Let $$\Omega = \{(x_1, x_2, x_3) \in \mathbb{R}^3 : \max(|x|_1, |x|_2, |x|_3) \leq 1\}$$ $$F_i(x) = \frac{x_i}{\|x\|^3}$$ and suppose $\varphi(y)$ be a continuously differentiable function of $y_i = x_i/\|x\|$, with $\varphi$ having average value $1$ over the unit sphere.

Calculate $$\int_{\partial \Omega} \varphi F \cdot n \; dS$$

The solution I get is zero by the divergence theorem. But this is not what the solution manual says. Am I wrong, or is the solution manual wrong? I never use the assumption on the average value of $\varphi$.

  • $\begingroup$ $\vec{\nabla}\cdot\vec{F}$ is singular at $\vec{x} = 0$. You cannot apply divergence theorem to the whole $\Omega$. Instead, you need to apply it to $\Omega \setminus \{ \vec{0} \}$. $\endgroup$ – achille hui Jun 28 '14 at 6:09

If $\mathbf{F}(\mathbf{x})=\frac{\mathbf{x}}{\|\mathbf{x}\|^3}$, then $\mathrm{d}\sigma=\mathbf{F}\cdot\mathrm{d}\mathbf{S}=\mathbf{F}\cdot\mathbf{n}\,\mathrm{d}S$ is the differential solid angle. The solid angle surface integral of a function over an arbitrary region is then equal to the surface integral over the unit sphere

$$\begin{align} \oint_{\partial\Omega}\phi\mathbf{F}\cdot\mathrm{d}\mathbf{S} &=\oint_{\partial\Omega}\phi\,\mathrm{d}\sigma\\ &=\oint_{S_2}\phi\,\mathrm{d}\sigma\\ &=4\pi\cdot\frac{1}{4\pi}\oint_{S_2}\phi\,\mathrm{d}\sigma\\ &=4\pi, \end{align}$$

where in the last line we've used the given information that the average value of $\phi$ over the unit sphere is $\frac{1}{4\pi}\oint_{S_2}\phi\,\mathrm{d}\sigma=1$.


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