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# 12.01 - Section 13.8 wrap-up Let F = xz i yz j xy k as in...

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Section 13.8 wrap-up: Let F = xz i + yz j + xy k as in Example 2, and let S be the sphere of radius 1 centered at (1,2,3). Find ∫∫ S curl F d S . ..?.. Surface integrals over closed surfaces: If S is a closed surface, (*) ∫∫ S curl F d S = 0. Proof #1: If S is a closed surface, its boundary is the empty set, so ∫∫ S curl F d S = S F d r = 0. Proof #2: Let C be a piecewise-smooth simple closed curve that divides S into two pieces S + and S , so that C is positively oriented with respect to S + and negatively oriented with respect to S . [Draw picture.] Then ∫∫ S curl F d S = ∫∫ S + curl F d S + ∫∫ S curl F d S = C F d r + C F d r = C F d r C F d r = 0. Hybrid proof: Draw a curve C , but let it shrink down to a point. Suppose we have to evaluate C F d r where C is the unit circle in the x , y plane, and we suspect that Stokes’ Theorem would be helpful. What are some surfaces S for which S = C ?

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