This is a question from classical electrodynamics, but it's the maths of it I need some help in.
I have fields $\rho\left(\vec{r},t\right)$, $\vec{J}\left(\vec{r},t\right)$, $\vec{E}\left(\vec{r},t\right)$ and $\vec{B}\left(\vec{r},t\right)$ in $\mathbb{R}^3$. The fields $\rho$ and $\vec{J}$ are related by the Continuity Equation: $$ -\dfrac {\partial \rho (\vec{r}, t)} {\partial t} = \nabla \cdot \vec{J} (\vec{r}, t) $$ Let us say $\mathbb{V}_s \subset \mathbb{R}^3$ is the smallest spherical volume such that: $$ \forall t, \qquad \forall \vec{r}\notin \mathbb{V}_s: \qquad\qquad \rho\left(\vec{r},t\right) = \vec{J}\left(\vec{r},t\right) = 0 $$ $$ \forall t, \qquad \forall \vec{r}\in \partial\mathbb{V}_s: \qquad\qquad \rho\left(\vec{r},t\right) = \vec{J}\left(\vec{r},t\right) = 0 $$ Since $\mathbb{V}_s$ is a sphere, I say that its center is at $\vec{r}_0$, and its diameter id $\mathcal{D}$.
Now given the definitions: $$ \vec{r}_s \in \mathbb{V}_s,\qquad \vec{R} = \vec{r} - \vec{r}_s,\qquad R=\left|\vec{R}\right|, \qquad \hat{R}=\frac {\vec{R}} {R}, \qquad t_r = t - \frac {R} {c} $$ The fields $\vec{E}$ and $\vec{B}$ are as follows: $$ \vec{E}\left(\vec{r},t\right) = \frac {1} {4 \pi \epsilon_0} \iiint_{\mathbb{V}_s} {\left[ \frac {\rho (\vec{r}_s, t_r)} {R^2} \hat{R} + \frac {1} {R c} \frac {\partial \rho (\vec{r}_s, t_r) } {\partial t} \hat{R} - \frac {1} {R c^2} \frac {\partial \vec{J} (\vec{r}_s, t_r) } {\partial t} \right]} \space dV\left(\vec{r}_s\right) $$ $$ \vec{B}\left(\vec{r},t\right) = \frac {\mu_0} {4 \pi} \iiint_{\mathbb{V}_s} {\left[ \frac {\vec{J} (\vec{r}_s, t_r)} {R^2} \times \hat{R} + \frac {1} {R c} \frac {\partial \vec{J} (\vec{r}_s, t_r) } {\partial t} \times \hat{R} \right]} \space dV\left(\vec{r}_s\right) $$ (Physically, $\vec{E}$, $\vec{B}$. $\vec{J}$ & $\rho$ have the standard meanings from Electrodynamics. The equations above are the Jefimenko's equations)
Now the paper "The Relation Between Expressions for Time-Dependent Electromagnetic Fields Given by Jefimenko and by Panofsky and Phillips" by Kirk T. McDonald shows that the expression for $\vec{E}$ can be transformed into: $$ \scriptsize{ \vec{E}\left(\vec{r},t\right) = \frac {1} {4 \pi \epsilon_0} \iiint_{\mathbb{V}_s} {\left[ \frac {\rho (\vec{r}_s, t_r)} {R^2} \hat{R} + \frac { \left(\vec{J} (\vec{r}_s, t_r) \cdot \hat{R}\right)\hat{R} + \left(\vec{J} (\vec{r}_s, t_r) \times\hat{R}\right) \times \hat{R} } {R^2 c} + \frac {1} {R c^2} \left( \frac {\partial \vec{J} (\vec{r}_s, t_r) } {\partial t} \times \hat{R} \right) \times \hat{R} \right]} \space dV\left(\vec{r}_s\right) } $$ Now I want to compute the quantity $\vec{S}\left(\vec{r},t\right) = \dfrac {\vec{E}\left(\vec{r},t\right) \times \vec{B}\left(\vec{r},t\right)} {\mu_0}$ for $\vec{r}$ for which $R \gg \mathcal{D}$. I figured that for such values of $\vec{r}$, the maximum possible difference between the values of $\dfrac 1 R$ for any two points $\vec{r}_s \in \mathbb{V}_s$ is: $$\frac{1}{R} - \frac{1}{R+\mathcal{D}} = \frac{\mathcal{D}}{R\left(R+\mathcal{D}\right)} = \frac{1}{R}\cfrac{ \color{red}{\frac{\mathcal{D}}{R}}}{1+\color{red}{\frac{\mathcal{D}}{R}}}$$ Similarly, the maximum possible angle between the values of $\hat{R}$ for any two points $\vec{r}_s \in \mathbb{V}_s$ is: $$ 2\;{\tan}^{-1}\left(\frac12 \color{red} {\frac{\mathcal{D}}{R}}\right) $$ When $R \gg \mathcal{D}$, $\dfrac {\mathcal{D}} R \approx 0$, and each of the two expressions above are $\approx 0$ as well. So $\dfrac 1 R$ and $\hat{R}$ are essentially independent of $\vec{r}_s$. We can, therefore, bring out all $\dfrac 1 R$ and $\hat{R}$ factors outside the integral sign.
