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May 5

line integrals with respect to arc length versus x or y

I'm a little confused how to picture these as opposed to line integrals with respect to arc length. Suppose you have a line through a field that "curls back" or even forms a circle. Normally what I see happen is that an integral with a dx term gets added to an integral with a dy term. But what if you just evaluated one of them? Is it possible? And how would you set its bounds? (You would start at x=0, for example, and end back at x=0.) John Riemann Soong (talk) 21:18, 5 May 2010 (UTC)[reply]

If (x(t),y(t)), 0<t<1 is a curve, the line integral with respect to arc length of the function f(x,y) is

\int _{0}^{1}f(x(t),y(t)){\sqrt {\left({\frac {dx}{dt}}\right)^{2}+\left({\frac {dy}{dt}}\right)^{2}}}dt

while the integral with respect to x is

\int _{0}^{1}f(x(t),y(t)){\frac {dx}{dt}}dt

Bo Jacoby (talk) 23:22, 5 May 2010 (UTC).[reply]

1D projection

It is really easy to make a 2D projection of a 3D shape, but what does a 1D projection of a 2D shape look like? —Preceding unsigned comment added by 93.96.113.87 (talk) 21:56, 5 May 2010 (UTC)[reply]

A 1D projection of any shape in any number of dimensions is either a point or a line segment or a collection of points and/or line segments. That's all you can have in 1D space - it's not very exciting. Gandalf61 (talk) 22:23, 5 May 2010 (UTC)[reply]

—————————————————— — Knowledge Seeker দ 17:26, 7 May 2010 (UTC)[reply]