$\displaystyle f=\frac{E}{4\pi}(\frac{x}{(x^2+y^2+z^2)^{3/2}}dy\wedge dz+\frac{y}{(x^2+y^2+z^2)^{3/2}}dz\wedge dx+\frac{x}{(x^2+y^2+z^2)^{3/2}}dx\wedge dy)$
over a sphere $\displaystyle x^2+y^2+z^2 \ge \mathbb{R}^3$
$\displaystyle f=\frac{E}{4\pi}(\frac{x}{(x^2+y^2+z^2)^{3/2}}dy\wedge dz+\frac{y}{(x^2+y^2+z^2)^{3/2}}dz\wedge dx+\frac{x}{(x^2+y^2+z^2)^{3/2}}dx\wedge dy)$
over a sphere $\displaystyle x^2+y^2+z^2 \ge \mathbb{R}^3$
I suppose you mean the sphere $\displaystyle S\equiv x^2+y^2+z^2=R^2$. In such case, apply the Divergence Theorem to $\displaystyle V-V_r$ where $\displaystyle 0<r<R$ . That is,
$\displaystyle \iint_Sw+\iint_{S_r}w=\iiint_{V-V_r}dw=0$
So,
$\displaystyle \iint_Sw=-\iint_{S_r}w$