Line integral & logarithmic spiral

[irycio]

Homework Statement

Calcualte the value of \int\limits_L \sqrt{x^2+y^2}dl, where L is an arc of a logarithmic spiral r=ae^{m\phi} between points A(0,a) and B(-\infty,0).

Problem: I can't find a value of \phi where x=-\infty or y=a.

Homework Equations

We parametrise and get:
x=ae^{m\phi}\cos(\phi)
y=ae^{m\phi}\sin(\phi)

The Attempt at a Solution

Well, I guess I can't do much without the boundaries. I typed the equation for y=a into mathematica and got error messages, more or less the same for function Solve and Reduce (that the equation can not be solved using algebraic methods).
Obviously, the equation x=-\infty doesn't give any result either. Help, please!

 

[Mark44]

Homework Statement

Calcualte the value of \int\limits_L \sqrt{x^2+y^2}dl, where L is an arc of a logarithmic spiral r=ae^{m\phi} between points A(0,a) and B(-\infty,0).

Problem: I can't find a value of \phi where x=-\infty or y=a.

They don't make sense to me, either. If phi = 0, r = a, so in Cartesian coordinates, this is (a, 0), and it's the same in polar coordinates. As phi approaches -infinity, r approaches 0 (assuming m is a positive number). In polar coordinates, this is (0, -infinity).

Homework Equations

We parametrise and get:
x=ae^{m\phi}\cos(\phi)
y=ae^{m\phi}\sin(\phi)

The Attempt at a Solution

Well, I guess I can't do much without the boundaries. I typed the equation for y=a into mathematica and got error messages, more or less the same for function Solve and Reduce (that the equation can not be solved using algebraic methods).
Obviously, the equation x=-\infty doesn't give any result either. Help, please!

This problem seems well-suited for converting to polar coordinates. Arc length in polar form is
dl = \sqrt{r^2 + (\frac{dr}{d\theta})^2} d\theta

The resulting integral is straightforward but improper, so you'll need to take the limit as phi approaches -infinity.

 

[irycio]

Well, polar coordinates are pretty obvious, but I still don't like the boundaries. Not only do they not satisfy the conditions given, but I also don't catch how a phi, which is an angle in spherical coordinates, can approach -infinity. Although i believe, that there is a limit of functions like e^(phi)*sin(phi) and it's equal 0, regardless the periodic function that doesn't have a boundary.

 

[Mark44]

Why do you think this has anything to do with spherical coordinates? The points are given as ordered pairs, not ordered triples, so it seems to me that you are working in the plane, not in three dimensions.

It is true that phi is used in spherical coordinates, but the presence of phi doesn't necessarily have any special meaning.

 

[irycio]

My bad, I meant polar coordinates. But I guess you are right anyway, that it doesn't have to be limited to [0, 2pi), we just do it because it's usually enough.

Nevertheless, after moving from carthesian to polar coordinates, I would really appreciate having any boundaries, but apparently they do not exist, unless A and B are pairs of points already in spherical coordinates, like P(radius, phi). That would make more sense, but it doesn't say anything like that.

Anyway, cheers for help.

 

[Mark44]

Points A and B, as given, seem incorrect to me. Instead of (0, a) and (-inf, 0), it seems more reasonable to me that they would be given as A(a, 0) and B(0, -inf), both in polar coordinates.

 

[irycio]

Well, I must have made a typo somewhere, foolish me. Using convention (radius, phi) it's (a,0) and (0, -inf), as you've written.

Sorry for my mistakes and thanks for help.