lecture notes 13
lecture notes 13
lecture notes 13
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.Example. Let F(x, y, z) = (x, y, z), and T be the surface defined by{ (x, y, z) ∈ R 3 | |x| + |y| + |z| = 1 }, with outward ∫∫ pointing unitnormal n on T. Evaluate the surface integral F · ndS..TSolution. Let D be the solid bounded by the surface T, i.e.D = { (x, y, z) ∈ R 3 | |x| + |y| + |z| ≤ 1 }. One can check that thecondition of the divergence theorem holds, and hence∫∫∫∫∫∫∫∫F · ndS = divFdV = ( ∂xTDD ∂x + ∂x∂x + ∂x∂x ) dV∫∫∫= 3 × dV = 3Vol(D) = 3 × 8 × 1 3! × 1 × 1 × 1 = 4.D. . . . . .