 Solution

ID: past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2021-resit/8/b/solution

Take

0 < r_{1} < r_{2}

\nabla \cdot F = \frac{1}{r ^{2}} \frac{d}{d r} (r^{2} r^{- β}) = (2 - β) r^{- β - 1} .

d V = r^{2} sin θ d r d θ d ϕ

, direct integration gives

∭_{V} \nabla \cdot F d V = \int_{r_{1}}^{r_{2}} \int_{0}^{π} \int_{0}^{2 π} (2 - β) r^{1 - β} sin θ d ϕ d θ d r = 4 π (r_{2}^{2 - β} - r_{1}^{2 - β}) .

For the independent boundary calculation, the outer sphere has outward unit normal vector

+ \hat{r}

and the inner sphere has outward unit normal vector

- \hat{r}

. Their fluxes are

Φ_{outer} = 4 π r_{2}^{2} r_{2}^{- β}, Φ_{inner} = - 4 π r_{1}^{2} r_{1}^{- β} .

Their sum equals the direct volume integral, verifying the divergence theorem. The restriction

β \neq = 2

permits the displayed power-antiderivative step. At

β = 2

, treated separately, both the shell volume integral and net boundary flux are zero; the excluded origin must still not be filled in.

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