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By virtue of a result of Kazarnovskii's (stated in [2] and proved in [1] Theorem 8.1) one has \begin{equation}\label{k} \mu_1(B_4)= 2 \sum_{\Delta\in{\mathcal B}(\Gamma,1)} {\mathop{\rm vol\,}\nolimits}_1(\Delta) {\mathop{\rm vol\,}\nolimits}_3(K_\Delta\cap B_4)\,.\tag{3} \end{equation} The preceding formula reveals an interesting fact: $\mu_1(B_4)$ is invariant under orthogonal transformations of $\Gamma$ as a subset of $\mathbb R^4$. If $m\in\mathbb N^*$, let \begin{equation*} \varsigma_m(z)= \begin{cases} 1 & \textrm{if}\quad \Vert z\Vert \leq 1\,,\\ \exp\left(1+\dfrac{1}{m^2(1-\Vert z\Vert)^2-1}\right) &\textrm{if}\quad1< \Vert z\Vert <1+\dfrac{1}{m}\,,\\ 0 &\textrm{if}\quad \Vert z\Vert \geq 1+\dfrac{1}{m}\,. \end{cases} \end{equation*} The function $\varsigma_m$ is radial, smooth, its support is the full-dimensional ball of radius $1+(1/m)$ about the origin, it takes on the value 1 on $B_4$, $0\leqslant\varsigma_m\leqslant 1$ and it converges (point-wise) to the characteristic function of $B_4$. So \begin{equation} \mu_1(B_4) = \lim_{m\to\infty} \sum_{\Delta\in{\mathcal B}(\Gamma,1)} {\mathop{\rm vol\,}\nolimits}_1(\Delta) \int_{K_\Delta} \iota_\Delta^*(\upsilon_{iE_\Delta}\wedge h_{B_4}\mathrm {dd}^{\mathrm c} \varsigma_m)\label{s}\tag{4} \end{equation} and in view of (4), it would be enough to prove that, for any $\Delta\in{\mathcal B}(\Gamma,1)$, \begin{equation} { \lim_{m\to\infty} \int_{K_\Delta} \iota_\Delta^*(\upsilon_{iE_\Delta}\wedge h_{B_4}\mathrm {dd}^{\mathrm c} \varsigma_m) = 2{\mathop{\rm vol\,}\nolimits}_3(K_\Delta\cap B_4).\label{u}\tag{5} } \end{equation} Let $\Delta$ be fixed; up to an orthogonal trasformation, we may supposte that $E_\Delta=\{z\in\mathbb C^2\mid {\mathop{\rm Im\,}\nolimits} z_1=z_2=0\}$. In such a case, $K_\Delta\subset E_\Delta^\perp=\{z\in\mathbb C^2\mid x_1=0\}$, $\iota_\Delta^*(\upsilon_{iE_\Delta})={\mathrm d}y_1$ and \begin{align} &\iota_\Delta^*(\upsilon_{iE_\Delta}\wedge h_{B_4}\mathrm {dd}^{\mathrm c}\varsigma_m)\tag{6} \\ &= {\mathrm d}y_1\wedge \iota_\Delta^*\left(h_{B_4}2i\dfrac{\partial^2\varsigma_m}{\partial z_2\partial\bar z_2}{\mathrm d}z_2\wedge{\mathrm d}\bar z_2\right)\tag{7} \\ &= {\mathrm d}y_1\wedge\iota_\Delta^*\left(h_{B_4}2i\left[\dfrac{1}{2}\left(\dfrac{\partial}{\partial x_2}-i\dfrac{\partial}{\partial y_2}\right)\dfrac{1}{2}\left(\dfrac{\partial\varsigma_m}{\partial x_2}+i\dfrac{\partial\varsigma_m}{\partial y_2}\right)\right](-2i){\mathrm d}x_2\wedge{\mathrm d}y_2\right)\tag{8} \\ &= \iota_\Delta^*\left[h_{B_4}\left(\dfrac{\partial^2\varsigma_m}{\partial x_2^2}+\dfrac{\partial^2\varsigma_m}{\partial y_2^2}\right) {\mathrm d}y_1\wedge{\mathrm d}x_2\wedge {\mathrm d}y_2\right]\tag{9} \end{align} so that (5) becomes \begin{equation}\label{w} { \lim_{m\to\infty} \int_{K_\Delta} \iota_\Delta^*\left[h_{B_4}\left(\dfrac{\partial^2\varsigma_m}{\partial x_2^2}+\dfrac{\partial^2\varsigma_m}{\partial y_2^2}\right) {\mathrm d}y_1\wedge{\mathrm d}x_2\wedge {\mathrm d}y_2\right] = 2{\mathop{\rm vol\,}\nolimits}_3(K_\Delta\cap B_4).\tag{10} } \end{equation} In order to prove this equality let us pass to spherical coordinates in $E_\Delta^\perp$. Set $y_1=\rho\cos\vartheta_1$, $x_2=\rho\sin\vartheta_1\cos\vartheta_2$, $y_2=\rho\sin\vartheta_1\sin\vartheta_2$, with $\rho\geqslant0$, $\vartheta_1\in[0,\pi)$ and $\vartheta_2\in[0,2\pi)$. In these coordinates, on $K_\Delta$ one has $h_{B_4}=\rho$, \begin{equation} \dfrac{\partial \varsigma_m}{\partial x_2} = \dfrac{\partial\varsigma_m}{\partial \rho} \dfrac{\partial\rho}{\partial x_2} + \dfrac{\partial\varsigma_m}{\partial \vartheta_1} \dfrac{\partial\vartheta_1}{\partial x_2} + \dfrac{\partial\varsigma_m}{\partial \vartheta_2} \dfrac{\partial\vartheta_2}{\partial x_2} = \dfrac{\partial\varsigma_m}{\partial \rho} \dfrac{x_2}{\rho} = \dfrac{\partial\varsigma_m}{\partial \rho} \sin\vartheta_1\cos\vartheta_2\tag{11} \end{equation} and similarly \begin{align} \dfrac{\partial \varsigma_m}{\partial y_2} &= \dfrac{\partial\varsigma_m}{\partial \rho} \sin\vartheta_1\sin\vartheta_2\tag{12} \end{align} whence \begin{equation} \dfrac{\partial^2 \varsigma_m}{\partial x_2^2} = \dfrac{\partial^2 \varsigma_m}{\partial \rho^2} \sin^2\vartheta_1\cos^2\vartheta_2 + \dfrac{\partial \varsigma_m}{\partial \rho} \left( \dfrac{1-\sin^2\vartheta_1\cos^2\vartheta_2}{\rho} \right)\tag{13} \end{equation} and \begin{align} \dfrac{\partial^2 \varsigma_m}{\partial y_2^2} = \dfrac{\partial^2 \varsigma_m}{\partial \rho^2} \sin^2\vartheta_1\sin^2\vartheta_2 + \dfrac{\partial \varsigma_m}{\partial \rho} \left( \dfrac{1-\sin^2\vartheta_1\sin^2\vartheta_2}{\rho} \right)\tag{14} \end{align} so that \begin{align} & \lim_{m\to\infty} \int_{K_\Delta} h_{B_4}\left(\dfrac{\partial^2\varsigma_m}{\partial x_2^2}+\dfrac{\partial^2\varsigma_m}{\partial y_2^2}\right) {\mathrm d}y_1\wedge{\mathrm d}x_2\wedge {\mathrm d}y_2\tag{15} \\ &= \lim_{m\to\infty} \int_{K_\Delta} \rho \left[ \dfrac{\partial^2 \varsigma_m}{\partial \rho^2}\sin^2\vartheta_1 + \dfrac{\partial \varsigma_m}{\partial \rho}\left(\dfrac{2-\sin^2\vartheta_1}{\rho}\right) \right] \rho^2\sin\vartheta_1{\mathrm d}\rho\wedge{\mathrm d}\vartheta_1\wedge{\mathrm d}\vartheta_2\tag{16} \\ &= \lim_{m\to\infty} \int_{K_\Delta} \rho^3 \dfrac{\partial^2 \varsigma_m}{\partial \rho^2}\sin^3\vartheta_1 {\mathrm d}\rho\wedge{\mathrm d}\vartheta_1\wedge{\mathrm d}\vartheta_2\label{1}\tag{17} \\ &+ \lim_{m\to\infty} \int_{K_\Delta} 2\rho^2 \dfrac{\partial \varsigma_m}{\partial \rho}\sin\vartheta_1 {\mathrm d}\rho\wedge{\mathrm d}\vartheta_1\wedge{\mathrm d}\vartheta_2\label{2}\tag{18} \\ &- \lim_{m\to\infty} \int_{K_\Delta} \rho^2 \dfrac{\partial \varsigma_m}{\partial \rho}\sin^3\vartheta_1 {\mathrm d}\rho\wedge{\mathrm d}\vartheta_1\wedge{\mathrm d}\vartheta_2\label{3}.