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$$\frac{\nabla'\times \vec{J}}{R}=\nabla'\times(\frac{\vec{J}}{R})-\frac{\vec{J}\times(\nabla'R)}{R^2}$$ Thanks to Prof. Y. F. Chen I was able to figure it out. While in the integral the first term on the RHS can be converted into a surface integral as below: $$\int\nabla'\times(\frac{\vec{J}}{R})d^3x^{'}=\oint(\vec{n}\times\frac{\vec{J}}{R})d^2x^{'}$$ ...

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The square bracket transformation This is just the application of chain rule. The LHS means a derivative over the primed spacial coordinates while keeping unprimed spacial and time coordinates fixed. $$\nabla'[ \rho(\mathbf{x'},t')]_{ret} = \left(\sum_i \frac{\partial }{\partial x_i'} \hat{i}\right)[\rho(x_i',x_j',x_k',t')]_{ret}\\$$ But the $\rho$ is a ...

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I'm not completely sure what you want, but honestly the entirety of Spivak's Calculus on manifolds is devoted to exactly that. If you want something that feels familiar, you can simply find $\nabla$ in various coordinate systems in Wikipedia, but if you want a less coordinate-centric view then you're probably going to need to step outside of your comfort ...

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I would say that the Wikipedia page on curvilinear coordinates and the article Mathematical Physics Lessons - Gradient, Divergence and Curl in Curvilinear Coordinates by James Foadi are enough to understand what is going on.

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Basically, you need a higher dimensional version of the Leibniz rule to differentiate under the integral sign. Suppose you have an integral of this kind $$\int_{\Omega(s)} f(s,\vec x) \, d\vec x$$ That's it, your region of integration and your function depends of a parameter $s$. (in your case $s = x_i$). Then if you want to calculate $$... 1 Your question is "Is there an infinite series of higher derivatives of position for this to work?" Answer: No. Acceleration can jump from zero to something. When it does, its derivative is not defined, so the series of position derivatives stops after the second one. From the question: "A change is velocity is acceleration, so the value of the ... 3 Stokes' theorem needs no physical reason to be true. However, there is a nice intuitive description of the two-dimensional case. Tesselate the surface with little (infinitesimal) oriented squares and consider the integral as the sum of the curl on all these little squares: The inner sides of the squares have no contribution to this sum at all, because ... 1 It follows from the definitions. Velocity is defined as the time-derivative of displacement,$$ v=\frac{dx}{dt} $$from which we get$$ x=\int v(t)\,dt  which you may recall is the area under the $v(t)$ graph. For the elastic energy case, start from the definition of force, $f=-dU/dx$, and follow the same argument.

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1) They are slightly different in the sense that the function $C(x)$ (or $\vec C(\vec r)$) has different values at different places in space, and so even though the two opposite faces of the cube are very close the value of the function $C$ at the two faces will be slightly different. 2) Only if the two points are infinitesimally close or the function $C$ ...

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