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I am trying to apply the divergence theorem, but I need to find an outward pointing normal vector on the unit sphere. The answer gives $\hat n= (x_1,x_2,x_3)$.

Is the person who wrote up the solution just saying that any ordered triple on the sphere is an outward pointing normal - and this is all we need for the nds part of the integral?

Thanks,

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The outward pointing normal, for a sphere, is in the same direction as the radius. That is because both are perpendicular to the tangent plane.
The position vector has unit length because it is a unit sphere. So the normal, which is the unit vector in the same direction as the position vector, must equal the position vector.

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  • $\begingroup$ Very nice explanation, @Michael - thanks a lot :) $\endgroup$ – User001 Dec 29 '14 at 7:47

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