Fdtxjr

I am trying to understand this answer.

The answer is the following:

I'll assume that $V'$ is bounded and that $mathbfM'$ is continuously differentiable. To your primary question:

Is [the] divergence theorem applicable to 𝐿𝐻𝑆 of equation (4)?

The answer is, strictly speaking, no. The hypotheses of the divergence theorem include that the vector field in question is continuously differentiable, and $fracmathbfM'(r)r$ is not (in general).
However, one can still rigorously employ the divergence theorem to yield your expression by removing a ball $B_epsilon(P)$ from $V'$. Writing $B_epsilon '(P) equiv B_epsilon(P) cap V' $,

$$iiint_V' left[nabla' cdot left( fracmathbfM(r')r' right) right]dV' =
iiint_V' backslash B_epsilon'(P) left[nabla' cdot left( fracmathbfM(r')r' right) right]dV' + iiint_B_epsilon'(P) left[nabla' cdot left( fracmathbfM(r')r' right) right]dV' $$
The divergence Theorem may be applied to the first term for any $epsilon>0$, and one can show that the second term goes to $0$ in the limit $epsilon to 0$ via your chain of equalities in equation $(4)$, which hold $a.e.$ in $B_epsilon'(P)$, since $mathbfM'$, being continuous on the compact set $overlineV'$, is bounded, say by $M>0$.

With some computation, one can show that the integral over $partial(V' backslash B_epsilon '(P))$ arising from the first term approaches the integral over $S' = partial V'$ in the same limit, assuming $S'$ is smooth in a neighborhood of $P$. To be explicit,

$$ left( unicodex222F_partial V' - unicodex222F_partial(V' backslash B_epsilon '(P)) right)left[ fracmathbfM'(r') cdot hatmathbfnr' right]dS'
= left( iint_partial B_epsilon '(P) backslash partial B_epsilon(P) - iint_partial B_epsilon '(P) backslash partial V' right)left[ fracmathbfM'(r') cdot hatmathbfnr' right]dS' $$

That is, the integrals are the same up to the difference of the integrals over the piece of $partial B_epsilon '(P)$ from $partial V'$ and that from $partial B_epsilon(P)$. Since $S'$ is smooth near $P$, it is locally the graph of a smooth function $f(x,y)$ on the tangent plane at $P$ with standard linear coordinates $(x,y)$. So, for $epsilon$ sufficiently small, we may compute the first integral on the RHS above as an integral in the tangent plane, with area form $sqrt1 + f_x^2 + f_y^2dxdy=sqrt1 + f_x^2 + f_y^2sdsdtheta$ in polar coordinates $(s,theta)$. The term $sqrt1 + f_x^2 + f_y^2$ is continuous and therefore bounded, say by $C$ (independent of $epsilon$ for $epsilon$ sufficiently small), and further $r'(s,theta)=sqrts^2+f(s,theta)^2 geq s$. Hence

$$left|iint_partial B_epsilon '(P) backslash partial B_epsilon(P)left[ fracmathbfM'(r') cdot hatmathbfnr' right]dS'right| leq C int_0^epsilon int_0^2pi|mathbfM(s,theta)| fracssqrts^2+f^2dsdtheta leq 2pi MC epsilon$$

Note that the exact integration here may not range over all of $0 leq s leq epsilon$, but the bound still holds. Finally, in the second term on the RHS above, $r'=epsilon$, so that integral is bounded by $4pi Mepsilon$, and we have

$$left| left( unicodex222F_partial V' - unicodex222F_partial(V' backslash B_epsilon '(P)) right)left[ fracmathbfM'(r') cdot hatmathbfnr' right]dS' right| leq 2pi M(2+C)epsilon $$

which goes to $0$ as $epsilon to 0$. Since my first equation holds for all $epsilon >0$, we may take the limit of both sides as $epsilon to 0$ to obtain

$$iiint_V' left[nabla' cdot left( fracmathbfM(r')r' right) right]dV' = unicodex222F_partial V' left[ fracmathbfM'(r') cdot hatmathbfnr' right]dS'$$
As a naive application of the divergence theorem would have suggested.

Note that that "singularity" in the RHS integration does not cause the integral to diverge for essentially the same reason that it didn't in the volume integral: by passing to an appropriately defined set of coordinates in a neighborhood of $P$, we see that the area form gets small fast enough in this neighborhood for the integral to converge, just as the volume form did in your computations.

My misunderstandings are the following:

(1) What does $$ left( unicodex222F_partial V' - unicodex222F_partial(V' backslash B_epsilon '(P)) right)left[ fracmathbfM'(r') cdot hatmathbfnr' right]dS'
= left( iint_partial B_epsilon '(P) backslash partial B_epsilon(P) - iint_partial B_epsilon '(P) backslash partial V' right)left[ fracmathbfM'(r') cdot hatmathbfnr' right]dS' $$ mean? Why are they equal?

My interpretation:

$unicodex222F_partial V'$ means the integral over the whole of the surface of $V'$.

$text $

$unicodex222F_partial(V' backslash B_epsilon '(P))$ means the integral over the surface of "$B_epsilon '(P)$ removed from $V'$.

I have no idea about $displaystyleiint_partial B_epsilon '(P) backslash partial B_epsilon(P)$ and $displaystyleiint_partial B_epsilon '(P) backslash partial V'$ Please explain.

Original answerer here. You are correct regarding what the two integrations mean individually. The notation on the LHS of the equation under consideration simply means the difference of the surface integrals over the two domains you've identified (of the same integrand $fracmathbfM'(mathbfr') cdot mathbfhatnr'$). Clearly the result will be an integral of this same integrand; the question is the domain.

The integrations over the intersection of the two original domains cancel, so our resulting integration won't contain any of the boundary of your rectangular prism except the bit above the removed ball, and the only other surface we need to include is the portion of the boundary of $B_epsilon '(P)$ inside the rectangular prism. Note that $B_epsilon'(P)$ in your drawn scenario is a half-ball, the boundary of which is a hemisphere (part of $partial B_epsilon(P)$, which is the full sphere of radius $epsilon$ centered at $P$) plus a circular disc (part of $partial V'$), and these two pieces are exactly what we've identified as what we want to integrate over. Note each of these pieces came from a different term on our LHS, so they'll be integrated over on the RHS with different signs. In particular, the disc should be integrated over with a positive sign, while the hemisphere should be integrated over with a negative sign. That is, our integral should be
$$ left( iint_textdisc - iint_texthemisphere right)left[ fracmathbfM'(r') cdot hatmathbfnr' right]dS'$$

All that remains is to check that my original notation agrees with this. But $partial B_epsilon '(P) backslash partial B_epsilon(P)$ just means "the hemisphere and disc, remove anything contained in the full sphere", which is just the disc, while $partial B_epsilon '(P) backslash partial V'$ just means "the hemisphere and disc, remove anything contained in the rectangular prism's boundary", which is just the hemisphere, so the above integral is the same as
$$ left( iint_partial B_epsilon '(P) backslash partial B_epsilon(P) - iint_partial B_epsilon '(P) backslash partial V' right)left[ fracmathbfM'(r') cdot hatmathbfnr' right]dS'$$
As our equation claims. The advantage of using this notation is that it represents the correct domain no matter what the shape of $V'$ is, and it emphasizes that the difference only depends on a region within distance $epsilon$ of $P$.