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**physics friday**! (carpe diem etc, what else) :)

1. Homework Statement

1. Homework Statement

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

[itex]\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}][/itex]

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

What to do with it? This

[itex]\int_C \vec{F} d\vec{r}=?[/itex]

Where C is given by the intersection between

[itex]S_1: x^2+4y^2 = 12a^2 + 8ay[/itex] and

[itex]S_2: x^2+y^2=4az-2ay-a^2[/itex].

## 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:

[itex]\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})[/itex]

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

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

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

The intersection is a curve whose z-value depends on y like so:

[itex]z=\frac{13a}{4}-\frac{3y^2}{4a}+\frac{5y}{2}[/itex] (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?