1
$\begingroup$

I am trying to find the flux of a given field ($\vec{F} = (x+yz^3)\hat{\textbf{i}} + (xz^3)\hat{\textbf{j}} -z\hat{\textbf{k}}$ over a surface defined by $x^2 +y^2 + z^2 = 9$ between the planes $z=1$ and $z=2$.

My vector calculus is quite rusty. I think the optimal way to do this is use the divergence theorem over the solid region, and I want to check that I'm setting up the integral correctly and also see if there are other ways to go about this (to build more intuition for the geometry of these situations).

My work:

I think setting up the volume integral will be easiest in cylindrical coordinates. For a given $z$, $x^2+y^2+z^2 = 9 \Rightarrow ~r^2 = 9-z^2 \Rightarrow r = \sqrt{9-z^2}$.

So I should be able to set up my integral as:

$$\int _1 ^2 \int_0 ^{\sqrt{9-z^2}} \int_0^{2\pi} \nabla \cdot \vec{F}~ r d\theta dr dz$$

Then I can simply use the standard substitution $x = rcos \theta ~ y = rsin \theta, z = z$ on the vector field to get

$\vec{F} = (r~cos \theta+ r~sin \theta~ z^3) \hat{\textbf{i}} + r~cos \theta~z^3 \hat{\textbf{j}} - z~\hat{\textbf{k}}$.

a) Is this correct? b) Is this the optimal way to have done it (in terms of ease) c) Are there other set ups that would work without complicating things significantly?

  • 0
    Ah, right. I didn't even look at the field to be honest. Though, procedurally, this is a good point; it is much easier to find the divergence first in cartesian coordinates and then change to cylindrical (since taking derivatives in non rectangular coordinates is so messy). – 2017-01-06
  • 0
    You're going probably to have problems in cylindrical since the projections of the solid in the planes $\;z=1,\,z=2\;$ are different : in the first one you get the circle $\;x^2+y^2=8\;$ , whereas in the second one you get the circle $\;x^2+y^2=5\;$ . You may want to consider spherical coordinates... – 2017-01-06
  • 0
    @DonAntonio, how would you set up the limits of integration for such a solid in spherical coordinates? Obviously it can be done, but it would be a mess. And I'm not sure why cylindrical won't work; I've expressed the integral over $r$ in terms of z, so from my vantage point there shouldn't be an issue. – 2017-01-06
  • 0
    Your limits say: the vector $\;r\;$ runs from between $\;1\;$ and $\;2\;$ (why??), while the plane angle is from zero to $\;2\pi\;$ (this one I understand), and $\;z\;$ is between zero and the whole sphere $\;\sqrt{9-r^2}\;$ (again, why?). But your supposed to integrate over a sphere of radius $\;3\;$ *cut* by the plane $\;z=1\;$ and then by the plane $\;z=2\;$ you're not even close to the plane $\;z=0\;$ , for example! – 2017-01-06
  • 0
    @DonAntonio no... that isn't what it says at all. R (which is a spherical coordinate and *not* a vector) goes from 0 to $\sqrt{9-z^2}$, and z runs from 1 to 2. You aren't actually reading what my post says. – 2017-01-07
  • 0
    That doesn't make sense at all: if you meant to use spherical coordinates you, first of all should explicitly say so (and **you say** you're using *cylindrical coordinates*!), but then: where's the azimut angle $\;\theta\;$ in that integral of yours ?? You *also* use $\;z\;$ as upper limit in one of those integrals... All this makes it clear, imo, that something very wrong is in that integral and that $\;r\;$ in that integral is the length of vector's coordinate. – 2017-01-07
  • 0
    Both saying "spherical" and the use of $/phi$ were typos. $\theta$ is the azimuthal angle throughout – 2017-01-07
  • 0
    Even then, what about that $\;z\;$ in the middle integral? And how come your azimutal angle goes all the way until $\;2\pi\;$, when we all know it is **always** between zero and $\;\pi\;$ ? Anyway, if you want to use spherical, make clear which one is the azimutal angle and which one the plane one. – 2017-01-07
  • 0
    There is a z in the middle integral because r has an implicit z dependence. Also, the azimuthal angle always runs 0 to $2 \pi$. The polar angle drops down from the positive z axis and runs 0 to $\pi$. – 2017-01-07
  • 0
    @Be Do ass you wish: if you really believe you can have $\;z\;$ *after* you passed to spherical coordinates then there is no point to continue, no matter how you call the angles. Good luck to you – 2017-01-07

2 Answers 2

2

Divergence theorem can not be used right away as the surface is not closed.

