 
$$\vec \psi(\vec r;a)=\frac{\vec r}{(r^2+a^2)^{3/2}} \tag 3$$
Note that in $(3)$, $\vec \psi(\vec r;a=0)=\frac{\hat r}{r^2}$ represents the electric field from a point charge $q=4\pi \epsilon_0$.  
Hence, we interpret $(2)$ as a representation of $\nabla \cdot \left(\frac{\hat r}{r^2}\right)$ in the sense of distributions.  The symbol $\sim$ in $(2)$ means that for any suitable test function $\phi(\vec r)$, we have 
$$\lim_{a\to 0}\int_{V} \phi(\vec r)\,\nabla \cdot \vec \psi(\vec r;a)\,dV=4\pi \phi(0)$$
whenever the region $V$ contains the origin $\vec r=0$.
