Curl and Divergence β€” Question 9

Section 19-1 Β· Calculus pages 3497–3498

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Question 9

On D=ℝ3\{𝟎}D=\mathbb R^3\setminus\{\mathbf 0\}, define 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}}.

Tasks

  1. Compute βˆ‡β‹…π‘­\nabla\cdot\mathbf F on DD.

  2. Compute or otherwise justify βˆ‡Γ—π‘­\nabla\times\mathbf F on DD.

  3. Explain why the excluded origin must be stated.

Question 9 – Solution

Strategy. Write r=(x2+y2+z2)1/2r=(x^2+y^2+z^2)^{1/2} and differentiate xrβˆ’3xr^{-3}, yrβˆ’3yr^{-3}, and zrβˆ’3zr^{-3}.

Step 1: Divergence Since βˆ‚βˆ‚x(xrβˆ’3)=rβˆ’3βˆ’3x2rβˆ’5,\frac{\partial}{\partial x}(xr^{-3})=r^{-3}-3x^2r^{-5}, with analogous formulas for yy and zz, βˆ‡β‹…π‘­=3rβˆ’3βˆ’3(x2+y2+z2)rβˆ’5=3rβˆ’3βˆ’3r2rβˆ’5=0,r>0.\begin{align*} \nabla\cdot\mathbf F &=3r^{-3}-3(x^2+y^2+z^2)r^{-5}\\ &=3r^{-3}-3r^2r^{-5}=\boxed{0},\qquad r>0. \end{align*}

Diagram shown in the original SVG worksheet.

Step 2: Curl Because 𝑭=βˆ‡(βˆ’1r)\mathbf F=\nabla\!\left(-\frac 1r\right) on DD, the curl-of-a-gradient identity gives βˆ‡Γ—π‘­=𝟎.\boxed{\nabla\times\mathbf F=\mathbf 0}.

Step 3: Domain caveat The formula is undefined at the origin, and the derivatives used above do not exist there. The conclusions are pointwise statements on DD, not statements at 𝟎\mathbf 0.

Verification Direct cross differentiation gives, for example, Ryβˆ’Qz=βˆ’3yzrβˆ’5βˆ’(βˆ’3yzrβˆ’5)=0R_y-Q_z=-3yzr^{-5}-(-3yzr^{-5})=0, and the other two components cancel similarly.