Stokes' theorem application

vendredi 29 novembre 2013

The problem statement, all variables and given/known data.

Let ##F## be the vector field defined by ##F(x,y,z)=(-y,yz^2,,x^2,z)## and ##S \subset \mathbb R^3## the surface defined as ##S=\{x^2+y^2+z^2=4, z\geq 0\}##, oriented according to the exterior normal vector. Calculate:

##\iint_S (\nabla\times F).dS##.



The attempt at a solution.



I've calculated the curl, it's not an easy integral to calculate.



I can't apply Stokes' theorem because it is not a closed surface, but if I consider the surface ##S^{*}=\{x^2+y^2+z^2=4, z\geq 0\} \cup \{ x^2+y^2\leq 4, z=0\}##, then this is a closed surface and ##F## is of class ##C^1##, so Stokes'theorem says that:



##\iint_S^{*} (\nabla\times F).dS=\int_CF.ds## where ##C## is the boundary of the surface ##S^{*}##.



Now, my original integral is



##\iint_S (\nabla\times F).dS=\iint_S^{*} (\nabla\times F).dS-\iint_D (\nabla\times F).dS##, where ##D=\{ x^2+y^2\leq 4, z=0\}##. But as ##D## is a closed surface, I can also apply Stokes' theorem, so



##\iint_D (\nabla\times F).dS=\int_{C'} F.ds##, where ##C'## is the boundary of ##D##.



Now, my question is: isn't ##C=C'##?, I mean, the curve boundary of ##S^{*}## is the same boundary than the one of ##D##. If this is the case



##\iint_S (\nabla\times F).dS=\int_C F.ds-\int_{C'} F.ds=\int_C F.ds-\int_{C} F.ds=0##.





Could someone tell me if I am doing something wrong?






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