Learning Stoke’s Theorem by example ! Reply


Manmohan Dash<g6pontiac@gmail.com>


Calculate CurlF and then use Stokes’ theorem to compute the flux of CurlF through the given surface as a line integral. F  =   < y,  x,  x2 + y2 > , the upper hemisphere; x2 + y2 + z2 = 1,  z ≥ 0.

Answer; curlF =

i j k      
x y z
Fx Fy Fz

Now expanding the terms, the first component of curl of F is, (curlF)x = ∂yFz − ∂zFy and the 2nd term here has to be zero, as there is no z dependence. Thus the first term is 2y.

Similarly the y component of curl of F, is: (curlF)y = ∂xFz − ∂zFx = 2x.

But the 3rd term of the curl, (curlF)z = ∂xFy − ∂yFx is zero, since x-partial of 2nd component, ∂xFy, and y-partial of the 1st component of F, ∂yFx, are both 1.

So, CurlF =  < 2y,  2x,  0 > .

Now Stoke’s Theorem is the following; ∫S curlF.dS = ∮F.dl.

The line integral of F is over the contour of the circle on the z = 0 plane, as S on the surface integral on LHS is the upper hemisphere.

So, the dot product of F and dl is this: (yi + xj).(dxi + dyj) = ydx + xdy.

Consequently the integral is ∮ydx + xdy where the limits of integration vary from (0, 1).

So RHS = ∫01dx√1 - x2 + ∫01dy√1 − y2 = 2∫01dx√1 − x2.

We replace x = sinθ so that dx =  − cosθdθ.

We then have RHS =  − 2∫0 π/2dθcos2θ =  − 2[1/2(θ + sinθcosθ)]0 π/2 =  − π/2.

So the flux of the CurlF through the given surface is  − π/2.

Document generated by eLyXer 1.2.5 (2013-03-10) on 2015-05-30T22:42:34.204000

Leave a Reply

Please log in using one of these methods to post your comment:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s