Size of the Observable Universe I wanted to know what the observable universe is so I was thinking and I thought, it must be age of the universe times 2.
Well I was wrong. I found on one website that it is 46B LY across in each direction. How does this make sense?
I get how the universe has expanded since then, but we should only be able to see light that is 13.7 billion LY old. Does this mean that the Universe is expanding faster than the speed of light? Or light from other objects is travelling to us faster than speed of light?
 A: Explanatory Framework: Foliated Spacetime
In General Relativity, which is used to describe the universe on cosmological length scales, spatial and temporal distances are no longer absolute quantities. Furthermore, in astronomy, there are several methods to determine a distance, which could disagree on cosmological length scales. 
Therefore, it is convenient to initially visualize the problem in a framework that makes it easier to understand the involved quantities and the concrete question.
In the 3+1 formalism of General Relativity, spacetime is described as a foliation of spatial, i.e. 3-dimensional, hypersurfaces along the time-axis. In the figure, each slice $\Sigma_t$ represents 3-dimensional space at a given time $t$. 

Looking into the past
Since light is locally always traveling with the speed of light, $\mathrm{c}$, on large length scales it becomes important that light that reaches the observer on Earth at a time $t_0$ ("today") has been emitted at its origin at a time $t_e:<t_0$, i.e. earlier than $t_0$. 
Expansion of the universe
Experimentally, by measuring the speed of distant objects relative to us and by measuring their distance, one finds that objects which are farther away move away from us with a greater velocity. This is called Hubble’s Law.
$$
v=H_0 \cdot D  \tag{Hubble’s Law}
$$
$\displaystyle v:\equiv\frac{dD}{dt}$ is the velocity of the object relative to us. $D$ is the proper distance to the object, which is the distance within one spatial slice at a given time (as seen in the figure). And $H_0$, for historical reasons, is called Hubble constant. But, actually, $H(t)$ is a function of time:
Parametrizing an expansion of the universe, where the proper distance $D(t)$ between two objects is some time-fixed distance, the comoving distance, $x$, multiplied with time-dependant scale factor $R(t)$, (i.e. all distances grow as the universe expands with growing scale factor $R(t)$),
$$
D(t) = x \cdot R(t) ,
$$
one finds, that the Hubble parameter $H(t)$ is actually the relative rate of expansion at time $t$:
$$
v(t) \equiv \frac{dD(t)}{dt} = \frac{d}{dt} \left( x \cdot R(t) \right) 
= x \cdot \frac{dR(t)}{dt} = \frac{D(t)}{R(t)} \cdot \frac{dR(t)}{dt}
\equiv \frac{\dot R(t)}{R(t)} \cdot D(t) \equiv H(t) \cdot D(t)
$$
Writing the very left and the very right expression together, $v(t) = H(t) \cdot D(t)$, one sees, that Hubble’s Law, $v = H_0 \cdot D$, describes a special case, namely the situation "today": $v(t_0) = H(t_0) \cdot D(t_0)$, where $H(t_0)\equiv H_0 = \dot R(t_0) / R(t_0)$.
Superluminal velocity

Does this mean that the Universe is expanding faster than the speed of light?

