Divergence Theorem β€” Question 9

Section 19-6 Β· Calculus pages 3597–3598

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Question 9

On ℝ3\{𝟎}\mathbb R^3\setminus\{\mathbf 0\}, consider 𝑭(x,y,z)=⟨x,y,z⟩(x2+y2+z2)3/2.\mathbf F(x,y,z)=\frac{\langle x,y,z\rangle}{(x^2+y^2+z^2)^{3/2}}. Let SS be the outward-oriented unit sphere.

Tasks

  1. Compute the flux through SS directly.

  2. Explain why zero divergence away from the origin does not permit applying the Divergence Theorem to the full unit ball.

  3. Reconcile the result using a punctured ball.

Question 9 – Solution

Strategy. The nonzero flux is caused by a singularity inside the sphere; remove a small ball so the theorem’s hypotheses hold.

Step 1: Direct flux On the unit sphere, 𝑭=𝒏\mathbf F=\mathbf n, so ∬S𝑭⋅𝒏dS=∬S1dS=4Ο€.\boxed{\iint_S\mathbf F\cdot\mathbf n\,dS =\iint_S1\,dS=4\pi}. Although βˆ‡β‹…π‘­=0\nabla\cdot\mathbf F=0 for r>0r>0, the field is undefined at the origin and is not smooth throughout the full ball.

Diagram shown in the original SVG worksheet.

Step 2: Puncture the ball Remove the ball r<Ξ΅r<\varepsilon. On the shell Ρ≀r≀1\varepsilon\le r\le 1, the divergence is zero, so the total outward flux is zero.

Step 3: Track both boundaries The outer flux is 4Ο€4\pi. On r=Ξ΅r=\varepsilon, the shell’s outward normal is βˆ’π’†r-\mathbf e_r, while 𝑭=𝒆r/Ξ΅2\mathbf F=\mathbf e_r/\varepsilon^2. Thus Ξ¦r=Ξ΅=βˆ’1Ξ΅2(4πΡ2)=βˆ’4Ο€.\Phi_{r=\varepsilon}=-\frac 1{\varepsilon^2}(4\pi\varepsilon^2)=-4\pi. Therefore 4Ο€βˆ’4Ο€=04\pi-4\pi=0, consistent with the theorem on the punctured region.

Verification The inner flux remains βˆ’4Ο€-4\pi as Ξ΅β†’0+\varepsilon\to 0^+, recording the enclosed point source rather than disappearing.