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Let $R$ be the rectangular box consisting of all points $(x,y,z)$ with $-1\leq x\leq 1$, $-1\leq y\leq 1$, and $-1\leq z\leq 1$. Define the vector field $$V= x(x^2+y^2+z^2)^{-3/2}\mathbf{i}+y(x^2+y^2+z^2)^{-3/2}\mathbf{j}+z(x^2+y^2+z^2)^{-3/2}\mathbf{k}.$$ Find the outward flux of the vector field $V$ through the boundary of the box $R$.

I tried using divergence theorem by converting to spherical coordinates but I'm not sure if I did it correctly, I got an answer of $2 \pi$. This hint was also included in the problem:

Keep in mind that the vector field V in this problem is not defined at the origin, so that you can not apply the Divergence Theorem to the rectangular box and its interior.

Which makes me assume that my method is possibly incorrect, however I'm not sure if converting to spherical coordinates makes the use of the Divergence Theorem valid.

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  • $\begingroup$ Both field and box symmetric with respect to each axis. $\endgroup$ – mathreadler Apr 20 '16 at 13:19
  • $\begingroup$ @b00nheT Glancing at the form of the vector field suggests it's not pleasant to integrate (though we could get away with using symmetry to reduce the number of integrals we'd actually have to compute). $\endgroup$ – Travis Willse Apr 20 '16 at 13:22
  • $\begingroup$ You actually end up with just one integral to compute, but I must admit that it is not the most elegant solution $\endgroup$ – b00n heT Apr 20 '16 at 13:25
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You're right: Since $\bf V$ is not defined at $0$ but $0 \in R$, the hypotheses of the Divergence Theorem do not hold for the desired flux integral, $$\iint_{\partial R} {\bf V} \cdot d{\bf A} .$$

Hint There are two key observations here: The first is that $\bf V$ is radially symmetric (that is, we can write it as $f(r) {\bf r}$ for some function $f$ (here, ${\bf r}$ is just the unit radial vector field). So, while it is difficult to evaluate the integral over $R$ directly (because the expression for the components of $\bf V$ are complicated), it would be much easier to find the flux of $\bf V$ across a sphere $S$ of radius $R$ centered at the origin, as we could write $$\iint_S {\bf V} \cdot d{\bf A} = \iint_S f(R) {\bf r} \cdot {\bf r} \,dA = f(R) \iint_S dA = 4 \pi f(R).$$

Since we suppose we might want to use the Divergence Theorem, we may as well compute $\operatorname{div} {\bf V}$, just to see what kind of function you'd end up integrating, and this yields our second key observation, namely that $$\operatorname{div} {\bf V} = 0 .$$ (To carry out this computation, it's best to use the spherical coordinate expression for $\bf V$ and then use the corresponding divergence formula.)

Remark In fact, we can use the above ingredients to see that $\bf V$ is actually quite a special vector field. We can ask which radial vector fields $\bf X$ (say, defined on $\Bbb R^3 - \{ 0 \}$) have zero divergence. Substituting ${\bf X} = g(r) {\bf r}$ into the divergence formula for the sphere gives that our condition is $$0 = \operatorname{div} {\bf X} = r^{-2} \frac{d}{dr} [r^2 g(r)] .$$ Multiplying both sides by $r^2$ and integrating gives $r^2 g(r) = C$ for some constant, and hence $g(r) = \frac{C}{r^2}$, but this is exactly the form of $\bf V$ for some $C$, so the only radial vector fields with zero divergence are the given vector field $\bf V$ and its constant multiples.

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  • $\begingroup$ Ok thank you, yeah I was somewhat confused when I first attempted the problem; after finding the divergence of V it made me assume the answer was 0, but I wasn't sure if that was too simplistic. $\endgroup$ – George Bencosme Apr 20 '16 at 13:33
  • $\begingroup$ To be clear, $\operatorname{div} {\bf V} = 0$, but the value of the integral in the question is not zero. $\endgroup$ – Travis Willse Apr 20 '16 at 13:48
  • $\begingroup$ Ok I think I understand. Just as clarification, for f(R) is there a way that V is used to find this function or am I approaching it the wrong way? $\endgroup$ – George Bencosme Apr 20 '16 at 14:03
  • $\begingroup$ Yes, and there's an easy way in fact: Since ${\bf V} = f(|{\bf r}|) {\bf r}$, taking the dot product with $\bf r$ gives $f(|{\bf r}|) = {\bf V} \cdot {\bf r} = f(|{\bf r}|) {\bf r} \cdot {\bf r} = f(|{\bf r}|)$. (Note that I've rewritten $r$ as $|{\bf r}|$ for emphasis.) $\endgroup$ – Travis Willse Apr 20 '16 at 14:39
  • $\begingroup$ Oh I see! So taking the dot product of $V$ and $r$ will give me the function to multiply by $4π$? $\endgroup$ – George Bencosme Apr 20 '16 at 14:45

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