Recent content by calculus-stud

Oh right, sorry... but then what is the surface? I'm so lost... I thought you were asking for flux of the vector field through any closed curve...

Ok, does it have to do with the orientation of the curve? I'm really stuck here ... any hints?

Hey, umm, I'm kinda lost here. Can you give some more clues?

Yeah it should be F = (-2y + x2...) sorry. So, curl is 3k. The flux is zero? So, we don't really need to integrate or Stokes' theorem here?

But if the end points of the curve are the same, isn't that the definition of a closed curve?

Curl F is 3 k. I'm having trouble figuring out dS... I know for \intcurlF.dS I need a surface with the given boundary curve for dS... but I'm stuck there.

So, using the definition of derivative --- f'(x) = lim h->0 f (x+h) - f(x)/h, if we have h = y-x then as you said from the growth condition, then the numerator which is f(y) - f(x) is much smaller than the denominator y-x, so f' = 0 for y-x close to 0. Is that right? Also, when I'm changing h...

I'm new to analysis, so I'm still trying to grapple with the concepts... there is one that has been bugging me forever now --- ||f(x) - f(y)|| <= ||x -y||a is the Holder-continuous equality. What happens if a becomes > 1, does that still remain Holder-continuous or is Holder-continuity valid...

Stokes' Theorem says that \intF.dr inside a boundary C = \intcurlF.dS over a surface S... but how can I use it here? And is dc = (-sin t, cos t, cos (t/2)/2) ??

Given c(t) = [cos t, sin t, 2 + sin (t/2)] where t \epsilon [0, 2pi] and F(x,y,z) = (2-y + x2, x + sin y, \sqrt{}z4+1) --- Find \intF.dS over c(0, 2pi). I've no idea how to do this... any help would be awesome! Thanks!