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I found this problem, I wanna do. Let $f$ be continuesly differentiable, let $r=\sqrt{x^2+y^2+z^2}$ Prove $\iint \limits _S \vec f (r) \cdot \vec n \ \Bbb dS = \iiint \limits _B \vec {f'(r)} \cdot \frac {\vec r} r \ \Bbb dV$ where left integral over sphere, right over the ball both of radius $R$.

I tried to apply divergence theorem using that $\vec n=(\frac x r,\frac y r, \frac z r)$. I could not get anything out of that.

I tried to then to work in case $f$ is a vector field, I got the right hand side: $\iiint \frac {f'(r)} {r} \cdot (-2x,-2y,-2z) \ \Bbb dV$.

Any help please. I am also interstate what case we can get if $f$ is scaler. I feel like there is a confusion with notation.

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  • $\begingroup$ if f is scalar you cannot use the divergence theorem. can you? $\endgroup$ – Sepideh Abadpour Aug 14 '15 at 19:16
  • $\begingroup$ Yes, I multiply f with constant vector $\endgroup$ – user65304 Aug 14 '15 at 19:19
  • $\begingroup$ I have modified part of your question in order for it to make sense mathematically. You put arrows above symbols by writing "\vec {my_symbol}". $\endgroup$ – Alex M. Aug 14 '15 at 19:19
  • $\begingroup$ @Alex is f scalar or vector field. if it's scalar, you cannot write $\vec{f'(r)}$ you should write $f'(\vec r)$ $\endgroup$ – Sepideh Abadpour Aug 14 '15 at 19:20
  • $\begingroup$ @sepideh: Arrow over the wrong symbol, thank you for noting. Now it's fine ($\vec f$ is a vector function with spherical symmetry, therefore depending only on $r$, not on $\vec r$). $\endgroup$ – Alex M. Aug 14 '15 at 19:26
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You solve this by applying the divergence theorem. Note that $\vec {f(r)}$ is a vector-valued function with spherical symmetry (that is, it does not depend on $\vec r = (x,y,z)$, but only on its norm $r$). By the chain rule, the divergence here will be

$$\operatorname{div} \vec f = \operatorname{div} {(f_x, f_y, f_z)} = f_x ' (r) \frac {\partial r} {\partial x} + f_y ' (r) \frac {\partial r} {\partial y} + f_z ' (r) \frac {\partial r} {\partial z} = (f_x ', f_y ', f_z ') \cdot (\frac x r, \frac y r, \frac z r) = \vec {f'(r)} \cdot \frac {\vec r} r$$

which is precisely the integrand in the right-hand side integral.

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  • $\begingroup$ Thanks Alex, your notations corrections makes the problem obvious. $\endgroup$ – user65304 Aug 14 '15 at 19:40
  • $\begingroup$ Would the same result hold if $f$ is scaler, and multiply by constant vector.? $\endgroup$ – user65304 Aug 14 '15 at 19:41
  • $\begingroup$ if f is scalar and n vector, you can not write $f.n$ in left-hand side? what's the notation of the problem exactly? can you post a photo of the problem. Alex M's answer is for the vector case if F is scalar you can not define divergence $\endgroup$ – Sepideh Abadpour Aug 14 '15 at 19:55
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    $\begingroup$ @sepideh: The OP wonders whether the dot $\cdot$ means inner product (as in my solution above) or the product of a scalar function and a vector. $\endgroup$ – Alex M. Aug 14 '15 at 19:57
  • $\begingroup$ Sepideh, I agree with what you said. What I am trying to say, if I use the divergence theorm on $F(x,y,z)=f(x)(c_1,c_2,c_3), \ \ where \ c_i \ are \ constants$ Now apply the divergence theorm $\iint \limits _S cf (r). \ \Bbb dS = \iiint \limits _B( c.\bigtriangledown f + f \ (\bigtriangledown c))\ \Bbb dV$ second term on right hand side will vanish. then I did not carry out the calculation, as I am not sure if it will yield anything similar. $\endgroup$ – user65304 Aug 14 '15 at 21:14

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