Vector Calculus | Part IA, 2003

State the divergence theorem. By applying this to f(x)kf(\mathbf{x}) \mathbf{k}, where f(x)f(\mathbf{x}) is a scalar field in a closed region VV in R3\mathbb{R}^{3} bounded by a piecewise smooth surface SS, and k\mathbf{k} an arbitrary constant vector, show that

VfdV=SfdS\int_{V} \nabla f d V=\int_{S} f d \mathbf{S}

A vector field G\mathbf{G} satisfies

G=ρ(x) with ρ(x)={ρ0xa0x>a\begin{gathered} \boldsymbol{\nabla} \cdot \mathbf{G}=\rho(\mathbf{x}) \\ \text { with } \quad \rho(\mathbf{x})= \begin{cases}\rho_{0} & |\mathbf{x}| \leqslant a \\ 0 & |\mathbf{x}|>a\end{cases} \end{gathered}

By applying the divergence theorem to VGdV\int_{V} \nabla \cdot \mathbf{G} d V, prove Gauss's law

SGdS=Vρ(x)dV\int_{S} \mathbf{G} \cdot d \mathbf{S}=\int_{V} \rho(\mathbf{x}) d V

where SS is the piecewise smooth surface bounding the volume VV.

Consider the spherically symmetric solution

G(x)=G(r)xr\mathbf{G}(\mathbf{x})=G(r) \frac{\mathbf{x}}{r}

where r=xr=|\mathbf{x}|. By using Gauss's law with SS a sphere of radius rr, centre 0\mathbf{0}, in the two cases 0<ra0<r \leqslant a and r>ar>a, show that

G(x)={ρ03xraρ03(ar)3xr>a\mathbf{G}(\mathbf{x})= \begin{cases}\frac{\rho_{0}}{3} \mathbf{x} & r \leqslant a \\ \frac{\rho_{0}}{3}\left(\frac{a}{r}\right)^{3} \mathbf{x} & r>a\end{cases}

The scalar field f(x)f(\mathbf{x}) satisfies G=f\mathbf{G}=\nabla f. Assuming that f0f \rightarrow 0 as rr \rightarrow \infty, and that ff is continuous at r=ar=a, find ff everywhere.

By using a symmetry argument, explain why ()(*) is clearly satisfied for this ff if SS is any sphere centred at the origin.

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