I am familiar with how Noether's theorem is derived in some sources/books, the answer in 534699 is particularly clear. However, I'm reading A First Book of Quantum Field Theory by Pal, and although some people here may say to ignore it if I find it confusing, I still want to understand the logic in its derivation. I'm going to state its derivation including its exact words in describing things since it probably contributes to my confusion. Consider the infinitesimal transformation,
\begin{equation} x'^\mu = x^\mu + \delta x^\mu,\tag{1}\label{1} \end{equation}
under which the fields transform as,
\begin{equation} \Phi'^A(x') = \Phi^A(x) + \delta \Phi^A(x)\tag{2}\label{2}. \end{equation}
The change in the action from these transformations,
\begin{equation} \delta I = \int_{\Omega'} d^4x' L(\Phi'^A(x'), \partial'_\mu \Phi'^A(x')) - \int_{\Omega} d^4x L(\Phi^A(x), \partial_\mu \Phi^A(x)),\tag{3}\label{3} \end{equation}
where if $\delta I=0$ we say that the theory is invariant under the transformation. The $x'$ in the first integral is a dummy variable so we can replace it with $x$,
\begin{align} \delta I & = \int_{\Omega'} d^4x L(\Phi'^A(x), \partial'_\mu \Phi'^A(x)) - \int_{\Omega} d^4x L(\Phi^A(x), \partial_\mu \Phi^A(x))\\ & = \int_{\Omega} d^4x \left[ L(\Phi'^A(x), \partial'_\mu \Phi'^A(x)) - L(\Phi^A(x), \partial_\mu \Phi^A(x)) \right]\\ & \qquad + \int_{\Omega' - \Omega} d^4x L(\Phi'^A(x), \partial'_\mu \Phi'^A(x)). \tag{4}\label{4} \end{align}
The last term is an integral over the infinitesimal volume $\Omega' - \Omega$ so we replace it by an integral over the boundary $\partial \Omega$,
\begin{align} \int_{\Omega' - \Omega} d^4x L(\Phi'^A, \partial'_\mu \Phi'^A) & = \int_{\partial \Omega} dS_\lambda \delta x^\lambda L(\Phi^A, \partial_\mu \Phi^A)\\ & = \int_{\Omega} d^4x \partial_\lambda \left( \delta x^\lambda L(\Phi^A, \partial_\mu \Phi^A) \right), \tag{5}\label{5} \end{align}
where the spacetime index $x$ is suppressed since there's no longer a need to distinguish between $x$ and $x'$. In the first equation, $\Phi'^A$ is replaced by $\Phi^A$ as the difference is of higher order in $\delta x^\lambda$. It's convenient to define the variation for fixed $x$ which gives the change in the functional forms only. For any function $f(x)$ whose functional form changes to $f'(x)$,
\begin{align} \bar{\delta} f(x) \equiv f'(x) - f(x) & = [f'(x') - f(x)] - [f'(x') - f'(x)]\\ & = \delta f(x) - \partial_\mu f(x) \delta x^\mu \tag{6} \label{6} \end{align}
This property allows the simplification of the integrals in eq.(\ref{4}),
\begin{align} L(\Phi'^A(x), \partial'_\mu \Phi'^A(x)) - L(\Phi^A(x), \partial_\mu \Phi^A(x)) & = \frac{\partial L}{\partial \Phi^A} \bar{\delta} \Phi^A + \frac{\partial L}{\partial (\partial_\mu \Phi^A)} \bar{\delta} (\partial_\mu \Phi^A)\\ & = \partial_\mu \left( \frac{\partial L}{\partial (\partial_\mu \Phi^A)} \bar{\delta} \Phi^A \right)\tag{7}\label{7} \end{align}
using the Euler-Lagrange to get the last equality. Plugging this to eq.(\ref{4}) and using eq.(\ref{5}),
\begin{align} \delta I = & \int_{\Omega} d^4x \partial_\mu \left[ \frac{\partial L}{\partial (\partial_\mu \Phi^A)} \bar{\delta} \Phi^A + L \delta x^\mu \right]\\ & = \int_{\Omega} d^4x \partial_\mu \left[ \frac{\partial L}{\partial (\partial_\mu \Phi^A)} \delta \Phi^A - T^{\mu\nu} \delta x_\nu \right].\tag{8} \end{align}
where $T^{\mu\nu}$ is the stress-energy tensor (I didn't write it to save space).
First of all, eq.(\ref{2}) is already misleading since it looks like a total variation of the fields but it was written with a $\delta$. I think it should be $$\Phi'^A(x') = \Phi^A(x) + \Delta \Phi^A(x)$$ to be consistent with notations. I think what he meant by the change in action in eq.(\ref{3}) should also be $\Delta I$, plus I'm not sure why the first Lagrangian has no prime, I was thinking maybe he used $$L'(\Phi'^A(x'), \partial'_\mu \Phi'^A(x')) = L(\Phi'^A(x'), \partial'_\mu \Phi'^A(x')) + \delta L$$ which is a functional variation at $\Phi'^A$ and took only the first term. I'm not sure what's the logic in arguing that $x'$ is a dummy variable that leads to eq.{\ref{4}}. I know that an infinitesimal volume can be written as an area element $dS$ multiplied by $\delta x$, but I'm not sure how to obtain the first equation in eq.(\ref{5}) analytically, plus I don't know how $\Phi'^A$ is replaced by $\Phi^A$. I don't understand how eq.(\ref{6}) is used in eq.(\ref{7}). Can anyone please explain more clearly what's going on in each step?
