Question: If S is the surface defined by x^6 + y^6 + z^6 = 1 and F is a smooth vector field on S. Explain why
If S is the surface defined by x^6 + y^6 + z^6 = 1 and F is a smooth vector field on S. Explain why S curl(F) dS = 0. In general, if we replace S with any surface such that S = (i.e. it has no boundary) does the same result hold?
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