\tag{19} \end{align} For computing the latter integrals, let us observe that the derivatives of $\varsigma_m$ vanish at every $\rho\in[0,1]\cup[1+(1/m),+\infty)$ and that $$ 0\leqslant\int_1^{1+(1/m)}\rho\,\varsigma_m\,{\mathrm d}\rho\leqslant\left(1+\dfrac{1}{m}\right)\cdot1\cdot\left(1+\dfrac{1}{m}-1\right)=\dfrac{1}{m}+\dfrac{1}{m^2} $$ so \begin{equation}\label{z} \lim_{m\to\infty} \int_1^{1+(1/m)} \rho\,\varsigma_m\,{\mathrm d}\rho=0\,.\tag{20} \end{equation} Now by Fubini' theorem, (17) equals \begin{equation} \lim_{m\to\infty} \int_1^{1+(1/m)} \rho^3 \dfrac{\partial^2 \varsigma_m}{\partial \rho^2}{\mathrm d}\rho \cdot \int_{K_\Delta\cap\partial B_4} \sin^3\vartheta_1 {\mathrm d}\vartheta_1\wedge{\mathrm d}\vartheta_2\tag{21} \end{equation} with \begin{align} \lim_{m\to\infty} \int_1^{1+(1/m)} \rho^3 \dfrac{\partial^2 \varsigma_m}{\partial \rho^2}{\mathrm d}\rho &= \lim_{m\to\infty} \left[\rho^3\dfrac{\partial\varsigma_m}{\partial\rho}\right]_1^{1+(1/m)}-3\int_1^{1+(1/m)}\rho^2\dfrac{\partial\varsigma_m}{\partial\rho}\,{\mathrm d}\rho\tag{22} \\ &= \lim_{m\to\infty} -3\left[\rho^2\varsigma_m\right]_1^{1+(1/m)}+6\int_1^{1+(1/m)}\rho\,\varsigma_m\,{\mathrm d}\rho\tag{23} \\ &= 3\,,\tag{24} \end{align} next (18) equals \begin{align} \lim_{m\to\infty} \int_{K_\Delta} 2\rho^2 \dfrac{\partial \varsigma_m}{\partial \rho}\,{\mathrm d}\rho \cdot \int_{K_\Delta\cap\partial B_4} \sin\vartheta_1 {\mathrm d}\vartheta_1\wedge{\mathrm d}\vartheta_2\tag{25} \end{align} with \begin{align} \lim_{m\to\infty} \int_1^{1+(1/m)} 2\rho^2 \dfrac{\partial \varsigma_m}{\partial \rho}\,{\mathrm d}\rho &= \lim_{m\to\infty} 2\left[\rho^2\varsigma_m\right]_1^{1+(1/m)}-2\int_1^{1+(1/m)}\rho\,\varsigma_m\,{\mathrm d}\rho\tag{26} \\ &= -2\tag{27} \end{align} and finally (19) equals \begin{align} \lim_{m\to\infty} -\int_1^{1+(1/m)} \rho^2 \dfrac{\partial \varsigma_m}{\partial \rho}{\mathrm d}\rho \cdot \int_{K_\Delta\cap\partial B_4} \sin^3\vartheta_1 {\mathrm d}\vartheta_1\wedge{\mathrm d}\vartheta_2\tag{28} \end{align} with \begin{align} \lim_{m\to\infty} -\int_1^{1+(1/m)} \rho^2 \dfrac{\partial \varsigma_m}{\partial \rho}\,{\mathrm d}\rho =1.\tag{29} \end{align} It follows that the left hand side of (10) is equal to \begin{align} \int_{K_\Delta\cap\partial B_4} (4\sin^3\vartheta_1-2\sin\vartheta_1){\mathrm d}\vartheta_1\wedge{\mathrm d}\vartheta_2\tag{30} \end{align} whereas the right hand one equals \begin{align} 2{\mathop{\rm vol\,}\nolimits}_3(K_\Delta\cap B_4) &= 2\int_{K_\Delta\cap B_4} {\mathrm d}y_1\wedge{\mathrm d}x_2\wedge{\mathrm d}y_2\tag{31} \\ &= 2\int_0^1 \rho^2{\mathrm d}\rho \cdot \int_{K_\Delta\cap\partial B_4} \sin\vartheta_1{\mathrm d}\vartheta_1\wedge{\mathrm d}\vartheta_2\tag{32} \\ &= \dfrac{2}{3} \int_{K_\Delta\cap\partial B_4} \sin\vartheta_1{\mathrm d}\vartheta_1\wedge{\mathrm d}\vartheta_2\,,\tag{33} \end{align} hence (10) is equivalent to \begin{align} { \int_{K_\Delta\cap\partial B_4} \left(4\sin^3\vartheta_1-\dfrac{8}{3}\sin\vartheta_1\right){\mathrm d}\vartheta_1\wedge{\mathrm d}\vartheta_2 =0\label{x}\tag{34} } \end{align} where $K_\Delta\cap \partial B_4$ is a subset of the unit 2-dimensional sphere about the origin. In fact $\partial B_4$ is the unit 3-dimensional sphere about the origin, $K_\Delta\cap\partial B_4\subseteq E_\Delta^\perp\cap \partial B_4$ and $\dim E_\Delta^\perp\cap \partial B_4=2$, since $E_\Delta^\perp$ is a hyperplane transverse to $\partial B_4$.

By the arbitrary choice of $\Gamma$, equality (34) should always hold true, i.e. for every subset of the 2-sphere of the form $K_\Delta\cap \partial B_4$. The simplest case is when $\Gamma$ is a segment, so $\Gamma=\Delta$, $K_\Delta=E_\Delta^\perp$ and $K_\Delta\cap\partial B_4$ is the whole unit 2-sphere about the origin. The second simplest case is when $\Gamma$ is a polygon and, for each of its edges $\Delta$, $K_\Delta$ is a half-space and $K_\Delta\cap\partial B_4$ is a half of the unit 2-sphere about the origin. If $\Gamma$ è un polyhedron, for any of its edges $\Delta$, $K_\Delta\cap\partial B_4$ is a spherical polygon with two edges only, smaller than a hemisphere. If $\Gamma$ is a polychoron, for any of its edges $\Delta$, $K_\Delta$ is a 3-dimensional convex polyhedral cone with apex in the origin not including lines through the origin and $K_\Delta\cap\partial B_4$ is a spherical polygon.

