Transformation of coordinates in Noether's Theorem I am confused, in the proof of Noether's theorem, by the change of boundary in the action integral during the transformation of coordinates. I have seen on Wikipedia 
that along with the change of Field, they also change $\Omega$ to $\Omega'$, where $\Omega$ is the space-time boundary of the action integral.
If we change the fields and boundaries both due to coordinate transformations then wouldn't that constitute a zero change? (I am keeping the intrinsic changes
in the field apart)
Don't we consider a fixed region (arbitrary but unchanging during the flow) of space-time and then see the changes on Lagrangian due to only the flow of fields and some intrinsic change of fields, before and after the flow? (as shown below) The coordinates should be treated as dummy variables.
$$\int_\Omega \delta L\ d^4x$$
I don't think we should move our boundary with the flow, am I right?
Moreover, in the proof shown by joshphysics he didn't consider action at all. He worked only with the variation of Lagrangian and so there was no integral and hence, no boundary.
So, why do some proofs change the boundary and some do not? I mean how are these equivalent?
Another question: If we prove Noether's Theorem as  joshphysics did, using only Lagrangian and not action, do we miss some conservations compared to the proof done in  Wikipedia  using the action integral?
 A: *

*Note that the transformations$^1$ in Noether's (first) theorem are generally a combination of (vertical) transformations of target space fields $\phi^{\alpha}$ and (horizontal) transformations of spacetime coordinates $x^{\mu}$. 

*Recall that the action $S_{\Omega}[\phi] =\int_{\Omega} \! d^nx~{\cal L}$ is the Lagrangian density ${\cal L}$ integrated over a spacetime integration region $\Omega$. The more general formulation of Noether's theorem is in term of a (quasi)symmetry of the action rather than the Lagrangian density.

*The transformation of the spacetime integration region $\Omega$ is induced by the horizontal transformation $\delta x^{\mu}$. 

*The Phys.SE post that joshphysics answered does not consider horizontal transformations, and hence has no transformation of the integration region $\Omega$. 
--
$^1$ Noether's theorem can be formulated for finite transformations, but let us only consider infinitesimal transformations in this answer for simplicity. 
A: 
So, why do some proofs change the boundary and some do not? I mean how are these equivalent?

They aren't: Common proofs of 'Noether's theorem' often only consider certain limits of it. There's also a tendency to hide intricacies behind notation.
In  the following, I'll present an elementary proof of a simple version of Noether's first theorem in 1 dimension in a way that should generalize to what's called the the field-theoretic version on Wikipedia by going from $t$ to $x^\mu$ and $q$ to $\varphi^A$.
The Lagrangian will be a function
$$
L = L(x,v,t)
$$
and the action a functional
$$
S[q] = \int_{t_1}^{t_2} L(q(t),\dot q(t),t) \,dt
$$
Proposition. If the transformation
$$ t\to t'(t) = t + \epsilon T(t) $$
$$ x\to x'(x,t) = x + \epsilon X(t)$$
$$
q'(t') = q(t(t')) + \epsilon X(t(t'))
$$
is a quasi-symmetry of the action
$$
\delta S \approx \Delta K
$$
on-shell (ie assuming the equations of motion), then there is a conserved quantity
$$
\frac{d}{dt} \left( \frac{\partial L}{\partial v} (X - \dot q T) + LT - K \right) \approx 0
$$
Here,
$$ \delta S = \frac{d}{d\epsilon}\Big|_{\epsilon=0} S[q'] $$
$$ \Delta K = K(t_2)-K(t_1) = \int_{t_1}^{t_2} \frac{dK}{dt} dt $$
Proof.
\begin{align}
\delta S &= \frac{d}{d\epsilon}\Big|_{\epsilon=0} \int_{t'(t_1)}^{t'(t_2)} L(q'(t'),\frac{d}{dt'}q'(t'),t')\,dt'
\\&= \frac{d}{d\epsilon}\Big|_{\epsilon=0} \int_{t_1}^{t_2} L(q(t) + \epsilon X(t),\left( \frac{dt'}{dt} \right)^{-1}\frac{d}{dt}(q(t) + \epsilon X(t)),t + \epsilon T(t)) \,dt'(t)
\\&= \int_{t_1}^{t_2}\left[ \left( \frac{\partial L}{\partial x}X + \frac{\partial L}{\partial v}\frac{d}{d\epsilon}\Big|_{\epsilon=0}\frac{\dot q + \epsilon \dot X}{1 + \epsilon \dot T} + \frac{\partial L}{\partial t} T \right)\,dt + L \frac{d}{d\epsilon}\Big|_{\epsilon=0}d(t + \epsilon T) \right]
\\&= \int_{t_1}^{t_2}\left[ \frac{\partial L}{\partial x}X + \frac{\partial L}{\partial v} (\dot X - \dot q \dot T) + \frac{\partial L}{\partial t} T + L \dot T\right]\,dt
\end{align}
Using
$$
\frac{\partial L}{\partial v} \dot X = \frac{d}{dt}\left( \frac{\partial L}{\partial v} X \right) - \left( \frac{d}{dt} \frac{\partial L}{\partial v} \right) X
$$
$$
\frac{\partial L}{\partial t} T + L \dot T = \frac{d}{dt}\left( LT \right) - \frac{\partial L}{\partial x} \dot q T - \frac{d}{dt}\left( \frac{\partial L}{\partial v} \dot q T \right) + \left( \frac{d}{dt}\frac{\partial L}{\partial v} \right) \dot q T + \frac{\partial L}{\partial v}\dot q \dot T
$$
we arrive at
$$
\delta S = \int_{t_1}^{t_2}\left[ \left( \frac{\partial L}{\partial x} - \frac{d}{dt} \frac{\partial L}{\partial v} \right)(X - \dot q T) + \frac{d}{dt} \left( \frac{\partial L}{\partial v}(X - \dot q T) + LT \right) \right]\,dt
$$
The first term vanishes if we asume the Euler-Lagrange equations, the second term yields our conservation law once we move $K$ to this side of the equation. This concludes the proof. $\square$
Note the change of region of intergration in the first step. The new time coordinate was by no means a dummy variable - the transformation is 'active', a one-parameter group of diffeomorphisms.
