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In $\mathbb{R}^3 \setminus \{0\}$, it's given the vector field $$\vec{E}(\vec{r}) = g(r) \vec{r}$$ where $g$ is some function of class $\mathcal{C}^\infty$ defined in $[0, + \infty[$, $\vec{r} = (x,y,z)$ and $r = \sqrt{x^2 + y^2 + z^2}$. Also, $\mathrm{div} \hspace{.5pt}\vec{E}(\vec{r}) = r g'(r) + 3g(r)$. I have to compute the flux of $\vec{E}$ through the sphere $S_a^2$.

My 1st attempt: Denoting by $B_a$ the ball centered in the origin with radius $a$, and using the divergence theorem, I have: $$\int_{S_a^2} \vec{E} \cdot \vec{N} \hspace{.5pt}\mathrm{d}S = \int_{B_a} r g'(r) + 3g(r) \hspace{.5pt} \mathrm{d}V$$ Using spherical coordinates, I would calculate: $$\int_{0}^{2\pi} \int_{0}^{\pi} \int_{0}^{a} \left(r g'(r) + 3g(r) \hspace{.5pt}\right)r^2 \sin{\varphi} \mathrm{d}r \mathrm{d}\varphi \mathrm{d}\theta = 4\pi \int_{0}^{a} \left(r g'(r) + 3g(r) \hspace{.5pt}\right)r^2 \mathrm{d}r$$

My 2nd attempt: By definition. $$\int_{S_a^2} \vec{E} \cdot \vec{N} \hspace{.5pt}\mathrm{d}S = \int_{[0,2\pi] \times [0, \pi]} \vec{E}(X(\theta, \varphi))\cdot (-a^2 \sin \varphi \cos \varphi, -a^2 \sin^2 \varphi \sin \theta, -a^2 \sin^2 \varphi \cos \theta) \mathrm{d}\theta \mathrm{d}\varphi$$

where $X$ is a parametrization of the sphere using spherical coordinates. However, I'm having trouble computing $\vec{E}(X(\theta,\varphi))$. Does not make any sense to me. I would make $(x,y,z) = (r \sin \varphi \cos \theta, \cdots )$ or $(x,y,z) = (a \sin \varphi \cos \theta, \cdots )$? In the first case, $r$ is a variable but the integral is only on $\theta$ and $\varphi$.

There is no way to give an answer without it being in terms of integrals of $g$, is it? If someone know an easy way to solve this, it is also welcome. Thank you.

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It would be the latter $(x,y,z)=(a\sin\phi\cos\theta,...)$, here $r=a$ is fixed, because this is a surface integral $(dS)$, rather than a volume integral $(dV)$.

Now with regards to the first integral, I think were in luck:

$4\pi\int_0^a(rg'(r)+3g(r))r^2dr=4\pi\int_0^a r^3g'(r)+3r^2g(r)dr=4\pi\int_0^a \frac{d}{dr}(r^3g(r))dr=4\pi a^3g(a)$

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  • $\begingroup$ Ok. The same way, I would make $g(a)$ instead of $g(r)$? (after passing "$g(r)$" to inside the vector) $\endgroup$
    – Ivo Terek
    Jun 8, 2014 at 21:18
  • $\begingroup$ You would, but see my edit, I believe I have solved the first integral. $\endgroup$
    – Ellya
    Jun 8, 2014 at 21:23
  • $\begingroup$ Damn. I should have seen that. Thank you very much, you nailed it! (; $\endgroup$
    – Ivo Terek
    Jun 8, 2014 at 21:24
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    $\begingroup$ No worries, happy to help :) $\endgroup$
    – Ellya
    Jun 8, 2014 at 21:25

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