By virtue of a result of Kazarnovskii's (stated in [2] and proved in [1] Theorem 8.1) one has \begin{equation}\label{k} \mu_1(B_4)= 2 \sum_{\Delta\in{\mathcal B}(\Gamma,1)} {\mathop{\rm vol\,}\nolimits}_1(\Delta) {\mathop{\rm vol\,}\nolimits}_3(K_\Delta\cap B_4)\,.\tag{3} \end{equation} The preceding formula reveals an interesting fact: $\mu_1(B_4)$ is invariant under orthogonal transformations of $\Gamma$ as a subset of $\mathbb R^4$. If $m\in\mathbb N^*$, let \begin{equation*} \varsigma_m(z)= \begin{cases} 1 & \textrm{if}\quad \Vert z\Vert \leq 1\,,\\ \exp\left(1+\dfrac{1}{m^2(1-\Vert z\Vert)^2-1}\right) &\textrm{if}\quad1< \Vert z\Vert <1+\dfrac{1}{m}\,,\\ 0 &\textrm{if}\quad \Vert z\Vert \geq 1+\dfrac{1}{m}\,. \end{cases} \end{equation*} The function $\varsigma_m$ is radial, smooth, its support is the full-dimensional ball of radius $1+(1/m)$ about the origin, it takes on the value 1 on $B_4$, $0\leqslant\varsigma_m\leqslant 1$ and it converges (point-wise) to the characteristic function of $B_4$. So \begin{equation} \mu_1(B_4) = \lim_{m\to\infty} \sum_{\Delta\in{\mathcal B}(\Gamma,1)} {\mathop{\rm vol\,}\nolimits}_1(\Delta) \int_{K_\Delta} \iota_\Delta^*(\upsilon_{iE_\Delta}\wedge h_{B_4}\mathrm {dd}^{\mathrm c} \varsigma_m)\label{s}\tag{4} \end{equation} and in view of (4), it would be enough to prove that, for any $\Delta\in{\mathcal B}(\Gamma,1)$, \begin{equation} { \lim_{m\to\infty} \int_{K_\Delta} \iota_\Delta^*(\upsilon_{iE_\Delta}\wedge h_{B_4}\mathrm {dd}^{\mathrm c} \varsigma_m) = 2{\mathop{\rm vol\,}\nolimits}_3(K_\Delta\cap B_4).\label{u}\tag{5} } \end{equation} Let $\Delta$ be fixed; up to an orthogonal trasformation, we may suppose that $E_\Delta=\{z\in\mathbb C^2\mid {\mathop{\rm Im\,}\nolimits} z_1=z_2=0\}$. In such a case, $K_\Delta\subset E_\Delta^\perp=\{z\in\mathbb C^2\mid x_1=0\}$, $\iota_\Delta^*(\upsilon_{iE_\Delta})={\mathrm d}y_1$ and \begin{align} &\iota_\Delta^*(\upsilon_{iE_\Delta}\wedge h_{B_4}\mathrm {dd}^{\mathrm c}\varsigma_m)\tag{6} \\ &= {\mathrm d}y_1\wedge \iota_\Delta^*\left(h_{B_4}2i\dfrac{\partial^2\varsigma_m}{\partial z_2\partial\bar z_2}{\mathrm d}z_2\wedge{\mathrm d}\bar z_2\right)\tag{7} \\ &= {\mathrm d}y_1\wedge\iota_\Delta^*\left(h_{B_4}2i\left[\dfrac{1}{2}\left(\dfrac{\partial}{\partial x_2}-i\dfrac{\partial}{\partial y_2}\right)\dfrac{1}{2}\left(\dfrac{\partial\varsigma_m}{\partial x_2}+i\dfrac{\partial\varsigma_m}{\partial y_2}\right)\right](-2i){\mathrm d}x_2\wedge{\mathrm d}y_2\right)\tag{8} \\ &= \iota_\Delta^*\left[h_{B_4}\left(\dfrac{\partial^2\varsigma_m}{\partial x_2^2}+\dfrac{\partial^2\varsigma_m}{\partial y_2^2}\right) {\mathrm d}y_1\wedge{\mathrm d}x_2\wedge {\mathrm d}y_2\right]\tag{9} \end{align} so that (5) becomes \begin{equation}\label{w} { \lim_{m\to\infty} \int_{K_\Delta} \iota_\Delta^*\left[h_{B_4}\left(\dfrac{\partial^2\varsigma_m}{\partial x_2^2}+\dfrac{\partial^2\varsigma_m}{\partial y_2^2}\right) {\mathrm d}y_1\wedge{\mathrm d}x_2\wedge {\mathrm d}y_2\right] = 2{\mathop{\rm vol\,}\nolimits}_3(K_\Delta\cap B_4).\tag{10} } \end{equation} In order to prove this equality let us pass to spherical coordinates in $E_\Delta^\perp$. Set $y_1=\rho\cos\vartheta_1$, $x_2=\rho\sin\vartheta_1\cos\vartheta_2$, $y_2=\rho\sin\vartheta_1\sin\vartheta_2$, with $\rho\geqslant0$, $\vartheta_1\in[0,\pi)$ and $\vartheta_2\in[0,2\pi)$. In these coordinates, on $K_\Delta$ one has $h_{B_4}=\rho$, \begin{equation} \dfrac{\partial \varsigma_m}{\partial x_2} = \dfrac{\partial\varsigma_m}{\partial \rho} \dfrac{\partial\rho}{\partial x_2} + \dfrac{\partial\varsigma_m}{\partial \vartheta_1} \dfrac{\partial\vartheta_1}{\partial x_2} + \dfrac{\partial\varsigma_m}{\partial \vartheta_2} \dfrac{\partial\vartheta_2}{\partial x_2} = \dfrac{\partial\varsigma_m}{\partial \rho} \dfrac{x_2}{\rho} = \dfrac{\partial\varsigma_m}{\partial \rho} \sin\vartheta_1\cos\vartheta_2\tag{11} \end{equation} and similarly \begin{align} \dfrac{\partial \varsigma_m}{\partial y_2} &= \dfrac{\partial\varsigma_m}{\partial \rho} \sin\vartheta_1\sin\vartheta_2\tag{12} \end{align} whence \begin{equation} \dfrac{\partial^2 \varsigma_m}{\partial x_2^2} = \dfrac{\partial^2 \varsigma_m}{\partial \rho^2} \sin^2\vartheta_1\cos^2\vartheta_2 + \dfrac{\partial \varsigma_m}{\partial \rho} \left( \dfrac{1-\sin^2\vartheta_1\cos^2\vartheta_2}{\rho} \right)\tag{13} \end{equation} and \begin{align} \dfrac{\partial^2 \varsigma_m}{\partial y_2^2} = \dfrac{\partial^2 \varsigma_m}{\partial \rho^2} \sin^2\vartheta_1\sin^2\vartheta_2 + \dfrac{\partial \varsigma_m}{\partial \rho} \left( \dfrac{1-\sin^2\vartheta_1\sin^2\vartheta_2}{\rho} \right)\tag{14} \end{align} so that \begin{align} & \lim_{m\to\infty} \int_{K_\Delta} h_{B_4}\left(\dfrac{\partial^2\varsigma_m}{\partial x_2^2}+\dfrac{\partial^2\varsigma_m}{\partial y_2^2}\right) {\mathrm d}y_1\wedge{\mathrm d}x_2\wedge {\mathrm d}y_2\tag{15} \\ &= \lim_{m\to\infty} \int_{K_\Delta} \rho \left[ \dfrac{\partial^2 \varsigma_m}{\partial \rho^2}\sin^2\vartheta_1 + \dfrac{\partial \varsigma_m}{\partial \rho}\left(\dfrac{2-\sin^2\vartheta_1}{\rho}\right) \right] \rho^2\sin\vartheta_1{\mathrm d}\rho\wedge{\mathrm d}\vartheta_1\wedge{\mathrm d}\vartheta_2\tag{16} \\ &= \lim_{m\to\infty} \int_{K_\Delta} \rho^3 \dfrac{\partial^2 \varsigma_m}{\partial \rho^2}\sin^3\vartheta_1 {\mathrm d}\rho\wedge{\mathrm d}\vartheta_1\wedge{\mathrm d}\vartheta_2\label{1}\tag{17} \\ &+ \lim_{m\to\infty} \int_{K_\Delta} 2\rho^2 \dfrac{\partial \varsigma_m}{\partial \rho}\sin\vartheta_1 {\mathrm d}\rho\wedge{\mathrm d}\vartheta_1\wedge{\mathrm d}\vartheta_2\label{2}\tag{18} \\ &- \lim_{m\to\infty} \int_{K_\Delta} \rho^2 \dfrac{\partial \varsigma_m}{\partial \rho}\sin^3\vartheta_1 {\mathrm d}\rho\wedge{\mathrm d}\vartheta_1\wedge{\mathrm d}\vartheta_2\label{3}.\tag{19} \end{align} For computing the latter integrals, let us observe that the derivatives of $\varsigma_m$ vanish at every $\rho\in[0,1]\cup[1+(1/m),+\infty)$ and that $$ 0\leqslant\int_1^{1+(1/m)}\rho\,\varsigma_m\,{\mathrm d}\rho\leqslant\left(1+\dfrac{1}{m}\right)\cdot1\cdot\left(1+\dfrac{1}{m}-1\right)=\dfrac{1}{m}+\dfrac{1}{m^2} $$ so \begin{equation}\label{z} \lim_{m\to\infty} \int_1^{1+(1/m)} \rho\,\varsigma_m\,{\mathrm d}\rho=0\,.