Thread: Stokes Theorum

I'm not sure if this is the right place to put this?

Show that: curl(fu) = fcurlu + (gradf) x u

By using this in relation to Stoke's theorum, show that for a simple closed curve C,

(Line integral) fdS = -(surface integral) (gradf) x dS

Firstly, is this a typo? I've never heard of a line integral of a function wrt to surface element?


I think Stoke's Theorum is:
(Line integral)(A.dl) = (surface integral)((curlA).dS)

I did the first part fine. I then tried to do

(surface integral) (gradf)xdS = (surface integral) (curl(fdS) - fcurldS)

I then got stuck. Please help!

I think it is a typo.

I put fu=fdS.
Stick it into the identity you made up to find a value to curl(fdS).

Re-arrange to find an identity for (gradf)xdS.

Then stick (gradf)xdS into an integral sign, use your second identity and Stokes theorem, and note that curl(dS) = 0 (I think).

Then the solution should pop out.

Sorry, I was wrong, it is harder than that.

curl(uF) = u(curlF) + grad(u)xF

let F (x,y,z) = f(x,y,z)i + g(x,y,z)j + h(x,y,z)k
then uF= u fi + ugi + uhk


curl(uF) = {d/dy(uh) - d/dz(ug)}i + {d/dz(uf) - d/dy(uh)}j + {d/dx(ug) - d/dy(uf)}k
= [ u{dh/dy-dg/dz}i + u{df/dz-dh/dy}j + u(dy/dx - df/dy}k ] + [ {(du/dy)(h) - (du/dz)(g)}i + {(du/dz)(f) - (du/dy)(h)}j + {(du/dx)(y) - (du/dy)(f)}k ]
=u(curlF) + grad(u)xF