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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.

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  • $\begingroup$ Is that surface integral supposed to be $\int_A \vec F \cdot \,d\vec S$? $\endgroup$ – Mark Viola Feb 17 at 21:41
  • $\begingroup$ Yes @Mark Viola $\endgroup$ – st.math Feb 17 at 21:42
  • $\begingroup$ 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. $\endgroup$ – Mark Viola Feb 17 at 22:19
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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. $$\int_A \vec{F}(\vec{r})\cdot\mathrm{d}\vec{S}=\int_B \Delta \phi(\vec{r})\,\mathrm{d}V_\vec{r}=0$$

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  • $\begingroup$ Thanks! :) I tried using Gauß before, but it always resulted in extremely long calculations. After trying it again, it indeed results in the value 0 of the volume integral. $\endgroup$ – st.math Feb 17 at 22:00
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    $\begingroup$ @st.math. You're welcome: I am happy to have been of some help. $\endgroup$ – Daniele Tampieri Feb 17 at 22:02

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