\tag{20} \end{equation} Now by Fubini' theorem, (17) equals \begin{equation} \lim_{m\to\infty} \int_1^{1+(1/m)} \rho^3 \dfrac{\partial^2 \varsigma_m}{\partial \rho^2}{\mathrm d}\rho \cdot \int_{K_\Delta\cap\partial B_4} \sin^3\vartheta_1 {\mathrm d}\vartheta_1\wedge{\mathrm d}\vartheta_2\tag{21} \end{equation} with \begin{align} \lim_{m\to\infty} \int_1^{1+(1/m)} \rho^3 \dfrac{\partial^2 \varsigma_m}{\partial \rho^2}{\mathrm d}\rho &= \lim_{m\to\infty} \left[\rho^3\dfrac{\partial\varsigma_m}{\partial\rho}\right]_1^{1+(1/m)}-3\int_1^{1+(1/m)}\rho^2\dfrac{\partial\varsigma_m}{\partial\rho}\,{\mathrm d}\rho\tag{22} \\ &= \lim_{m\to\infty} -3\left[\rho^2\varsigma_m\right]_1^{1+(1/m)}+6\int_1^{1+(1/m)}\rho\,\varsigma_m\,{\mathrm d}\rho\tag{23} \\ &= 3\,,\tag{24} \end{align} next (18) equals \begin{align} \lim_{m\to\infty} \int_{K_\Delta} 2\rho^2 \dfrac{\partial \varsigma_m}{\partial \rho}\,{\mathrm d}\rho \cdot \int_{K_\Delta\cap\partial B_4} \sin\vartheta_1 {\mathrm d}\vartheta_1\wedge{\mathrm d}\vartheta_2\tag{25} \end{align} with \begin{align} \lim_{m\to\infty} \int_1^{1+(1/m)} 2\rho^2 \dfrac{\partial \varsigma_m}{\partial \rho}\,{\mathrm d}\rho &= \lim_{m\to\infty} 2\left[\rho^2\varsigma_m\right]_1^{1+(1/m)}-2\int_1^{1+(1/m)}\rho\,\varsigma_m\,{\mathrm d}\rho\tag{26} \\ &= -2\tag{27} \end{align} and finally (19) equals \begin{align} \lim_{m\to\infty} -\int_1^{1+(1/m)} \rho^2 \dfrac{\partial \varsigma_m}{\partial \rho}{\mathrm d}\rho \cdot \int_{K_\Delta\cap\partial B_4} \sin^3\vartheta_1 {\mathrm d}\vartheta_1\wedge{\mathrm d}\vartheta_2\tag{28} \end{align} with \begin{align} \lim_{m\to\infty} -\int_1^{1+(1/m)} \rho^2 \dfrac{\partial \varsigma_m}{\partial \rho}\,{\mathrm d}\rho =1.\tag{29} \end{align} It follows that the left hand side of (10) is equal to \begin{align} \int_{K_\Delta\cap\partial B_4} (4\sin^3\vartheta_1-2\sin\vartheta_1){\mathrm d}\vartheta_1\wedge{\mathrm d}\vartheta_2\tag{30} \end{align} whereas the right hand one equals \begin{align} 2{\mathop{\rm vol\,}\nolimits}_3(K_\Delta\cap B_4) &= 2\int_{K_\Delta\cap B_4} {\mathrm d}y_1\wedge{\mathrm d}x_2\wedge{\mathrm d}y_2\tag{31} \\ &= 2\int_0^1 \rho^2{\mathrm d}\rho \cdot \int_{K_\Delta\cap\partial B_4} \sin\vartheta_1{\mathrm d}\vartheta_1\wedge{\mathrm d}\vartheta_2\tag{32} \\ &= \dfrac{2}{3} \int_{K_\Delta\cap\partial B_4} \sin\vartheta_1{\mathrm d}\vartheta_1\wedge{\mathrm d}\vartheta_2\,,\tag{33} \end{align} hence (10) is equivalent to \begin{align} { \int_{K_\Delta\cap\partial B_4} \left(4\sin^3\vartheta_1-\dfrac{8}{3}\sin\vartheta_1\right){\mathrm d}\vartheta_1\wedge{\mathrm d}\vartheta_2 =0\label{x}\tag{34} } \end{align} where $K_\Delta\cap \partial B_4$ is a subset of the unit 2-dimensional sphere about the origin. In fact $\partial B_4$ is the unit 3-dimensional sphere about the origin, $K_\Delta\cap\partial B_4\subseteq E_\Delta^\perp\cap \partial B_4$ and $\dim E_\Delta^\perp\cap \partial B_4=2$, since $E_\Delta^\perp$ is a hyperplane transverse to $\partial B_4$.

By the arbitrary choice of $\Gamma$, equality (34) should always hold true, i.e. for every subset of the 2-sphere of the form $K_\Delta\cap \partial B_4$. The simplest case is when $\Gamma$ is a segment, so $\Gamma=\Delta$, $K_\Delta=E_\Delta^\perp$ and $K_\Delta\cap\partial B_4$ is the whole unit 2-sphere about the origin. The second simplest case is when $\Gamma$ is a polygon and, for each of its edges $\Delta$, $K_\Delta$ is a half-space and $K_\Delta\cap\partial B_4$ is a half of the unit 2-sphere about the origin. If $\Gamma$ is a polyhedron, for any of its edges $\Delta$, $K_\Delta\cap\partial B_4$ is a spherical polygon with two edges only, smaller than a hemisphere. If $\Gamma$ is a polychoron, for any of its edges $\Delta$, $K_\Delta$ is a 3-dimensional convex polyhedral cone with apex in the origin not including lines through the origin and $K_\Delta\cap\partial B_4$ is a spherical polygon.

For the sake of simplicity, let us set $n=2$ and in $\mathbb C^2$ consider the coordinates $z_1=x_1+iy_1$ e $z_2=x_2+iy_2$. If the convex polytope $\Gamma\subset\mathbb C^2$ is not reduced to a single point, its support function $h_\Gamma(z)=\max_{u\in\Gamma}{\mathop{\rm Re\,}\nolimits}\langle z,u\rangle$ is continuous, convex and 1-homogeneous, however it is not differentiable. The support function of $B_4$ is just the euclidean norm and the wedge product of $\mathrm {dd}^{\mathrm c} h_\Gamma$ and $\mathrm {dd}^{\mathrm c} h_{B_4}$ is defined as \begin{equation} \langle\!\langle \mathrm {dd}^{\mathrm c} h_\Gamma\wedge\mathrm {dd}^{\mathrm c} h_{B_4},\varsigma \rangle\!\rangle = \langle\!\langle \mathrm {dd}^{\mathrm c}(h_{B_4}\mathrm {dd}^{\mathrm c} h_\Gamma),\varsigma \rangle\!\rangle = \langle\!\langle \mathrm {dd}^{\mathrm c} h_\Gamma,h_{B_4}\mathrm {dd}^{\mathrm c}\varsigma \rangle\!\rangle.\tag{1} \end{equation} for any test function $\varsigma$ on $\mathbb C^2$. Since $h_\Gamma$ is convex, $\mathrm {dd}^{\mathrm c} h_\Gamma$ is a current of measure type, so its value against the continuous form with compact support $h_{B_4}\mathrm {dd}^{\mathrm c}\varsigma$ is well defined. Such a form is not smooth because $h_{B_4}$ is singular at the origin. The current $\mu_1=\mathrm {dd}^{\mathrm c} h_\Gamma\wedge\mathrm {dd}^{\mathrm c} h_{B_4}$, is just a positive measure and in order to compute $\mu_1(B_4)$ there are at least two different ways. The first one would involve a convolution of $h_\Gamma$ and $h_{B_4}$ with some regularizing kernel $\varrho_\epsilon$ thus getting $$ \mu(B_4)=\lim_{\varepsilon\to0}\int_{B_4}\mathrm {dd}^{\mathrm c}(h_\Gamma*\varrho_\varepsilon)\wedge \mathrm {dd}^{\mathrm c}(h_{B_4}*\varrho_\varepsilon). $$ This first method seems quite cumbersome to me, but I am possibly wrong. A second possibility makes use of a regularization of the characteristic function of $B_4$ together with the representation \begin{equation}\label{c} \mathrm {dd}^{\mathrm c} h_\Gamma = \sum_{\Delta\in{\mathcal B}(\Gamma,1)} {\mathop{\rm vol\,}\nolimits}_1(\Delta) \,\lambda_\Delta\tag{2} \end{equation} (cf. [1] Theorem 9.1), where ${\mathcal B}(\Gamma,1)$ is the set of the edges of $\Gamma$ and, for every such edge $\Delta$, ${\mathop{\rm vol\,}\nolimits}_1\Delta$ is the length of $\Delta$, whereas $\lambda_\Delta$ is the positive current acting against a generic $(1,1)$-test form $\varphi$ as $$ \langle\!\langle \lambda_\Delta,\varphi \rangle\!\rangle = \int_{K_\Delta} \iota_\Delta^*(\upsilon_{E'_\Delta}\wedge\varphi). $$ Here $K_\Delta$ is the dual cone to the face $\Delta$ (i.e. the relatively open convex polyhedral cone spanned by the rays issuing from the origin and orthogonal to the affine hulls of the cells of $\Gamma$ admitting $\Delta$ as an edge), $\iota_\Delta:K_\Delta\to \mathbb C^2$ is the inclusion and $\upsilon_{E'_\Delta}$ is the volume form on the line $E'_\Delta=iE_\Delta$, where $E_\Delta$ denotes the line parallel to $\Delta$ and passing through the origin.