In a way, that is correct. The proper distance $D$ between a distant object and us can grow more rapidly than the speed of light. But, that is not because objects would move locally faster than the speed of light. 
$$
\frac{dD}{dt} = \frac{dx\,R}{dt} + \frac{x\,dR}{dt}
$$
In the formula above, the term $\displaystyle \frac{dx\,R}{dt}$ can be interpreted as the local velocity, or peculiar velocity, the term $\displaystyle \frac{x\,dR}{dt}$ as the part of the apparent velocity that is caused by the expansion of space.
In this formalism, the statement that nothing can move faster than the speed of light, would mean that nothing can locally move faster than the speed of light:
$$
v_\text{local} := \frac{dx\,R}{dt} \leq \mathrm{c}
$$
But nothing prevents the universe from expanding more rapidly than the speed of light, i.e. prevent the scale factor $R(t)$ from growing. 
Therefore, given that $v_\text{local}$ has to be smaller than the speed of light, the "velocity" $\displaystyle \frac{dD(t)}{dt}$, observed on Earth, which is corresponding to the proper distance, $D(t):=x\cdot R(t)$, can still be larger than the speed of light.
Not entirely accurate, but for demonstration, one can visualize a balloon with coins glued to its surface. As you pump the balloon up, the proper distance between the coins grows, but locally the size of the coins stays the same.
Age of the universe
In cosmology, one can parametrize the Hubble parameter $H(t)$ by cosmological parameters, which can be measured, experimentally, by various methods. Therefore, one knows $H$ as a function of the scale factor $R(t)$ and those cosmological parameters.
$$
H=H(t)=H(R(t), \text{several cosmological parameters})
$$
Discussing Hubble’s law above, we have seen that $\displaystyle H(t)=\dot R(t)/R(t) = \frac{dR(t)}{dt}/R(t)$. Solved by $dt$, that reads $\displaystyle dt=dR \cdot \frac{1}{HR}$.
Integrating over each infinitesimal time interval $dt$ from the big bang ($t=0$) to today ($t=t_0$) gives the age of the universe $t_0$:
$$
t_0 = \int_0^{t_0} dt
= \int_0^{R(t_0)} dR\,\frac{1}{R\,H(R, \text{cosm. param.})}
\approx 13.7\,\text{Gyr}
$$
Size of the observable universe
We can only observe the universe by looking at particles, e.g. photons, i.e. light, that reaches us. Since locally nothing can travel faster than the speed of light, the distance that light could have travelled within the age of the universe, $t_0$, determines the size of the observable universe.
The distance to the particle horizon, $r_p$, is the distance of an object that has emitted particles (light), which reach us today, and were emitted at $t=0$, i.e. one age of the universe ago.
Which kind of distance? Proper distance, $D:=R\,x$, or comoving distance, $x$? 
A reasonable thing would be to ask for the size of the universe as it is today, i.e. to ask for the proper distance, as shown in the figure.
But, most commonly, the scale factor $R(t)$ is defined such that $R(t_0)=1$. Therefore, if we ask for today’s particle horizon, $r_p$ ($t=t_0$), there is no difference.
$$
r_p := D_p(t_0) = x_p \, R(t_0), \ \ \ R(t_0)=1
$$
How to calculate the proper size of the universe

I found on one website that it is 46B LY across in each direction

This quantity refers to the distance to the particle horizon, $r_p$, i.e. to the radius of the observable universe. The diameter would be twice as large.
$$
\text{proper radius of the observable universe} = r_p \approx 46\,Gly
$$
This can be calculated similarly to calculating the age of the universe as shown above.
$$
r_p = D_p(t_0) = R(t_0)\cdot x_p = R(t_0)\int_0^{x_p} dx
$$
We have seen that $\displaystyle v_\text{local} := \frac{dx\,R}{dt} \leq \mathrm{c}$. For light, we know, $v_\text{local}=\mathrm{c}$. Therefore, for light, $\displaystyle dx = \frac{\mathrm{c}\,dt}{R}$.
$$
r_p = R(t_0)\int_0^{x_p} dx = R(t_0)\int_0^{t_0} \frac{\mathrm{c}\,dt}{R(t)}
= R(t_0)\int_0^{R(t_0)} dR \frac{\mathrm{c}}{R} \frac{1}{R\,H(R, \text{cosm. param.})}
\approx 46\,Gly
$$
Size of the rest of the universe
Since the particle horizon represents the limit from where any information can have reached us, we, in principle, cannot tell what is behind that horizon. Therefore, one cannot know the actual size of the rest of the universe. One cannot even know for sure whether there exists a rest of the universe---but it is a convenient assumption.
But models of the Inflationary Expansion of the universe of an early age suggest that the actual universe is considerably larger than the observable universe.
These models are strong in explaining structure formation of the universe, i.e. how and on which length scales clusters of galaxies etc. could have formed, despite the universe being homogeneous on larger scales. They also solve the cosmological fine-tuning problem, or flatness problem: 
The universe appears flat as far as we can see (i.e. not geometrically curved, i.e. the sum of angles in a triangle is 180 degrees) on large observable scales, despite the fact that it would be more probable that the universe is curved.
The models of inflation solve this problem by suggesting that the actual universe could actually be curved if it would be large enough: Then, the observable part of the universe would be sufficiently small to appear flat. Like you can describe Earth’s surface as being locally flat even though Earth is a sphere.
But from this analogy one can see that, if these inflationary models are correct, the rest of the universe has to be significantly larger than the observable universe.
Further Reading


*

*Schneider, Introduction to Extragalactic Astronomy and Cosmology, Springer, ch. 4.

*d'Inverno, Introducing Einstein’s Relativity, ch. 23.

*Hobson et al., General Relativity, An Introduction for Physicists, Cambridge University Press, ch. 14. In particular, chapter 14.11 for the different cosmological distance measures.

*Wikipedia article on distance measures in cosmology: Distance Measures (Cosmology)

*Wikipedia article on the size of the observable universe, including common misconceptions: Observable Universe, section Size.