So if I define the following: $$ W\left(\vec{r},t\right) = \iiint_{\mathbb{V}_s} \rho \left( \vec{r}_s, t_r \right) \space dV\left(\vec{r}_s\right) $$ $$ \vec{Y}\left(\vec{r},t\right) = \iiint_{\mathbb{V}_s} \vec{J} \left( \vec{r}_s, t_r \right) \space dV\left(\vec{r}_s\right) $$ $$ \vec{Z}\left(\vec{r},t\right) = \iiint_{\mathbb{V}_s} \frac {\partial \vec{J} \left( \vec{r}_s, t_r \right) } {\partial t} \space dV\left(\vec{r}_s\right) = \frac {\partial \vec{Y}\left(\vec{r},t\right)} {\partial t} $$ I can then say: $$ \vec{E}\left(\vec{r},t\right) = \frac {1} {4 \pi \epsilon_0} \left[ \frac {W (\vec{r}, t)} {R^2} \hat{R} + \frac {\vec{Y} (\vec{r}, t) \cdot \hat{R}} {R^2 c} \hat{R} + \frac {\left(\vec{Y}(\vec{r}, t) \times \hat{R}\right) \times \hat{R}} {R^2 c} + \frac {\left(\vec{Z}(\vec{r}, t) \times \hat{R}\right) \times \hat{R}} {R c^2} \right] $$ $$ \vec{B}\left(\vec{r},t\right) = \frac {\mu_0} {4 \pi} \left[ \frac {\vec{Y} (\vec{r}, t) \times \hat{R}} {R^2} + \frac {\vec{Z}(\vec{r}, t) \times \hat{R}} {R c} \right] $$ Applying $\color{blue}{\left(\vec{a} \times \hat{u}\right) \times \hat{u} \equiv \left(\vec{a}\cdot\hat{u}\right)\hat{u} - \vec{a}}$, we can then say: $$ \vec{E}\left(\vec{r},t\right) = \frac {1} {4 \pi \epsilon_0} \left[ \left( \frac {W (\vec{r}, t)} {R^2} + 2\frac {\vec{Y} (\vec{r}, t) \cdot \hat{R}} {R^2 c} + \frac {\vec{Z}\left( \vec{r}, t \right)\cdot\hat{R}} {R c^2} \right) \hat{R} - \frac 1 {Rc} \left( \frac {\vec{Y} (\vec{r}, t) } {R} + \frac {\vec{Z}(\vec{r}, t)} {c} \right) \right] $$ $$ \vec{B}\left(\vec{r},t\right) = \frac {\mu_0} {4 \pi} \left[ \frac 1 R {\left( \frac {\vec{Y} (\vec{r}, t)} {R} + \frac {\vec{Z}(\vec{r}, t)} {c} \right)} \times \hat{R} \right] $$ Thus, if I define: $$ \mathcal{U}\left(\vec{r},t\right) = \color{blue}{ \vec{W} \left(\vec{r},t\right) + \frac {\vec{Y} \left(\vec{r},t\right) \cdot \hat{R}} {c} } = \iiint_{\mathbb{V}_s} \left[ \rho \left( \vec{r}_s, t_r \right) + \frac { \vec{J} \left( \vec{r}_s, t_r \right) \cdot \hat{R} } {c} \right] \space dV\left(\vec{r}_s\right) $$ $$ \vec{\mathcal{X}} \left(\vec{r},t\right) = \color{blue}{ \frac {\vec{Y} \left(\vec{r},t\right)} {R} + \frac {\vec{Z} \left(\vec{r},t\right)} {c} } = \iiint_{\mathbb{V}_s} \left[ \frac { \vec{J} \left( \vec{r}_s, t_r \right) } {R} + \frac 1 c \frac {\partial \vec{J} \left( \vec{r}_s, t_r \right) } {\partial t} \right] \space dV\left(\vec{r}_s\right) $$ I get: $$ \vec{E}\left(\vec{r},t\right) = \frac {1} {4 \pi \epsilon_0} \left[ \frac {\mathcal{U}\left(\vec{r},t\right)} {R^2} \hat{R} + \frac {\left( \vec{\mathcal{X}}\left(\vec{r},t\right) \times \hat{R}\right) \times \hat{R}} {Rc} \right] $$ $$ \vec{B}\left(\vec{r},t\right) = \frac {\mu_0} {4 \pi} \left[ \frac {\vec{\mathcal{X}} \left(\vec{r},t\right) \times \hat{R}} {R} \right] $$ Finally giving me: $$ \vec{S}\left(\vec{r},t\right) = \mathcal{U} \left(\vec{r},t\right) \frac {\hat{R} \times \left( \vec{\mathcal{X}} \left(\vec{r},t\right) \times \hat{R} \right) } {16 \pi^2 \epsilon_0 R^3} + \frac {\left| \vec{\mathcal{X}} \left(\vec{r},t\right) \times \hat{R} \right|^2} {16 \pi^2 \epsilon_0 R^2 c} \hat{R} $$ Now my first question is: Is my math correct so far?