For the sake of simplicity, let us set $n=2$ and in $\mathbb C^2$ consider the coordinates $z_1=x_1+iy_1$ e $z_2=x_2+iy_2$. If the convex polytope $\Gamma\subset\mathbb C^2$ is not reduced to a single point, its support function $h_\Gamma(z)=\max_{u\in\Gamma}{\mathop{\rm Re\,}\nolimits}\langle z,u\rangle$ is continuous, convex and 1-homogeneous, however it is not differentiable. The support function of $B_4$ is just the euclidean norm and the wedge product of $\mathrm {dd}^{\mathrm c} h_\Gamma$ and $\mathrm {dd}^{\mathrm c} h_{B_4}$ is defined as \begin{equation} \langle\!\langle \mathrm {dd}^{\mathrm c} h_\Gamma\wedge\mathrm {dd}^{\mathrm c} h_{B_4},\varsigma \rangle\!\rangle = \langle\!\langle \mathrm {dd}^{\mathrm c}(h_{B_4}\mathrm {dd}^{\mathrm c} h_\Gamma),\varsigma \rangle\!\rangle = \langle\!\langle \mathrm {dd}^{\mathrm c} h_\Gamma,h_{B_4}\mathrm {dd}^{\mathrm c}\varsigma \rangle\!\rangle.\tag{1} \end{equation} for any test function $\varsigma$ on $\mathbb C^2$. Since $h_\Gamma$ is convex, $\mathrm {dd}^{\mathrm c} h_\Gamma$ is a current of measure type, so its value against the continuous form with compact support $h_{B_4}\mathrm {dd}^{\mathrm c}\varsigma$ is well defined. Such a form is not smooth because $h_{B_4}$ is singular at the origin. The current $\mu_1=\mathrm {dd}^{\mathrm c} h_\Gamma\wedge\mathrm {dd}^{\mathrm c} h_{B_4}$, is just a positive measure and in order to compute $\mu_1(B_4)$ there are at least two different ways. The first one would involve a convolution of $h_\Gamma$ and $h_{B_4}$ with some regularizing kernel $\varrho_\epsilon$ thus getting $$ \mu_1(B_4)=\lim_{\varepsilon\to0}\int_{B_4}\mathrm {dd}^{\mathrm c}(h_\Gamma*\varrho_\varepsilon)\wedge \mathrm {dd}^{\mathrm c}(h_{B_4}*\varrho_\varepsilon). $$ This first method seems quite cumbersome to me, but I am possibly wrong. A second possibility makes use of a regularization of the characteristic function of $B_4$ together with the representation \begin{equation}\label{c} \mathrm {dd}^{\mathrm c} h_\Gamma = \sum_{\Delta\in{\mathcal B}(\Gamma,1)} {\mathop{\rm vol\,}\nolimits}_1(\Delta) \,\lambda_\Delta\tag{2} \end{equation} (cf. [1] Theorem 9.1), where ${\mathcal B}(\Gamma,1)$ is the set of the edges of $\Gamma$ and, for every such edge $\Delta$, ${\mathop{\rm vol\,}\nolimits}_1\Delta$ is the length of $\Delta$, whereas $\lambda_\Delta$ is the positive current acting against a generic $(1,1)$-test form $\varphi$ as $$ \langle\!\langle \lambda_\Delta,\varphi \rangle\!\rangle = \int_{K_\Delta} \iota_\Delta^*(\upsilon_{E'_\Delta}\wedge\varphi). $$ Here $K_\Delta$ is the dual cone to the face $\Delta$ (i.e. the relatively open convex polyhedral cone spanned by the rays issuing from the origin and orthogonal to the affine hulls of the cells of $\Gamma$ admitting $\Delta$ as an edge), $\iota_\Delta:K_\Delta\to \mathbb C^2$ is the inclusion and $\upsilon_{E'_\Delta}$ is the volume form on the line $E'_\Delta=iE_\Delta$, where $E_\Delta$ denotes the line parallel to $\Delta$ and passing through the origin.

By virtue of a result of Kazarnovskii's (stated in [2] and proved in [1] Theorem 8.1) one has \begin{equation}\label{k} \mu_1(B_4)= 2 \sum_{\Delta\in{\mathcal B}(\Gamma,1)} {\mathop{\rm vol\,}\nolimits}_1(\Delta) {\mathop{\rm vol\,}\nolimits}_3(K_\Delta\cap B_4)\,.\tag{3} \end{equation} The preceding formula reveals an interesting fact: $\mu_1(B_4)$ is invariant under orthogonal transformations of $\Gamma$ as a subset of $\mathbb R^4$. If $m\in\mathbb N^*$, let \begin{equation*} \varsigma_m(z)= \begin{cases} 1 & \textrm{if}\quad \Vert z\Vert \leq 1\,,\\ \exp\left(1+\dfrac{1}{m^2(1-\Vert z\Vert)^2-1}\right) &\textrm{if}\quad1< \Vert z\Vert <1+\dfrac{1}{m}\,,\\ 0 &\textrm{if}\quad \Vert z\Vert \geq 1+\dfrac{1}{m}\,. \end{cases} \end{equation*} The function $\varsigma_m$ is radial, smooth, its support is the full-dimensional ball of radius $1+(1/m)$ about the origin, it takes on the value 1 on $B_4$, $0\leqslant\varsigma_m\leqslant 1$ and it converges (point-wise) to the characteristic function of $B_4$. So \begin{equation} \mu_1(B_4) = \lim_{m\to\infty} \sum_{\Delta\in{\mathcal B}(\Gamma,1)} {\mathop{\rm vol\,}\nolimits}_1(\Delta) \int_{K_\Delta} \iota_\Delta^*(\upsilon_{iE_\Delta}\wedge h_{B_4}\mathrm {dd}^{\mathrm c} \varsigma_m)\label{s}\tag{4} \end{equation} and in view of (4), it would be enough to prove that, for any $\Delta\in{\mathcal B}(\Gamma,1)$, \begin{equation} { \lim_{m\to\infty} \int_{K_\Delta} \iota_\Delta^*(\upsilon_{iE_\Delta}\wedge h_{B_4}\mathrm {dd}^{\mathrm c} \varsigma_m) = 2{\mathop{\rm vol\,}\nolimits}_3(K_\Delta\cap B_4).\label{u}\tag{5} } \end{equation} Let $\Delta$ be fixed; up to an orthogonal trasformation, we may supposte that $E_\Delta=\{z\in\mathbb C^2\mid {\mathop{\rm Im\,}\nolimits} z_1=z_2=0\}$. In such a case, $K_\Delta\subset E_\Delta^\perp=\{z\in\mathbb C^2\mid x_1=0\}$, $\iota_\Delta^*(\upsilon_{iE_\Delta})={\mathrm d}y_1$ and \begin{align} &\iota_\Delta^*(\upsilon_{iE_\Delta}\wedge h_{B_4}\mathrm {dd}^{\mathrm c}\varsigma_m)\tag{6} \\ &= {\mathrm d}y_1\wedge \iota_\Delta^*\left(h_{B_4}2i\dfrac{\partial^2\varsigma_m}{\partial z_2\partial\bar z_2}{\mathrm d}z_2\wedge{\mathrm d}\bar z_2\right)\tag{7} \\ &= {\mathrm d}y_1\wedge\iota_\Delta^*\left(h_{B_4}2i\left[\dfrac{1}{2}\left(\dfrac{\partial}{\partial x_2}-i\dfrac{\partial}{\partial y_2}\right)\dfrac{1}{2}\left(\dfrac{\partial\varsigma_m}{\partial x_2}+i\dfrac{\partial\varsigma_m}{\partial y_2}\right)\right](-2i){\mathrm d}x_2\wedge{\mathrm d}y_2\right)\tag{8} \\ &= \iota_\Delta^*\left[h_{B_4}\left(\dfrac{\partial^2\varsigma_m}{\partial x_2^2}+\dfrac{\partial^2\varsigma_m}{\partial y_2^2}\right) {\mathrm d}y_1\wedge{\mathrm d}x_2\wedge {\mathrm d}y_2\right]\tag{9} \end{align} so that (5) becomes \begin{equation}\label{w} { \lim_{m\to\infty} \int_{K_\Delta} \iota_\Delta^*\left[h_{B_4}\left(\dfrac{\partial^2\varsigma_m}{\partial x_2^2}+\dfrac{\partial^2\varsigma_m}{\partial y_2^2}\right) {\mathrm d}y_1\wedge{\mathrm d}x_2\wedge {\mathrm d}y_2\right] = 2{\mathop{\rm vol\,}\nolimits}_3(K_\Delta\cap B_4).