Learning Stoke’s Theorem by example !

Calculus

Manmohan Dash<g6pontiac@gmail.com>

Question.

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 =

Now expanding the terms, the first component of curl of F is, (curlF)_x = ∂_yF_z − ∂_zF_y 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 = ∂_xF_z − ∂_zF_x = 2x.

But the 3rd term of the curl, (curlF)_z = ∂_xF_y − ∂_yF_x is zero, since x-partial of 2nd component, ∂_xF_y, and y-partial of the 1st component of F, ∂_yF_x, 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.

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