Why does one work with the Lagrangian density in field theory? Why is it necessary to introduce the Lagrangian density (integral of the Lagrangian over volume) when describing the dynamics of fields? Is there a specific reason for that or just for convenience?
 A: Remember that the fundamental thing is the action $S = \int dt L(q, \dot{q}, t)$. The Lagrangian should always be thought of as an ingredient for defining actions.
The Lagrangian density is the most natural way to describe a field, i.e. something that varies in space. A way of motivating this from the original Lagrangian formalism is to think of the field on a lattice:
$\phi( a n_i \mathbf{e}_i, t) \sim \phi_{n}(t)$, where $a$ is the lattice spacing and $n\in\mathbb{Z}^3$, and you should regard the $\phi_n$ as ordinary dynamical variables. If the fields extend through a 3x3 cube of side $d$, then there are $N = (d/a)^3$ variables. The Lagrangian is now a smooth function $\mathbb{R}^N \to \mathbb{R}$. This is a bit unwieldy, however we expect $L$ to take a somewhat special form on the basis of the universe's translation symmetry:
$L(\{\phi_n(t), \dot{\phi}_n(t)\}) = \sum_n \left[\mathcal{L}_0(\phi_n,  \dot{\phi}_n) + \mathcal{L}_1(\phi_n, \phi_{n+a\mathbf{e}_i}, \dot{\phi}_{n}, \dot{\phi_{n+a\mathbf{e}_i}}) + \ldots \right]$
where $\mathcal{L}_1$ depends on nearest neighbours, $\mathcal{L}_2$ depends on second nearest neighbours and so on. Crucially, none of the $\mathcal{L}$'s depend on where they are evaluated - there is no $n$ index on them - they only care about the value of the field at that point.
A basic assumption of field theory is that fields are local - the energy of a given field configuration should depend only on what that field is doing at a point $\mathbf{x}$, i.e. the energy at a point should depend only on the value of $\phi$ and its deformations.
Now observe that these dependencies on nearest neighbours can be thought of as discretised derivatives:
$$a \mathbf{e}_i \cdot \nabla \phi_n = \phi_{n+a\mathbf{e}_i} - \phi_n$$
It can be shown with a little work that you need to account for $n$th nearest neighbours to get the full $n$th order derivative. (Aside: This provides some intuition for why Taylor series work - if you know all [discrete] derivatives of an analytic function at a point, you can determine the function's value at any point and vice versa).
In the continuum $a \to 0$ limit, this should manifest as $\mathcal{L}(\phi, \nabla \phi, \dot{\phi}, \nabla \nabla \phi, \nabla \nabla \dot{\phi},...) $. In principle, there is no reason a priori to truncate the series, indicating dependence on all derivatives of $\phi$. Likewise, the sum becomes an integral: $\sum_n \mapsto \int d^3 \mathbf{x}$, and we end up with
$$ L(\phi \text{ at all points in space}) = \int d^3 \mathbf{x} \mathcal{L}(\phi(\mathbf(x)), \partial^\mu \phi(\mathbf{x}), \partial^\mu \partial^\nu \phi(\mathbf{x}),  ...)$$
and ultimately, the more relativity-friendly object
$$ S[\phi] = \int d^4\mathbf{x} \mathcal{L}(\phi(\mathbf(x)), \partial^\mu \phi(\mathbf{x}), \partial^\mu \partial^\nu \phi(\mathbf{x}),  ...) $$
In practice, Lagrangians in particle physics do not depend on higher order derivatives than $\dot{\phi}, \nabla \phi$, for various theoretical reasons. However, some classical Lagrangians in condensed matter do!
In short, the Lagrangian density is used to formalise the notion that the energy depends only on the fields, i.e. that the laws of physics are the same everywhere.
A: I guess, the question is about why in field theory the Lagrange density is preferred against the Lagrange function used in classical mechanics.
So if is the case, the reason is simple. The action as an integral over the Lagrangian has to be Lorentz invariant, using the Lagrange density this can be achieved, whereas with the Lagrange function of classical mechanics it cannot be achieved.
The goal is to get the integral of the Lagrangian (which defines the action)  to be Lorentz-invariant:
$$S=\int dt\, d^3x\, \mathcal{L}.$$
The 4-volume $dt\,d^3 x$ is Lorentz invariant (but a simple $dt$ is not Lorentz invariant), and if $\mathcal{L}$ is (and it can easily be) constructed as scalar built out of fields, the whole integral and therefore the action are Lorentz invariant.
The 4-volume $dt\,d^{3}x$ can be written in Lorentz-covariant way using the totally antisymmetrical Levi-Civita tensor $\epsilon_{\mu\nu\rho\sigma}$:
$$dt\,d^{3}x = \frac{1}{4!}\epsilon_{\mu\nu\rho\sigma}\,dx^{\mu}\,dx^{\nu}\,dx^{\rho}\,dx^{\sigma}.$$
This formalism is based on Minkowski space-time.  So curved space-time is not included here.
