# Surface Integral of Vector Field

Given the scalar field $$\phi(\vec{r})=\frac{1}{|\vec{r}-\vec{a}|},$$ where $$\vec{a}=(-2,0,0)$$, and the corresponding vector field $$\vec{F}(\vec{r})=\operatorname{grad}\phi,$$ as well as the surface $$A$$ of the unit circle, $$A=\{(x,y,z)\in\mathbb{R}^3\, |\, x^2+y^2+z^2=1\},$$ how can you solve the surface integral $$\int_A \vec{F}(\vec{r})\cdot\mathrm{d}\vec{S},$$ where $$\mathrm{d}\vec{S}=\vec{\mathrm{e}}_\mathrm{r}\sin\vartheta\,\mathrm{d}\vartheta\,\mathrm{d}\varphi$$ is the usual surface element in spherical coordinates?

For $$\vec{a}=(0,0,0)$$, this would be pretty simple. Then, $$\vec{F}(\vec{r})=-r^{-2}\,\vec{\mathrm{e}}_\mathrm{r}$$ and the integral would be $$\int_A (-1)\,\vec{\mathrm{e}}_\mathrm{r}\cdot\vec{\mathrm{e}}_\mathrm{r}\,\sin\vartheta\,\mathrm{d}\vartheta\,\mathrm{d}\varphi=-4\pi$$. This would result in $$\Delta\phi=-4\pi\,\delta(\vec{r})=-4\pi\,\delta(x)\delta(y)\delta(z)$$ after applying Gauß and using the Dirac delta distribution $$\delta$$. The upper choice of $$\vec{a}$$ seems to make this more complicated, however.

• Is that surface integral supposed to be $\int_A \vec F \cdot \,d\vec S$? – Mark Viola Feb 17 at 21:41
• Yes @Mark Viola – st.math Feb 17 at 21:42
• Then, edit to make that clear. There is a difference between $\vec F\,d\vec S$, which is a tensor, $\vec F\times \,d\vec S$, which is a vector, and $\vec F\cdot \,d\vec S$, which is a scalar. – Mark Viola Feb 17 at 22:19

By applying the Gauss-Green theorem to your integral, you get $$\int_A \vec{F}(\vec{r})\cdot\mathrm{d}\vec{S}=\int_B \operatorname{div}\operatorname{grad} \phi(\vec{r})\,\mathrm{d}V_\vec{r}=\int_B \Delta \phi(\vec{r})\,\mathrm{d}V_\vec{r}$$ where $$B=\{\vec{r}=(x,y,z)\in\Bbb R^3\,|\,x^2+y^2+z^2\le 1\}$$ is the unit sphere centered in $$(x,y,z)=(0,0,0)$$. This last observation is crucial, since you can see by calculations that $$\Delta\phi(\vec{r})=\Delta\frac{1}{|\vec{r}-\vec{a}|}=0\quad\forall\vec{r}\neq \vec{a}$$ Then, since $$\vec{a}\notin B$$, the above integral is $$0$$, i.e. $$\int_A \vec{F}(\vec{r})\cdot\mathrm{d}\vec{S}=\int_B \Delta \phi(\vec{r})\,\mathrm{d}V_\vec{r}=0$$