\tag{10} } \end{equation} In order to prove this equality let us pass to spherical coordinates in $E_\Delta^\perp$. Set $y_1=\rho\cos\vartheta_1$, $x_2=\rho\sin\vartheta_1\cos\vartheta_2$, $y_2=\rho\sin\vartheta_1\sin\vartheta_2$, with $\rho\geqslant0$, $\vartheta_1\in[0,\pi)$ and $\vartheta_2\in[0,2\pi)$. In these coordinates, on $K_\Delta$ on has $h_{B_4}=\rho$, \begin{equation} \dfrac{\partial \varsigma_m}{\partial x_2} = \dfrac{\partial\varsigma_m}{\partial \rho} \dfrac{\partial\rho}{\partial x_2} + \dfrac{\partial\varsigma_m}{\partial \vartheta_1} \dfrac{\partial\vartheta_1}{\partial x_2} + \dfrac{\partial\varsigma_m}{\partial \vartheta_2} \dfrac{\partial\vartheta_2}{\partial x_2} = \dfrac{\partial\varsigma_m}{\partial \rho} \dfrac{x_2}{\rho} = \dfrac{\partial\varsigma_m}{\partial \rho} \sin\vartheta_1\cos\vartheta_2\tag{11} \end{equation} and similarly \begin{align} \dfrac{\partial \varsigma_m}{\partial y_2} &= \dfrac{\partial\varsigma_m}{\partial \rho} \sin\vartheta_1\sin\vartheta_2\tag{12} \end{align} whence \begin{equation} \dfrac{\partial^2 \varsigma_m}{\partial x_2^2} = \dfrac{\partial^2 \varsigma_m}{\partial \rho^2} \sin^2\vartheta_1\cos^2\vartheta_2 + \dfrac{\partial \varsigma_m}{\partial \rho} \left( \dfrac{1-\sin^2\vartheta_1\cos^2\vartheta_2}{\rho} \right)\tag{13} \end{equation} and \begin{align} \dfrac{\partial^2 \varsigma_m}{\partial y_2^2} = \dfrac{\partial^2 \varsigma_m}{\partial \rho^2} \sin^2\vartheta_1\sin^2\vartheta_2 + \dfrac{\partial \varsigma_m}{\partial \rho} \left( \dfrac{1-\sin^2\vartheta_1\sin^2\vartheta_2}{\rho} \right)\tag{14} \end{align} so that \begin{align} & \lim_{m\to\infty} \int_{K_\Delta} h_{B_4}\left(\dfrac{\partial^2\varsigma_m}{\partial x_2^2}+\dfrac{\partial^2\varsigma_m}{\partial y_2^2}\right) {\mathrm d}y_1\wedge{\mathrm d}x_2\wedge {\mathrm d}y_2\tag{15} \\ &= \lim_{m\to\infty} \int_{K_\Delta} \rho \left[ \dfrac{\partial^2 \varsigma_m}{\partial \rho^2}\sin^2\vartheta_1 + \dfrac{\partial \varsigma_m}{\partial \rho}\left(\dfrac{2-\sin^2\vartheta_1}{\rho}\right) \right] \rho^2\sin\vartheta_1{\mathrm d}\rho\wedge{\mathrm d}\vartheta_1\wedge{\mathrm d}\vartheta_2\tag{16} \\ &= \lim_{m\to\infty} \int_{K_\Delta} \rho^3 \dfrac{\partial^2 \varsigma_m}{\partial \rho^2}\sin^3\vartheta_1 {\mathrm d}\rho\wedge{\mathrm d}\vartheta_1\wedge{\mathrm d}\vartheta_2\label{1}\tag{17} \\ &+ \lim_{m\to\infty} \int_{K_\Delta} 2\rho^2 \dfrac{\partial \varsigma_m}{\partial \rho}\sin\vartheta_1 {\mathrm d}\rho\wedge{\mathrm d}\vartheta_1\wedge{\mathrm d}\vartheta_2\label{2}\tag{18} \\ &- \lim_{m\to\infty} \int_{K_\Delta} \rho^2 \dfrac{\partial \varsigma_m}{\partial \rho}\sin^3\vartheta_1 {\mathrm d}\rho\wedge{\mathrm d}\vartheta_1\wedge{\mathrm d}\vartheta_2\label{3}.\tag{19} \end{align} For computing the latter integrals, let us observe that the derivatives of $\varsigma_m$ vanish at every $\rho\in[0,1]\cup[1+(1/m),+\infty)$ and that $$ 0\leqslant\int_1^{1+(1/m)}\rho\,\varsigma_m\,{\mathrm d}\rho\leqslant\left(1+\dfrac{1}{m}\right)\cdot1\cdot\left(1+\dfrac{1}{m}-1\right)=\dfrac{1}{m}+\dfrac{1}{m^2} $$ so \begin{equation}\label{z} \lim_{m\to\infty} \int_1^{1+(1/m)} \rho\,\varsigma_m\,{\mathrm d}\rho=0\,.\tag{20} \end{equation} Now by Fubini' theorem, (17) equals \begin{equation} \lim_{m\to\infty} \int_1^{1+(1/m)} \rho^3 \dfrac{\partial^2 \varsigma_m}{\partial \rho^2}{\mathrm d}\rho \cdot \int_{K_\Delta\cap\partial B_4} \sin^3\vartheta_1 {\mathrm d}\vartheta_1\wedge{\mathrm d}\vartheta_2\tag{21} \end{equation} with \begin{align} \lim_{m\to\infty} \int_1^{1+(1/m)} \rho^3 \dfrac{\partial^2 \varsigma_m}{\partial \rho^2}{\mathrm d}\rho &= \lim_{m\to\infty} \left[\rho^3\dfrac{\partial\varsigma_m}{\partial\rho}\right]_1^{1+(1/m)}-3\int_1^{1+(1/m)}\rho^2\dfrac{\partial\varsigma_m}{\partial\rho}\,{\mathrm d}\rho\tag{22} \\ &= \lim_{m\to\infty} -3\left[\rho^2\varsigma_m\right]_1^{1+(1/m)}+6\int_1^{1+(1/m)}\rho\,\varsigma_m\,{\mathrm d}\rho\tag{23} \\ &= 3\,,\tag{24} \end{align} next (18) equals \begin{align} \lim_{m\to\infty} \int_{K_\Delta} 2\rho^2 \dfrac{\partial \varsigma_m}{\partial \rho}\,{\mathrm d}\rho \cdot \int_{K_\Delta\cap\partial B_4} \sin\vartheta_1 {\mathrm d}\vartheta_1\wedge{\mathrm d}\vartheta_2\tag{25} \end{align} with \begin{align} \lim_{m\to\infty} \int_1^{1+(1/m)} 2\rho^2 \dfrac{\partial \varsigma_m}{\partial \rho}\,{\mathrm d}\rho &= \lim_{m\to\infty} 2\left[\rho^2\varsigma_m\right]_1^{1+(1/m)}-2\int_1^{1+(1/m)}\rho\,\varsigma_m\,{\mathrm d}\rho\tag{26} \\ &= -2\tag{27} \end{align} and finally (19) equals \begin{align} \lim_{m\to\infty} -\int_1^{1+(1/m)} \rho^2 \dfrac{\partial \varsigma_m}{\partial \rho}{\mathrm d}\rho \cdot \int_{K_\Delta\cap\partial B_4} \sin^3\vartheta_1 {\mathrm d}\vartheta_1\wedge{\mathrm d}\vartheta_2\tag{28} \end{align} with \begin{align} \lim_{m\to\infty} -\int_1^{1+(1/m)} \rho^2 \dfrac{\partial \varsigma_m}{\partial \rho}\,{\mathrm d}\rho =1.\tag{29} \end{align} It follows that the left hand side of (10) is equal to \begin{align} \int_{K_\Delta\cap\partial B_4} (4\sin^3\vartheta_1-2\sin\vartheta_1){\mathrm d}\vartheta_1\wedge{\mathrm d}\vartheta_2\tag{30} \end{align} whereas the right hand one equals \begin{align} 2{\mathop{\rm vol\,}\nolimits}_3(K_\Delta\cap B_4) &= 2\int_{K_\Delta\cap B_4} {\mathrm d}y_1\wedge{\mathrm d}x_2\wedge{\mathrm d}y_2\tag{31} \\ &= 2\int_0^1 \rho^2{\mathrm d}\rho \cdot \int_{K_\Delta\cap\partial B_4} \sin\vartheta_1{\mathrm d}\vartheta_1\wedge{\mathrm d}\vartheta_2\tag{32} \\ &= \dfrac{2}{3} \int_{K_\Delta\cap\partial B_4} \sin\vartheta_1{\mathrm d}\vartheta_1\wedge{\mathrm d}\vartheta_2\,,\tag{33} \end{align} hence (10) is equivalent to \begin{align} { \int_{K_\Delta\cap\partial B_4} \left(4\sin^3\vartheta_1-\dfrac{8}{3}\sin\vartheta_1\right){\mathrm d}\vartheta_1\wedge{\mathrm d}\vartheta_2 =0\label{x}\tag{34} } \end{align} where $K_\Delta\cap \partial B_4$ is a subset of the unit 2-dimensional sphere about the origin. In fact $\partial B_4$ is the unit 3-dimensional sphere about the origin, $K_\Delta\cap\partial B_4\subseteq E_\Delta^\perp\cap \partial B_4$ and $\dim E_\Delta^\perp\cap \partial B_4=2$, since $E_\Delta^\perp$ is a hyperplane transverse to $\partial B_4$.

