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This question came in my physics test: What is the value of the surface integral $\oint_S\frac{\overrightarrow{r}}{r^3} \,\cdot\mathrm{d}\overrightarrow{A}$ for r>0? The professor says that the answer is $4\pi$, but I think that the answer will be either 0 or 4$\pi$ depending on if the origin is enclosed by the surface S. After all physically, the integrand is just the field of a point charge of magnitude $4\pi\epsilon_0$ placed at the origin, and therefore the integral its flux through S. So depending on if the surface encloses the point charge or not, the flux will be either 0 or $4\pi$ by Gauss's law. Is this reasoning correct?

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  • $\begingroup$ I think it's assumed the surface includes the origin, you should just ask the professor to clarify $\endgroup$
    – Triatticus
    Commented Jan 30, 2021 at 18:18
  • $\begingroup$ You are correct, but weren't really there other specifications in the question? $\endgroup$
    – Massimo Ortolano
    Commented Jan 30, 2021 at 18:19
  • $\begingroup$ @MassimoOrtolano No, I have written the exact statement of the question. $\endgroup$
    – user208090
    Commented Jan 30, 2021 at 18:39

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With the given vector field you are integrating over, it means the origin is enclosed. Otherwise the expression would not be defined for all r>0.

However, a single integral indicates a contour integral i.e. over a closed curve, not a closed surface. For the latter you should use ∯S

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