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Given is the surface S which is the part of the sphere $$x^2+y^2+z^2=1 $$above the XY-plane. (z>0)

$$u(x,y,z)=x^3-y^3+z^2$$ $$v(x,y,z)=x+y+z$$

Calculate the surface integral: $$ \int_S^\ (\nabla u\times\nabla v)\cdot \vec{n}\cdot d\sigma $$

Stokes theorem gives $$\int_S^\ (\nabla u\times\nabla v)\cdot \vec{n}\cdot d\sigma\ =\dfrac{3}{2}\pi$$ (with n = unity vector perpendicular to the surface facing outward)

Divergention theorem states the following:

$$\int_S^\ \vec{G}\cdot \vec{n}\cdot d\sigma\ =\int_V^\ \nabla \vec{G}\cdot dV $$

But then we get:

$$ \nabla G = \nabla(\nabla u\times \nabla v)=\vec{0}$$

$$\int_S^\ \vec{G}\cdot \vec{n}\cdot d\sigma\ =\int_V^\ \nabla \vec{G}\cdot dV = 0 $$

This is not the same result as Stokes theorem, while the conditions are met for both theorems. What is the problem here?

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    Are you sure $S$ is closed? If it is not you cannot apply the divergence theorem as is, you need to subtract the flux through the opened part of $S$ (i.e., the disc at $z=0$ that closes the hemisphere here). – 2017-01-31
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    @Kuifje Didn't check that, but even then the flux is 0 through the disc at z=0. So the result would still be 0? $$\int_{-1}^{1} \! \int_{-sqrt(1-y^2)}^{sqrt(1-y^2)} \! (-3y^2-2z^2,2z-3x^2,3x^2+3y^2) \cdot (2x,2y,2z) d\x d\y = 0$$ – 2017-02-02

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