I have to verify Stokes' Theorem to the field $$F=(3z,4x,2y)$$ and the surface $$z=10-x^2-y^2,\quad 1\leq z\leq9$$
For $z=1$ and $z=9$ I get two circles. I'd like to know which one I should choose to evaluate the line integral.
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1The boundary curve is the union of these two circles with opposite orientation. If orientation of surface is the one with normal vectors facing upwards, the induced orientation of the circle on $z = 1$ is counter-clockwise (when projected to xy-plane) while that one at $z = 9$ is clockwise. – achille hui Jun 13 '23 at 03:11