# Line integrals - ds, (dx, dy, dz)

Printable View

• Aug 3rd 2005, 06:05 AM
Cold
Line integrals - ds, (dx, dy, dz)
Am I right in thinking that line integrals of the form int_c f(x,y)dx arise by approximating the arc length delta s by the line segment delta x so that as delta x tends to 0 delta s / delta x = 1?

I can undertsand int_c f(x,y)ds as dividing the curve into smaller and smaller arcs.

Is this similar to resolving vecors into i components and j components (for int_c f(x,y)dy) . The problem is that I have a poorish mechanics and phyisics backround and find some examples difficult. Thanks for any help.
• Aug 13th 2005, 05:07 AM
Rebesques
Quote:

Am I right in thinking that line integrals of the form int_c f(x,y)dx arise by approximating the arc length delta s by the line segment delta x so that as delta x tends to 0 delta s / delta x = 1?
Actually, no. Let the curve be c(t)=(a(t),b(t)). In the expression int_c f(x,y)dx, you have x= the i-coordinate of the curve= a(t). As for ds, it equals sqrt{a'(t)^2+b'(t)^2}dt, with a little help from differential geometry. So,
ds/dx->1 cannot hold.

Quote:

I can undertsand int_c f(x,y)ds as dividing the curve into smaller and smaller arcs. Is this similar to resolving vecors into i components and j components (for int_c f(x,y)dy) .
The general setting for the curvilinear integral int_c f(x,y)ds, is calculating the signed area of the surface, formed by the curve c and the graph of f. Let's see this (half-vigorously); Consider a partition of the curve, that is divide it into minor segments of length ds. For a random point (x,y) on the minor segment, f(x,y)ds is the signed area of the parallelogram of sides ds and f(x,y). Adding these up, we get Sum{f(x,y)ds}, which is actually a riemannian sum, converging to int_c f(x,y)ds.

For practical calculation, if c(t)=(a(t),b(t)), then
int_c f(x,y)ds=int f(a(t),b(t)) sqrt{(a'(t))^2+(b'(t))^2}dt.

Hope this helps.
• Aug 13th 2005, 06:09 AM
Cold
Thanks. The James Stewart 5th edition (great book) just tags line integrals dx and dy after line integrals ds. I can visualise them ds but am having trouble seeing dx and dy geometrically if that makes sense. I'm going to think about the parallelogram and get back to you if that's ok? I know how to calculate line integrals dx dy but I want to understand them!
• Aug 14th 2005, 05:07 AM
Cold
Hi
If we replace delta s with delta x then we have.... (see attached file)
I'm struggling to understand why we would want to do this. Every book I've seen just says if we do it then we have line integral with respect to x or y. Fair enough but I can't see how this relates to the curve if we're not summing along its path.
Thanks for your help.
• Aug 18th 2005, 02:27 AM
Rebesques
Let's see... We replace ds with dx, meaning that we calculate int_c fx' dt. Alone, this does not mean much, but the integral W=int_c fdx+gdy=int_t (fx'+gy' )dt has a precise physical meaning: it represents the work of a force (f,g), under the influence of which, a particle moves along the curve c.

I suppose the purpose here, was to arrive at the last integral, as smoothly as possible. So, you could trouble yourself :p explaining why does the integral W resemble work?
That is,

a) If the force field is conservatory, i.e. f=-dh/dx, g=-dh/dy for some h, how do we calculate W?
b) Does W depend on the curve, if we fix the endpoints?
c) What happens to W on closed curves?

Hope this helps.
• Aug 18th 2005, 08:31 AM
Cold
Yeah kind of. I was looking at line integrals with vectors and linking them to integrals Pdx + Qdy - that makes sense in a backwards way. I'm embarrassed to say mechanics isn't my strong point, more pure maths. Can you ask me a specific question relating to work so that I can do it and ask questions? Thanks.
• Aug 18th 2005, 12:40 PM
Rebesques
Well, I am no mechanics expert myself, but you can try these already mentioned. Prove that

a) If there exists h=h(x,y), such that f=-θh/θx and g=-θh/θy, then W=int_c fdx+gdy = h(end point of c) - h(start point of c).

b) Generalize a), in the case of int_c f_1 dx_1+f_2 dx_2+...+f_n dx_n.

c) Under the same hypotheses as in a), W does not depend on the path of integration.

d) Under the same hypotheses as in a), W=0 on closed paths.

e) Forget a), and let the force field (f,g) be conservatory, that is (...alternatively) W=0 for every closed path c. Prove that there exists h, such that f=-θh/θx and g=-θh/θy.

f) Consider a portion of the Earth's surface. Is weight a conservatory force field?

Good Luck.
• Aug 26th 2005, 07:50 AM
Cold
Been on holiday but will have a go over the next couple of days.