Surface integral on unit circle Let $S$ be the unit sphere in $\mathbb{R}^3$ and write $F(x)=\nabla V(x)$ where $V(x)=1/|x|$
Evaluate $$\iint_S F\cdot n dS$$
Without using divergence theorem, we can evaluate it straightforwardly, $$\iint_S-\frac{x}{|x|^3}\cdot \frac{x}{|x|}=\iint_S-1dS=-surface \ area (S)=-4\pi$$ I used the fact that $|x|=1$.
However, that is not the correct answer. Can anyone explain?
 A: We begin by noting that:
$$\nabla U = \left(\frac{\partial U}{\partial r}\right)\hat{\boldsymbol{r}}+\left(\frac{1}{r}\frac{\partial U}{\partial \theta}\right)\hat{\boldsymbol{\theta}}+\left(\frac{1}{r\sin(\theta)}\frac{\partial U}{\partial \phi}\right)\hat{\boldsymbol{\phi}}$$
We can immediately see that if $U = V(r) = \frac{1}{r}$, then: $\frac{\partial U}{\partial \theta}=\frac{\partial U}{\partial \phi}=0$, and so:
$$\nabla U=-\frac{\hat{\boldsymbol{r}}}{r^{2}}$$
So your integral becomes (noting that at any point on the surface of the unit sphere has normal $\hat{\boldsymbol{r}}$):
$$\iint_{S}-\frac{\hat{\boldsymbol{r}}}{r^{2}}\cdot \hat{\boldsymbol{r}}\:\mathrm{d}S = -\iint_{S}\:\mathrm{d}S=-4\pi$$
We can verify this is correct by using the divergence theorem:
$$\iint_{S}(\nabla U)\cdot \mathrm{d}\boldsymbol{S}=\iiint_{B}\nabla \cdot\left(\nabla U\right)\:\mathrm{d}V$$
Where $B$ represents the unit ball in $\mathbb{R}^{3}$. Evaluating, we have:
$$\nabla \cdot(\nabla U) = -4\pi \delta(\boldsymbol{r})$$
Where $\delta(\cdot)$ represents the Dirac delta distribution. Integrating, we therefore have:
$$\iint_{S}\nabla\left(\frac{1}{r}\right)\cdot \mathrm{d}\boldsymbol{S}=-\iiint_{B}4\pi \delta (\boldsymbol{r})\:\mathrm{d}V=-4\pi$$
So you have got the correct answer, what makes you think that it is wrong?
