Circulation of a vector field through a surface.

Question

I have the field $F=\biggr(\frac{x}{(x^x+y^2+z^2)^2},\frac{y}{(x^x+y^2+z^2)^2},\frac{z}{(x^x+y^2+z^2)^2}\biggr)$ (with $\operatorname{div} F= -\frac{1}{(x^2+y^2+z^2)^2}$) and the region $E_t= \{(x,y,z) \in R^3\ : \ \varepsilon \le x \le 1,\ \varepsilon \le y \le 1, \ \epsilon \le z \le 1\}$ where $0 < \varepsilon \lt 1$ i was asked to find $I_\varepsilon = \iiint_{E_t} \operatorname{div} F \ dV$ and $\lim_{\varepsilon\to 0} I_\varepsilon$ to latter "interpret the results".

My main doubt is about the region $E_t$ since i think the best way to solve this is with the Gauss theorem but i dont know how to evaluate it.

In any case, a hint to solve this will be appreciated.

Answer 1

No need for the Gauss theorem to get the integral since you already know the divergence to be $\frac{-1}{r^4}$.

Note that you want to evaluate this on a spherical shell intersected with an octant. So, by spherical symmetry of the integrand, $8I_\epsilon=V$ where V is the integral of the divergence over a spherical shell of inner radius $\epsilon$ and outer radius 1.

$$V=\int_{\epsilon}^1 4\pi r^2 \frac{-1}{r^4} dr = 4 \pi (\frac{1}{\epsilon}-1) $$ This gives

$$I_{\epsilon}= \frac{\pi}{2}(\frac{1}{\epsilon}-1)$$

In the limit, this integral blows up.

Answer 2

$$I_{\epsilon}= \iiint_{E_t} \operatorname{div} F \ dV = \int_0^{2\pi} \int_0^{\pi} \int_\varepsilon^1 \frac 1 {\rho^4}\cdot\rho^2\sin\theta\,d\rho \, d\theta \,d\phi$$

$$I_\varepsilon = 4\pi\left(\frac{1}{\epsilon}-1)\right)$$

$$\lim_{\varepsilon\to 0} I_\varepsilon \text{goes to infinity}$$