By virtue of a result of Kazarnovskii's (stated in [2] and proved in [1] Theorem 8.1) one has \begin{equation}\label{k} \mu_1(B_4)= 2 \sum_{\Delta\in{\mathcal B}(\Gamma,1)} {\mathop{\rm vol\,}\nolimits}_1(\Delta) {\mathop{\rm vol\,}\nolimits}_3(K_\Delta\cap B_4)\,.\tag{3} \end{equation} The preceding formula reveals an interesting fact: $\mu_1(B_4)$ is invariant under orthogonal transformations of $\Gamma$ as a subset of $\mathbb R^4$. If $m\in\mathbb N^*$, let \begin{equation*} \varsigma_m(z)= \begin{cases} 1 & \textrm{if}\quad \Vert z\Vert \leq 1\,,\\ \exp\left(1+\dfrac{1}{m^2(1-\Vert z\Vert)^2-1}\right) &\textrm{if}\quad1< \Vert z\Vert <1+\dfrac{1}{m}\,,\\ 0 &\textrm{if}\quad \Vert z\Vert \geq 1+\dfrac{1}{m}\,. \end{cases} \end{equation*} The function $\varsigma_m$ is radial, smooth, its support is the full-dimensional ball of radius $1+(1/m)$ about the origin, it takes on the value 1 on $B_4$, $0\leqslant\varsigma_m\leqslant 1$ and it converges (point-wise) to the characteristic function of $B_4$. So \begin{equation} \mu_1(B_4) = \lim_{m\to\infty} \sum_{\Delta\in{\mathcal B}(\Gamma,1)} {\mathop{\rm vol\,}\nolimits}_1(\Delta) \int_{K_\Delta} \iota_\Delta^*(\upsilon_{iE_\Delta}\wedge h_{B_4}\mathrm {dd}^{\mathrm c} \varsigma_m)\label{s}\tag{4} \end{equation} and in view of (4), it would be enough to prove that, for any $\Delta\in{\mathcal B}(\Gamma,1)$, \begin{equation} { \lim_{m\to\infty} \int_{K_\Delta} \iota_\Delta^*(\upsilon_{iE_\Delta}\wedge h_{B_4}\mathrm {dd}^{\mathrm c} \varsigma_m) = 2{\mathop{\rm vol\,}\nolimits}_3(K_\Delta\cap B_4).\label{u}\tag{5} } \end{equation} Let $\Delta$ be fixed; up to an orthogonal trasformation, we may supposte that $E_\Delta=\{z\in\mathbb C^2\mid {\mathop{\rm Im\,}\nolimits} z_1=z_2=0\}$. In such a case, $K_\Delta\subset E_\Delta^\perp=\{z\in\mathbb C^2\mid x_1=0\}$, $\iota_\Delta^*(\upsilon_{iE_\Delta})={\mathrm d}y_1$ and \begin{align} &\iota_\Delta^*(\upsilon_{iE_\Delta}\wedge h_{B_4}\mathrm {dd}^{\mathrm c}\varsigma_m)\tag{6} \\ &= {\mathrm d}y_1\wedge \iota_\Delta^*\left(h_{B_4}2i\dfrac{\partial^2\varsigma_m}{\partial z_2\partial\bar z_2}{\mathrm d}z_2\wedge{\mathrm d}\bar z_2\right)\tag{7} \\ &= {\mathrm d}y_1\wedge\iota_\Delta^*\left(h_{B_4}2i\left[\dfrac{1}{2}\left(\dfrac{\partial}{\partial x_2}-i\dfrac{\partial}{\partial y_2}\right)\dfrac{1}{2}\left(\dfrac{\partial\varsigma_m}{\partial x_2}+i\dfrac{\partial\varsigma_m}{\partial y_2}\right)\right](-2i){\mathrm d}x_2\wedge{\mathrm d}y_2\right)\tag{8} \\ &= \iota_\Delta^*\left[h_{B_4}\left(\dfrac{\partial^2\varsigma_m}{\partial x_2^2}+\dfrac{\partial^2\varsigma_m}{\partial y_2^2}\right) {\mathrm d}y_1\wedge{\mathrm d}x_2\wedge {\mathrm d}y_2\right]\tag{9} \end{align} so that (5) becomes \begin{equation}\label{w} { \lim_{m\to\infty} \int_{K_\Delta} \iota_\Delta^*\left[h_{B_4}\left(\dfrac{\partial^2\varsigma_m}{\partial x_2^2}+\dfrac{\partial^2\varsigma_m}{\partial y_2^2}\right) {\mathrm d}y_1\wedge{\mathrm d}x_2\wedge {\mathrm d}y_2\right] = 2{\mathop{\rm vol\,}\nolimits}_3(K_\Delta\cap B_4).\tag{10} } \end{equation} In order to prove this equality let us pass to spherical coordinates in $E_\Delta^\perp$. Set $y_1=\rho\cos\vartheta_1$, $x_2=\rho\sin\vartheta_1\cos\vartheta_2$, $y_2=\rho\sin\vartheta_1\sin\vartheta_2$, with $\rho\geqslant0$, $\vartheta_1\in[0,\pi)$ and $\vartheta_2\in[0,2\pi)$. In these coordinates, on $K_\Delta$ one has $h_{B_4}=\rho$, \begin{equation} \dfrac{\partial \varsigma_m}{\partial x_2} = \dfrac{\partial\varsigma_m}{\partial \rho} \dfrac{\partial\rho}{\partial x_2} + \dfrac{\partial\varsigma_m}{\partial \vartheta_1} \dfrac{\partial\vartheta_1}{\partial x_2} + \dfrac{\partial\varsigma_m}{\partial \vartheta_2} \dfrac{\partial\vartheta_2}{\partial x_2} = \dfrac{\partial\varsigma_m}{\partial \rho} \dfrac{x_2}{\rho} = \dfrac{\partial\varsigma_m}{\partial \rho} \sin\vartheta_1\cos\vartheta_2\tag{11} \end{equation} and similarly \begin{align} \dfrac{\partial \varsigma_m}{\partial y_2} &= \dfrac{\partial\varsigma_m}{\partial \rho} \sin\vartheta_1\sin\vartheta_2\tag{12} \end{align} whence \begin{equation} \dfrac{\partial^2 \varsigma_m}{\partial x_2^2} = \dfrac{\partial^2 \varsigma_m}{\partial \rho^2} \sin^2\vartheta_1\cos^2\vartheta_2 + \dfrac{\partial \varsigma_m}{\partial \rho} \left( \dfrac{1-\sin^2\vartheta_1\cos^2\vartheta_2}{\rho} \right)\tag{13} \end{equation} and \begin{align} \dfrac{\partial^2 \varsigma_m}{\partial y_2^2} = \dfrac{\partial^2 \varsigma_m}{\partial \rho^2} \sin^2\vartheta_1\sin^2\vartheta_2 + \dfrac{\partial \varsigma_m}{\partial \rho} \left( \dfrac{1-\sin^2\vartheta_1\sin^2\vartheta_2}{\rho} \right)\tag{14} \end{align} so