Flux and scaling on balls

The sphere is a C1 boundary: near any point one nonzero component of $x-a$ lets its equation be solved as a smooth square-root graph, with the ball on the inner side. Its outward normal is $\nu =(x-a)\mathrm{/}R$. For $F=x-a$, each ${\mathrm{\partial }}_i{F}_i=1$, so $\mathrm{div}F=n$, and on the boundary $F\cdot \nu =R$. F1 yields $n\mathrm{\mid }{B}_R(a)\mathrm{\mid }=R\mathrm{\mid }\mathrm{\partial }{B}_R(a)\mathrm{\mid }$; dividing by R and using F2 gives both stated area formulas. The ball has positive volume since it contains a cube of positive side length. In dimension three a spherical chart $X(\theta ,\varphi )=(\cos \theta \sin \varphi ,\sin \theta \sin \varphi ,\cos \varphi )$, $0<\theta <2\pi$, $0<\varphi <\pi$, has tangent squared lengths ${\sin }^2\varphi$ and 1 and zero cross inner product, hence density $\sin \varphi$. Rotated charts cover its omitted meridian and poles.

For a constant vector b all partial derivatives vanish. Applying F1 gives ${\int }_{\mathrm{\partial }{B}_R(a)}b\cdot \nu dS={\int }_{B_R(a)}0dx=0$. For the radial field and $r=\mathrm{\mid }x-a\mathrm{\mid }>0$, ${\mathrm{\partial }}_i{F}_i=f(r)+f^{\mathrm{\prime }}(r)(x_i-a_i{)}^2\mathrm{/}r$. Summing gives $\mathrm{div}F=nf(r)+r{f}^{\mathrm{\prime }}(r)$. The stipulated C1 extension supplies the value at the centre and the hypotheses of F1; no assertion about a singular f at zero is needed. On the sphere the flux density is the constant $f(R)R$, proving the claimed flux.

For the explicit polynomial field $F=\mathrm{\mid }x-a{\mathrm{\mid }}^2(x-a)$ the extension is automatic. Its divergence is $(n+2)\mathrm{\mid }x-a{\mathrm{\mid }}^2$ also at the centre by direct differentiation. Thus its flux is $R^3\mathrm{\mid }\mathrm{\partial }{B}_R(a)\mathrm{\mid }$, and F1 gives ${\int }_{B_R(a)}\mathrm{\mid }x-a{\mathrm{\mid }}^{2}dx=R^3\mathrm{\mid }\mathrm{\partial }{B}_R(a)\mathrm{\mid }\mathrm{/}(n+2)$. In particular at R=1, a=0 this moment is $n\mathrm{\mid }{B}_1\mathrm{\mid }\mathrm{/}(n+2)$ by F2.