Curve integral, singularity, and parametrization

[S. Moger]

Well, it's physics friday! (carpe diem etc, what else) :)

1. Homework Statement

I present to you this (not so) pleasant expression that seemingly appeared on a page out of nowhere.

$\vec{F}(r, \theta, \varphi) = \frac{F_0}{ar \sin\theta}[(a^2 + ar \sin\theta \cos\varphi)(\sin\theta \hat{r} + \cos\theta \hat{\theta}) - (a^2 + ar \sin\theta \sin\varphi - r^2 \sin^2 \theta)\hat{\varphi}]$

Getting it right on paper, on one line, is in and of itself a super great challenge.

What to do with it? This

$\int_C \vec{F} d\vec{r}=?$

Where C is given by the intersection between
$S_1: x^2+4y^2 = 12a^2 + 8ay$ and
$S_2: x^2+y^2=4az-2ay-a^2$.

The Attempt at a Solution

By doing a transformation into cylindrical base vectors, I get an expression that's less detrimental to people's health:

$\vec{F}(\rho, \varphi, z)=F_0 \frac{a}{\rho} (\hat{\rho} - \hat{\varphi}) + F_0 (\cos\varphi \hat{\rho} - (\sin\varphi + \frac{\rho}{a})\hat{\varphi})$

Where I split the z-axis singularity from the rest.

$S_1$ is an elliptic cylinder if I'm not mistaken: $\sqrt{x^2 +4(y-a)^2}=4a$, with the semimajor axis being 4a and the semiminor 2a. Origin at x=0, y=a.

$S_2$ is elliptical cone-like but rounded by the square root : $4az=x^2 + (y+a)^2$. Origin at x=0, y=-a.

The intersection is a curve whose z-value depends on y like so:
$z=\frac{13a}{4}-\frac{3y^2}{4a}+\frac{5y}{2}$ (If I got it right).

The z-axis is contained within, however what worries me is the parametrization of the curve so that I can integrate over it. Stokes would be nice, but the geometry isn't that easy (it seems). Or am I wrong? What would be the best approach here?

 

[andrewkirk]

I think $S_2$ is a paraboloid of revolution, ie a parabola in the y-z plane rotated around the line (x=0, y=-a) which passes through the parabola's minimum and focus.

The curve C is symmetric in x and has its lowest point at y=-a and its highest at y=3a. It will be vaguely elliptic-ish.

One approach would be to split the integral into the sum of two parts, each parametrised by y - the first from y=-a to 3a with $x\geq 0$ and the other from y = 3a to -a with $x\leq 0$.

An alternative would be to parameterise over the angle the perpendicular from the point on C to the line (x=0, y=a) makes with the y-z plane, from 0 to 2 pi. But my hunch is that that might be messier because it's using a different centre for angles to what is used for the spherical coordinates in the function $\vec{F}$.

 

[S. Moger]

Yes ok, a paraboloid of revolution.

I've tried a couple of approaches, but they all seem to result in a mess.

I've transformed the expression of the field to cartesian coordinates to perform the dot product with dr in the same system (or is this unnecessarily cumbersome?)

$\vec{F}(x,y,z) = a F_0 ( \frac{x+y}{x^2+y^2} \hat{x} + \frac{y-x}{x^2+y^2} \hat{y}) + F_0 ( (1 - \frac{y}{a} ) \hat{x} + \frac{x}{a} \hat{y})$

There's a hint that says that transformation to cylindrical coordinates should be done (maybe there's no need for further conversions after all?).

$d\vec{r}=dx \hat{x}+ dy \hat{y} + dz \hat{z}$

And just to check that I'm doing this right. I compute the dx, dy, dz in terms of dy to perform the dot product of F and dr. After having replaced all x's with the expression for x(y) given by the intersection of S1 and S2, I then integrate over y (and the suggested limits). The contributions that arise from the non-singular part seem to resolve to zero. I think the singular part can be evaluated without the cartesian conversion by checking if it's inside C, but I'm not entirely sure about how to treat both components in that case. At least I can't see how one arrives at $\pm 14 \pi a F_0$.

 

[S. Moger]

I'm trying a different approach now,

The field
$\vec{F}(\rho, \varphi, z)=F_0 \frac{a}{\rho} (\hat{\rho} - \hat{\varphi}) + F_0 (\cos\varphi \hat{\rho} - (\sin\varphi + \frac{\rho}{a})\hat{\varphi})$

Integral of the singular part:
$a F_0 \oint_C \frac{1}{\rho}(\hat{\rho}-\hat{\varphi}) d\vec{r}$

with
$d\vec{r} = d\rho \hat{\rho} + \rho d\phi \hat{\phi} + dz\hat{z}$

I think the intersection can be parametrized like this:

$\begin{cases} x = 4a \cos \varphi \\ y = 2a \sin \varphi + a \\ z = \frac{13 a^2 +10 ay - 3 y^2}{4a} \end{cases}$

However, z doesn't seem to be very interesting judging from the field expression.

To get rho in phi I do this $\rho^2 = x^2 +y^2$.

Anyway, after computing the above I get $-2 \pi a F_0$ from the singular part. And $-22 a \pi F_0$ from the other, while integrating from 0 to 2 pi.

In total that would give me $-24 \pi a F_0$. Which is ten respectively thirty less than the result I should get.

Something is obviously wrong, but what? And why should I get two answers out of this integral?