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As defined in spherical coordinates, $p =$ the distance from the origin to a point P in space.

In Stewart P1092 16.7.47 (below), $p$ is fixed by the solution to be $a$.
Yet in Stewart P1103 16.9.13, $p$ is a variable. How does one determine whether $p$ varies or is fixed?

I realise that in 13, the Divergence Theorem effects a triple integral so 3 variables must be integrated. In the interest of spherical coordinates, $\phi \, \& \, \theta$ are used, so $p$ may fit. This is desultory guesswork, so I'd like to learn a more definitive explanation.

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  • In example 47, $p$ is fixed because you're finding the flux across a sphere. A sphere is defined as the set of all points that are a fixed distance (in this case $a$) from the origin. – Eric Apr 03 '14 at 15:23
  • @Eric: Thanks, but question $13$ also involves a sphere? Yet $p$ varies therein? –  Apr 05 '14 at 16:35
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    Ahh, you're right - my explanation was inadequate. Here's the way I might think about it: In problem 47, you're evaluating the flux as an integral over the surface of the sphere. Think about traversing the surface of the sphere - the distance from the origin doesn't change. On the other hand, in problem 13, you're integrating the divergence throughout the volume enclosed by the sphere. Different point in the ball enclosed by the sphere will be different distances from the origin. Does this make any sense? – Eric Apr 09 '14 at 15:40
  • @Eric: Thank you. Yes, it does! Still, how would you determine whether a question needs an integral over the surface or throughout the volume? –  Apr 11 '14 at 08:38
  • If you're using the Divergence Theorem, integrate $div F$ throughout the volume. Otherwise, integrate $F\bullet dr$ over the surface. – Eric Apr 14 '14 at 15:08

2 Answers2

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Both problems, 13 and 47, are about the divergence theorem. This theorem reads as follows: Given a flow field ${\bf v}$ in some domain $\Omega\subset{\mathbb R}^3$, and in addition a "body" $B\subset\Omega$ with surface $\partial B$ (oriented outwards) the "integral formula" $$\int_{\partial B}{\bf v}\cdot d\vec\omega=\int_B{\rm div}({\bf v}){\rm d}({\bf x})\tag{1}$$ is valid. Here the left hand side is a surface integral (involving two variables $u$, $v$, or, e.g., $\phi$, $\theta$, when it has to be computed "the hard way"), and the right hand side is a volume integral (involving three variables $x$, $y$ $z$, or, e.g., $r$, $\phi$, $\theta$, when it has to be computed "the hard way").

In problem 47 one has $\Omega={\mathbb R}^3\setminus{\bf 0}$. Therefore we cannot apply the divergence theorem to balls centered at ${\bf 0}$. But we can apply it to spherical shells $B:=\{{\bf x}\>|\>a\leq|{\bf x}|\leq b\}$ with boundary $\partial B=\partial S_b-\partial S_a$, where the minus sign takes care of the fact that the standard orientation of the inner sphere $S_a$ has to inverted when $S_a$ is considered as part of $\partial B$.

Since it is easily verified that ${\rm div}({\bf v})\equiv0$ in $\Omega$ we obtain $$\int_{S_b}{\bf v}\cdot d\vec\omega -\int_{S_a}{\bf v}\cdot d\vec\omega=\int_{\partial B}{\bf v}\cdot d\vec\omega=0\ ,$$ which is saying that the integrals $\int_{S_b}{\bf v}\cdot d\vec\omega$ and $\int_{S_a}{\bf v}\cdot d\vec\omega$ over the outward oriented spheres have the same value.

In problem 13 the field ${\bf v}$ is defined on all of ${\mathbb R}^3$, so that $(1)$ may be applied to the ball $B:=\{{\bf x}\>|\>|{\bf x}|\leq R\}$. You have computed ${\rm div}({\bf v})=12(x^2z+y^2z+z^3)$, whence $(1)$ gives $$\int_{S_R}{\bf v}\cdot d\vec\omega =12 \int_B (x^2z+y^2z+z^3)\ {\rm d}(x,y,z)\ .$$ Now $B$ is symmetric with respect to ${\bf x}\mapsto -{\bf x}$, and each term in the last integral is odd. It follows that the integral is $0$, by symmetry.

Christian Blatter
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It is essentially a matter of convenience: sometimes it looks easier to do the integration over the surface, especially when the field is proportional to $\overrightarrow r$.

Other times it is easier to use Gauss' theorem and switch to $div$ and volume integral. And every time you will have to do some context-related guessing what the author intends you to do, obviously :-).

As you can easily see, $13$ is designed for applying $div$, as by magic a term $2z\rho^2$ appears which makes integrating over a volume easier. Trying to do the sphere surface integral is much more difficult, because one has to calculate the surface integral of $z(4x^4+4y^4+3z^4)$ and substitute $x=Rcos\theta sin\phi$ etcetera, as above.

$47$ is clearly fit for just doing the surface integral, but there's really no problem going the other route here and apply $div$: $$F_x = \frac{cx}{r^3} = \frac{cx}{(x^2+y^2+z^2)^{3/2}}$$ etcetera, so \begin{equation} \begin{split} div F &= \frac{\partial F_x}{\partial x} + \frac{\partial F_y}{\partial y} +\frac{\partial F_z}{\partial z} \\ &= \frac{c}{r^3} - \frac{3c}{2}\frac{x}{r^5}\cdot 2x + \frac{c}{r^3} - \frac{3c}{2}\frac{y}{r^5}\cdot 2y + \frac{c}{r^3} - \frac{3c}{2}\frac{z}{r^5}\cdot 2z \\ &= \frac{3c}{r^3} - \frac{3c(x^2+y^2+z^2)}{r^5} \\ &= \frac{3c}{r^3} - \frac{3c}{r^3} \\ &= 0 \end{split} \end{equation} so a volume integral of $div F$ seems to always equal zero - but we have to take into account the value of $div F$ at the origin. As you may know, one of Maxwell's equations states $$div \overrightarrow E = \frac{\rho}{\epsilon _0}$$ where $\rho$ is the electrical charge density; essentially it says that electrical field lines originate from charges. Here this means that we have to know whether $div F$ has a value in $O$. Obviously it has to: otherwise where are all these field lines coming from? Assuming the $\overrightarrow F$ mentioned is actually just the electrical field, we see that at the origin a (positive) point charge $4\pi \epsilon_0 c$ has to be present (with infinite charge density but obviously related to the point-likeness in such a way as to exactly equal the correct charge), and thus the surface integral of $\overrightarrow F$ over $any$ surface enclosing the origin (not just spheres) equals $4\pi \epsilon_0 c$.

Maestro13
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