Using the divergence theorem,

Close the surface by including the $z=1$ and $z=2$ that would make top and bottom parts as parts of the surface. So then the theorem can be applied.

$$\iint_{S} \vec F \cdot d \vec S=\iint \text{div} \vec F dV$$

Clearly $\text{div} F=0$ in this example, so the surface integral is $0$.

But $S=S_{\text{top}} \bigcup S_{\text{interest}} \bigcup S_{\text{bottom}}$.

So,

$$\iint_{S_{\text{top}}}+\iint_{S_{\text{interest}}}+\iint_{S_{\text{bottom}}}=0$$

And hence the surface integral of interest is equal to,

$$\iint_{S_{\text{interest}}}=-\iint_{S_{\text{bottom}}}-\iint_{S_{\text{top}}}$$

Note they all must be computed using an outward orientation, outward relative to the sphere.

For $z=1$ we have a downward outward orientation, so use $d \vec S=\langle 0,0,-1 \rangle dA$. For $z=2$ we have an upward outward orientation, so use $d \vec S= \langle 0,0,1 \rangle dA$. Note for $z=1$ we have $\vec F=\langle ...,...,-1 \rangle$. And for $z=2$ we have $\vec F= \langle ...,....,-2 \rangle$.

So the integral we need to evaluate is

$$-\iint_{D_1} \langle ...,...,-1 \rangle \cdot \langle 0,0,-1 \rangle dA -\iint_{D_2} \langle ...,....,-2 \rangle \cdot \langle 0,0,1 \rangle dA$$

Or,

$$-A(D_1)+2A(D_2)$$

$$=-\pi(\sqrt{8})^2+2\pi(\sqrt{5})^2$$

$$=10\pi-8\pi=2\pi$$

  • 0
    I think I expressed myself poorly in the initial question. The request flux was over the surface defined including the planes at $z=1,~z=2$, which is, in fact, zero. However, this was a really helpful walk. I still would appreciate feedback on whether my parameterization is correct given that. – 2017-01-07
  • 0
    You didn't take the divergence then substitute $x=r \cos (\theta)$ and $y=r \sin (\theta)$, you must do that. Otherwise, Your set up is correct. You parametrization for $F$ is correct but what we are interested in is the parametrization for $div F$. – 2017-01-07
  • 0
    Well I'm only interested in the set up for the integrals, really. Applying a differential operator to a vector field isn't so bad/ – 2017-01-07
  • 1
    The bounds for the integral is correct @BenL – 2017-01-07
0

On a sphere of radius $a$,

$$\vec n=\frac{1}{a}$$

Then,

$$\iint_{S} \vec F \cdot d \vec S=\iint_{S} \frac{1}{a} \vec F \cdot dS$$

In your case we have,

$$\iint_{S} \vec F \cdot d \vec S=\frac{1}{a}\iint_{S} (x^2+2xyz^3-z^2) dS$$

Note $ds=a^2 \sin (\phi) dA$. This can be directly calculated by parametrizing the sphere by $r(\theta, \phi)$ and calculating $|r_\theta \times r_\phi|$. But can also be easily seen by the fact that we have to factor in a Jacobian $\rho^2 \sin (\phi)$ whenever switching to spherical coordinates.

What are the bounds for $\theta$ and $\phi$?

It's pretty clear $\theta \in [0,2\pi]$

The bounds for $\phi$ however take some work.

Note we have $r^2+z^2=9$ so when $z=1$ we have $r^2=8$ and hence we have $r=\sqrt{8}$.

Note that have $r^2+2^2=9$ when $z=2$. So $r=\sqrt{5}$ there.

$\frac{r}{z}=\tan (\phi)$

So it follows,

$$\phi \in [\arctan (\frac {\sqrt{5}}{2}), \arctan (\sqrt{8})]$$

Messy but this is how you would set it up using spherical coordinates.