Is it possible to prove that
$$\displaystyle c = \frac 1 {\sqrt{\epsilon_0 \mu_0}}$$
using Maxwell's equation in integral form? Recently, I saw this kind of proof by Professor Walter Lewin in one of his 8.02x lectures. But I could not understand the derivation in the video lecture.
Here is the link to the lecture (Derivation part starts from 13:24) https://youtu.be/D3tnZzhSISo?feature=shared
Admittedly the derivation in the lecture was quite terse, So it may have been hard to follow.
The lecture makes this approach for a plane EM wave travelling in $+z$ direction $$\vec{E}(z,t)=E_o\hat{x}\cos(kz-\omega t) \tag{1}$$ $$\vec{B}(z,t)=B_o\hat{y}\cos(kz-\omega t) \tag{2}$$ with some constants ($E_0,B_0,k,\omega$).
Like every EM field this EM wave needs to satisfy the Ampere-Maxwell law $$\oint\vec{B}\cdot d\vec{l} = \mu_0\epsilon_0\frac{d}{dt}\underbrace{\int\vec{E}\cdot d\vec{A}}_{\Phi_E}. \tag{3}$$ We can use this law to deduce a relation between the constants in (1) and (2). For applying the Ampere-Maxwell law we choose as the closed loop this rectangle in the $yz$ plane.
Then we can calculate the change rate of the electric flux $\Phi_E$ appearing on the right side of (3) with the $\vec{E}$ field from (1). $$\begin{align} \frac{d\Phi_E(t)}{dt} &=\frac{d}{dt}\int\vec{E}(z,t)\cdot d\vec{A} \\ &=\frac{d}{dt}\int_0^l dy \int_0^{\lambda/4}dz\ E_0\cos(kz-\omega t) \\ &=\int_0^l dy \int_0^{\lambda/4}dz\ E_0\frac{d}{dt}\cos(kz-\omega t) \\ &=\int_0^l dy \int_0^{\lambda/4}dz\ E_0\omega\sin(kz-\omega t) \end{align}$$ For simplicity we consider this especially for $t=0$: $$\begin{align} \frac{d\Phi_E(t=0)}{dt} &=\int_0^l dy \int_0^{\lambda/4}dz E_0\omega\sin(kz) \\ &=lE_0\omega\left[-\frac{1}{k}\cos(kz)\right]_{z=0}^{z=\lambda/4} \\ &=lE_0\frac{\omega}{k}\left[-\cos(kz)\right]_{z=0}^{z=\lambda/4} &\text{remember }k=\frac{2\pi}{\lambda} \\ &=lE_0\frac{\omega}{k}\left(-\cos\left(\frac{\pi}{2}\right)+\cos(0)\right) \\ &=lE_0\frac{\omega}{k}(0+1) &\text{remember }c=\frac{\omega}{k} \\ &=lE_0c \tag{4} \end{align}$$
The loop integral on the left side of (3) is easier to calculate. Again for simplicity we calculate it only for $t=0$. Then only the left edge (at $z=0$) of the rectangular loop contributes to the integral. $$\begin{align} &\oint\vec{B}(z,t=0)\cdot d\vec{l} \\ &=\int_0^l dy\ B_0 \cos(kz-\omega t) \\ &=\int_0^l dy\ B_0 \\ &=lB_0 \tag{5} \end{align}$$ Inserting the results (4) and (5) into the Ampere-Maxwell law (3) we get $$lB_0=\mu_0\epsilon_0 lE_0c$$ or $$B_0=\mu_0\epsilon_0 E_0c \tag{6}$$
The EM wave also needs to satisfy Faraday's law. $$\oint\vec{E}\cdot d\vec{l} = -\frac{d}{dt}\underbrace{\int\vec{B}\cdot d\vec{A}}_{\Phi_B}. \tag{7}$$ We can use this law to deduce one more relation between the constants in (1) and (2). For applying Faraday's law we choose as the closed loop this rectangle in the $xz$ plane.
Then calculations very similar to the above yield $$\frac{d\Phi_B(t=0)}{dt}=-lB_0c \tag{8}$$ and $$\oint\vec{E}(z,t=0)\cdot d\vec{l}=lE_0 \tag{9}$$ Inserting the results (8) and (9) into Faraday's law (7) we get $$lE_0=lB_0c$$ or $$E_0=B_0c \tag{10}$$
Now we can combine (6) and (10) and find $$c=\frac{1}{\sqrt{\epsilon_0\mu_0}}$$
