Go with the Flux!

Calculus Level 5

Find the absolute value $\Phi$ of the flux of the field $\vec{F}(\vec{r})=\frac{\vec{r}}{||\vec{r}||^3}$ through the surface $S$ given by $\frac{x^2}{25}+y^2+z^2=1$ , $3\leq x \leq 4$ .

As your answer, return the integer nearest to $1000\Phi$ .

The answer is 143.

2 solutions

Steven Chase
Dec 11, 2018

The vector field projects out radially from the origin, as though there were a point-charge there. It would be rather difficult to compute the flux through the given surface directly, so luckily there is a shortcut. Make a closed surface out of the defined portion of the ellipsoid surface, plus two circular disks parallel to the $yz$ plane, located at $(x = 3)$ and $(x=4)$ .

By Gauss's law, since there is no enclosed "charge" within the surface, the total flux through the closed surface must be zero. This means that the sum of the fluxes through the two disks must be equal and opposite to the flux through the specified portion of the ellipsoid surface.

A few logistical notes. The radii of the two disks are the following:

$R_1^2 = 1 - \frac{9}{25} \\ R_2^2 = 1 - \frac{16}{25}$

It is also important to use the same surface normal convention for both disks (either outward or inward). Using an outward convention gives $(n_{1x},n_{1y},n_{1z}) = (-1,0,0)$ for disk 1 and $(n_{2x},n_{2y},n_{2z}) = (1,0,0)$ for disk 2. Otherwise, the flux calculations are straightforward but mechanically demanding, so I used a computer for those.

The magnitude of the total flux through the disks (and therefore through the ellipsoid portion) is $\approx -0.1427$ . Taking the absolute value, multiplying by $1000$ , and rounding to the nearest integer gives $\boxed{143}$ .

Very nicely (and quickly) done! Thank you! Now, as a mathematician (and a bit of a Luddite), I must ask: Can we do it without resorting to a computer? Can we write the answer in closed form? (I'm still using my Swiss logarithm table sometimes and the Swiss slide rule.)

Otto Bretscher - 2 years, 6 months ago

Thanks. On second thought, I suppose we could say that the flux through the disk is the same as the flux through a spherical cap. Then we could calculate the total flux through the entire sphere using Gauss's law, and find the fraction going through the cap based on the surface area of the cap relative to the surface area of the total sphere. This would only depend on the "phi" range, since "theta" would still go from zero to 2 pi. So yes, I think it could be done. And then you could use a tabulated sine / cosine lookup table. I wrote some pretty low-level code in Python which divides each disk into an $r, \theta$ polar coordinate space in the $y z$ plane, and then iterates through nested $r$ and $\theta$ loops to calculate the surface integrals. It's a matter of taste, I suppose.

Steven Chase - 2 years, 6 months ago

Yes, exactly! That's how I did it; I will post my solution when I find the time.

Otto Bretscher - 2 years, 6 months ago

Otto Bretscher
Dec 11, 2018

Let me suggest an alternative to Steven's fine solution.

Considering the points $(3,0.8,0)$ and $(4,0.6,0)$ on the boundary of $S$ , we see that the azimuthal angle (relative to the positive $x$ -axis) on $S$ is between $\phi_1=\arctan(3/20)$ and $\phi_2=\arctan(4/15)$ . Projecting $S$ radially down onto the unit sphere, we consider the portion $S'$ of the unit sphere with $\phi_1\leq \phi \leq \phi_2$ . The outward flux through $S$ is the same as the outward flux through $S'$ , which in turn is the same as the area of $S'$ , namely, $2\pi\left(\cos(\phi_1)-\cos(\phi_2)\right)=2\pi\left(\frac{20}{\sqrt{20^2+3^3}}-\frac{15}{\sqrt{15^2+4^2}}\right)\approx 0.1426$ .

The required answer is $\boxed{143}$ .

Go with the Flux!

The answer is 143.

2 solutions

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