that \begin{align} & \lim_{m\to\infty} \int_{K_\Delta} h_{B_4}\left(\dfrac{\partial^2\varsigma_m}{\partial x_2^2}+\dfrac{\partial^2\varsigma_m}{\partial y_2^2}\right) {\mathrm d}y_1\wedge{\mathrm d}x_2\wedge {\mathrm d}y_2\tag{15} \\ &= \lim_{m\to\infty} \int_{K_\Delta} \rho \left[ \dfrac{\partial^2 \varsigma_m}{\partial \rho^2}\sin^2\vartheta_1 + \dfrac{\partial \varsigma_m}{\partial \rho}\left(\dfrac{2-\sin^2\vartheta_1}{\rho}\right) \right] \rho^2\sin\vartheta_1{\mathrm d}\rho\wedge{\mathrm d}\vartheta_1\wedge{\mathrm d}\vartheta_2\tag{16} \\ &= \lim_{m\to\infty} \int_{K_\Delta} \rho^3 \dfrac{\partial^2 \varsigma_m}{\partial \rho^2}\sin^3\vartheta_1 {\mathrm d}\rho\wedge{\mathrm d}\vartheta_1\wedge{\mathrm d}\vartheta_2\label{1}\tag{17} \\ &+ \lim_{m\to\infty} \int_{K_\Delta} 2\rho^2 \dfrac{\partial \varsigma_m}{\partial \rho}\sin\vartheta_1 {\mathrm d}\rho\wedge{\mathrm d}\vartheta_1\wedge{\mathrm d}\vartheta_2\label{2}\tag{18} \\ &- \lim_{m\to\infty} \int_{K_\Delta} \rho^2 \dfrac{\partial \varsigma_m}{\partial \rho}\sin^3\vartheta_1 {\mathrm d}\rho\wedge{\mathrm d}\vartheta_1\wedge{\mathrm d}\vartheta_2\label{3}.\tag{19} \end{align} For computing the latter integrals, let us observe that the derivatives of $\varsigma_m$ vanish at every $\rho\in[0,1]\cup[1+(1/m),+\infty)$ and that $$ 0\leqslant\int_1^{1+(1/m)}\rho\,\varsigma_m\,{\mathrm d}\rho\leqslant\left(1+\dfrac{1}{m}\right)\cdot1\cdot\left(1+\dfrac{1}{m}-1\right)=\dfrac{1}{m}+\dfrac{1}{m^2} $$ so \begin{equation}\label{z} \lim_{m\to\infty} \int_1^{1+(1/m)} \rho\,\varsigma_m\,{\mathrm d}\rho=0\,.\tag{20} \end{equation} Now by Fubini' theorem, (17) equals \begin{equation} \lim_{m\to\infty} \int_1^{1+(1/m)} \rho^3 \dfrac{\partial^2 \varsigma_m}{\partial \rho^2}{\mathrm d}\rho \cdot \int_{K_\Delta\cap\partial B_4} \sin^3\vartheta_1 {\mathrm d}\vartheta_1\wedge{\mathrm d}\vartheta_2\tag{21} \end{equation} with \begin{align} \lim_{m\to\infty} \int_1^{1+(1/m)} \rho^3 \dfrac{\partial^2 \varsigma_m}{\partial \rho^2}{\mathrm d}\rho &= \lim_{m\to\infty} \left[\rho^3\dfrac{\partial\varsigma_m}{\partial\rho}\right]_1^{1+(1/m)}-3\int_1^{1+(1/m)}\rho^2\dfrac{\partial\varsigma_m}{\partial\rho}\,{\mathrm d}\rho\tag{22} \\ &= \lim_{m\to\infty} -3\left[\rho^2\varsigma_m\right]_1^{1+(1/m)}+6\int_1^{1+(1/m)}\rho\,\varsigma_m\,{\mathrm d}\rho\tag{23} \\ &= 3\,,\tag{24} \end{align} next (18) equals \begin{align} \lim_{m\to\infty} \int_{K_\Delta} 2\rho^2 \dfrac{\partial \varsigma_m}{\partial \rho}\,{\mathrm d}\rho \cdot \int_{K_\Delta\cap\partial B_4} \sin\vartheta_1 {\mathrm d}\vartheta_1\wedge{\mathrm d}\vartheta_2\tag{25} \end{align} with \begin{align} \lim_{m\to\infty} \int_1^{1+(1/m)} 2\rho^2 \dfrac{\partial \varsigma_m}{\partial \rho}\,{\mathrm d}\rho &= \lim_{m\to\infty} 2\left[\rho^2\varsigma_m\right]_1^{1+(1/m)}-2\int_1^{1+(1/m)}\rho\,\varsigma_m\,{\mathrm d}\rho\tag{26} \\ &= -2\tag{27} \end{align} and finally (19) equals \begin{align} \lim_{m\to\infty} -\int_1^{1+(1/m)} \rho^2 \dfrac{\partial \varsigma_m}{\partial \rho}{\mathrm d}\rho \cdot \int_{K_\Delta\cap\partial B_4} \sin^3\vartheta_1 {\mathrm d}\vartheta_1\wedge{\mathrm d}\vartheta_2\tag{28} \end{align} with \begin{align} \lim_{m\to\infty} -\int_1^{1+(1/m)} \rho^2 \dfrac{\partial \varsigma_m}{\partial \rho}\,{\mathrm d}\rho =1.\tag{29} \end{align} It follows that the left hand side of (10) is equal to \begin{align} \int_{K_\Delta\cap\partial B_4} (4\sin^3\vartheta_1-2\sin\vartheta_1){\mathrm d}\vartheta_1\wedge{\mathrm d}\vartheta_2\tag{30} \end{align} whereas the right hand one equals \begin{align} 2{\mathop{\rm vol\,}\nolimits}_3(K_\Delta\cap B_4) &= 2\int_{K_\Delta\cap B_4} {\mathrm d}y_1\wedge{\mathrm d}x_2\wedge{\mathrm d}y_2\tag{31} \\ &= 2\int_0^1 \rho^2{\mathrm d}\rho \cdot \int_{K_\Delta\cap\partial B_4} \sin\vartheta_1{\mathrm d}\vartheta_1\wedge{\mathrm d}\vartheta_2\tag{32} \\ &= \dfrac{2}{3} \int_{K_\Delta\cap\partial B_4} \sin\vartheta_1{\mathrm d}\vartheta_1\wedge{\mathrm d}\vartheta_2\,,\tag{33} \end{align} hence (10) is equivalent to \begin{align} { \int_{K_\Delta\cap\partial B_4} \left(4\sin^3\vartheta_1-\dfrac{8}{3}\sin\vartheta_1\right){\mathrm d}\vartheta_1\wedge{\mathrm d}\vartheta_2 =0\label{x}\tag{34} } \end{align} where $K_\Delta\cap \partial B_4$ is a subset of the unit 2-dimensional sphere about the origin. In fact $\partial B_4$ is the unit 3-dimensional sphere about the origin, $K_\Delta\cap\partial B_4\subseteq E_\Delta^\perp\cap \partial B_4$ and $\dim E_\Delta^\perp\cap \partial B_4=2$, since $E_\Delta^\perp$ is a hyperplane transverse to $\partial B_4$.