How to show $\unicode{x222F} \dfrac{\hat{r} \times \vec{dS}}{r^2}=0$ I can do the following derivation using solid angle:
$$\unicode{x222F} \dfrac{\hat{r} \cdot \vec{dS}}{r^2}
=\unicode{x222F} \dfrac{dS \cos\alpha}{r^2}
=\int^{2\pi}_0 \int^\pi_0 \sin\theta\ d\theta\ d\phi=4\pi$$
However I have no clue how to show that:
$$\unicode{x222F} \dfrac{\hat{r} \times \vec{dS}}{r^2}=0$$
Please give some hint to solve this problem.
 A: By Prove $ \oint_{\partial V} (\mathbf{\hat{n}} \times \mathbf{A}) \; \mathrm{d}S = \int_V (\nabla \times \mathbf{A}) \; \mathrm{d}V   $ (see also wiki-link)
$$\oint_S \dfrac{\hat{r} \times \vec{dS}}{r^2}=\oint_S \left(\dfrac{\hat{r}  }{r^2}\times \hat{n}\right) dS=-\int_V\left(\nabla \times\left(\dfrac{\hat{r}  }{r^2}\right)\right)dV=0$$
where at the last step we used the fact that
$$\dfrac{\hat{r}}{r^2}=\nabla\left(\frac{1}{r}\right)$$
and the curl of a gradient is identically zero, i.e. $\nabla \times \nabla (1/r)=0$.
A: Just "dot" the integral with constant vector $\vec{k}$ and apply divergence theorem.
Let $V$ be the volume bounded by closed surface $S$. 
Let us first consider the case $V$ doesn't contain origin. We have
$$\vec{k} \cdot \int_{S} \frac{\hat{r} \times \vec{dS}}{r^2} = \int_{S} \frac{\vec{k} \times \hat{r} }{r^2} \cdot \vec{dS} = \int_V \nabla \cdot \frac{\vec{k} \times \hat{r} }{r^2} dV = -\int_V \vec{k}\cdot \left(\nabla \times \frac{\hat{r}}{r^2}\right)dV\\
= \vec{k} \cdot \int_{V} \nabla \times \nabla \frac{1}{r} dV
= \vec{k} \cdot \int_{V} \vec{0} dV = \vec{k}\cdot \vec{0}$$
Since this is true for all $\vec{k}$, we get
$$\int_{S} \frac{\hat{r} \times \vec{dS}}{r^2} = \vec{0}$$
When $V$ contains origin, apply above result to volume $V \setminus B(0,\epsilon)$ 
where $B(0,\epsilon)$ is a small ball centered at origin with radius $\epsilon$ which
lies completely inside $V$. We obtain
$$\int_{\partial( V \setminus B(0,\epsilon))}\frac{\hat{r} \times \vec{dS}}{r^2} = \vec{0}
\implies
\int_{S}  \frac{\hat{r} \times \vec{dS}}{r^2}
= \int_{|r|=\epsilon} \frac{\hat{r} \times \vec{dS}}{r^2}
$$
On the sphere $|r| = \epsilon$, $\hat{r}$ and $\vec{dS}$ is pointing towards
same direction. This means 
$\displaystyle\;\frac{\hat{r} \times \vec{dS}}{r^2} = \vec{0}$ there and the integral on RHS vanishes.