Now, I define a sphere $\mathbb{V}$ with its center at $\vec{r}_0$ (the center of the source volume $\mathbb{V}_s$.) such that point $\vec{r}$ is on its surface. Then the unit vector $\hat{R}$ will actually be the unit normal $\hat{n}\left(\vec{r}\right)$ at $\vec{r}$.
I want to compute the power passing the surface of $\mathbb{V}$, which should be: $$ \oint_{\partial\mathbb{V}} \vec{S}\left(\vec{r},t\right) \cdot \hat{R} \;ds\left(\vec{r}\right) = \frac 1 {16 \pi^2 \epsilon_0 R^2 c} \oint_{\partial\mathbb{V}} \left| \frac {\vec{Y}\left(\vec{r},t\right) \times \hat{R}} {R} + \frac {\vec{Z}\left(\vec{r},t\right) \times \hat{R}} {c} \right|^2 \;ds\left(\vec{r}\right) $$
My second question is: How do I parameterize and compute this surface integral? I have got to the following expression (it seems to me that when I am integrating over $\vec{r}$, I can still take $\dfrac 1 R$ out from under the integral, but not $\hat{R}$), but I'm stuck after this: $$ \color{blue}{\dfrac {\oint_{\partial\mathbb{V}} \left|\vec{Y}\left(\vec{r},t\right) \times \hat{R}\right|^2 \;ds\left(\vec{r}\right)} {16 \pi^2 \epsilon_0 R^4 c} + \dfrac {\oint_{\partial\mathbb{V}} \left(\vec{Y}\left(\vec{r},t\right) \times \hat{R}\right)\cdot\left(\vec{Z}\left(\vec{r},t\right) \times \hat{R}\right) \;ds\left(\vec{r}\right)} {8 \pi^2 \epsilon_0 R^3 c^2}} \color{red}{+ \dfrac {\oint_{\partial\mathbb{V}} \left|\vec{Z}\left(\vec{r},t\right) \times \hat{R}\right|^2 \;ds\left(\vec{r}\right)} {16 \pi^2 \epsilon_0 R^2 c^3}} $$ I expected to prove that the term in $\color{red}{red}$ is independent of $R$, and the term in $\color{blue}{blue}$ is 'reversible', by which I mean if there are two time instants $t_1$ and $t_2$ such that: $$\forall \vec{r}_s \in \mathbb{V}_s:\qquad \rho\left(t_1\right) = \rho\left(t_2\right), \qquad\vec{J}\left(t_1\right) = \vec{J}\left(t_2\right)$$ ... then I hoped to prove that: $$ \int_{t_1}^{t_2} \dfrac {\oint_{\partial\mathbb{V}} \left|\vec{Y}\left(\vec{r},t\right) \times \hat{R}\right|^2 \;ds\left(\vec{r}\right)} {16 \pi^2 \epsilon_0 R^4 c} + \dfrac {\oint_{\partial\mathbb{V}} \left(\vec{Y}\left(\vec{r},t\right) \times \hat{R}\right)\cdot\left(\vec{Z}\left(\vec{r},t\right) \times \hat{R}\right) \;ds\left(\vec{r}\right)} {8 \pi^2 \epsilon_0 R^3 c^2}\;dt = 0$$ ... but so far I have not been able to make any progress. Would appreciate any help...