How can a density be larger than $1$? From Frank Morgan: Geometric Measure Theory, Fourth Edition: A Beginner's Guide, page 13,the $2$-dimensional density of the cone $x^2+y^2=z^2$ at $0$ is $\sqrt{2}$.  I feel strange of that，roughly speaking，the density is 
$$\lim \frac{m(E \cap B)}{m(B)}$$
how is it larger than $1$? 
 A: I just checked the book and I found out that the author defines the $m$-dimensional density of a set $E\subset \mathbb{R}^n$ near a point $x$ for $m\le n$ by
$$\lim_{r\to 0} \frac{\mathcal{H}^m(E\cap B^n(x,r))}{|B^m(0,r)|}$$
Note that in the numerator, the $n$-dimensional ball appears, whereas in the denominator it's the $m$-dimensional ball! For $m=n$ the density becomes
$$\lim_{r\to 0} \frac{|E\cap B(x,r)|}{|B(x,r)|}$$
and this is indeed $\le 1$, but for $m<n$ it may be greater than 1. This occurs for example if several surfaces intersect at some point, and in this case the density counts the number of such surfaces. The situation for the vertex of the cone is similar.
A: Roughly speaking, the answer is "because you have a (bi)cone". 
Any smooth embedded manifold (in fact any $C^2$ manifold) will have the property that the appropriately defined density is precisely equal to $1$. But if you have something that is not smooth, you can have other values. 
If you have just a single cone $z = \sqrt{x^2 + y^2}$, near the origin you see that locally there is a constant angle defect: a circle of radius $r$ from the origin on the cone has circumference $\sqrt{2}\pi r < 2 \pi r$. For smooth manifolds the angle defect is related to the integral of the curvature inside the circle, so if the curvature is a continuous function (which can be guaranteed by the manifold being $C^2$), the angle defect must become zero as the circle shrinks. The single cone, on the other hand, has a curvature singularity at the origin, which allows it to have an angle defect even as the circle shrinks to a point. 
Notice that a computation will show you that the single cone has density $1/\sqrt{2}$ at the origin. (The double cone, then, will have twice that.)
Remark 1 It is not necessary to have a multisheeted surfaces to get density $>1$. Consider $\mathbb{R}^3$ in spherical coordinates $(r,\phi,\theta)$. Define a surface by $\theta = f(\phi)$ with $|f(\phi)| < \pi/2$, this will by definition be a graph over $\mathbb{R}^2$ and thus single sheeted. Choose $f(\phi) = \sin (k\phi)$ for $k\in\mathbb{Z}$. The circumference of the unit circle then is given by 
$$ \int_0^{2\pi}\sqrt{1 + k^2\cos^2(k\phi)}\cos\circ\sin(k\phi)d\phi > 2\pi $$
if $|k| > 1$. So one gets that the graph defines a $C^0$ embedded surface with density $> 1$.
Remark 2 Very roughly, the density is a statement of how "crunched up" our set is in an infinitesimal setting.  You should compare the notion, and the statements given in Florian's answer, to the notion of fractal dimension. 
A: Probably what is going on (I do not have access to the book) is the author is using a different definition of density.  In geometric measure theory, we care about sets $E \subset \mathbb{R}^d$ which have, for example, positive (upper) Lebesgue density:
$$
\overline{D}(E) = \limsup_{R \to \infty} \frac{m(E \cap B_R)}{m(B_R)}.
$$
However, some authors (and I suspect Morgan is among them) define positive upper Lebesgue density as
$$
\overline{d(E)} = \limsup_{R \to \infty} \frac{m(E \cap B_R)}{R^d}.
$$
Notice that $\overline{D}(E) > 0$ if and only if $\overline{d}(E) > 0$.  Also, $\overline{D}(E) \in [0,1]$, whereas $\overline{d}(E)$ can in fact be larger than $1$.
