Let sigma denote the surface of a solid G with n the outward unit normal vector

Question

Let sigma denote the surface of a solid G with n the outward unit normal vector field to sigma. Assume that F is a vector field with continuous first-order partial derivatives on sigma. Prove that (curl F) n dS = 0 [Hint: Let C denote a simple closed curve on sigma that separates the surface into two subsurfaces sigma1 and sigma2 that share C as their common boundary. Apply Stokes' Theorem to sigma 1, and to sigma2 and add the results.] The vector field curl(F) is called the curl field of F. In words, interpret the formula in part (a) as a statement about the flux of the curl field.

Explanation / Answer

??s F · dS = ??? div F dV, by Divergence Theorem = ??? (3x^2 + 3y^2 - 6z) dV. Now, convert this to cylindrical coordinates: ?(? = 0 to 2p) ?(r = 0 to 2) ?(z = 0 to 2) (3r^2 - 6z) * (r dz dr d?) = 2p ?(r = 0 to 2) (3r^3 z - 3rz^2) {for z = 0 to 2} dr = p ?(r = 0 to 2) (12r^3 - 24r) dr = p(3r^4 - 12r^2) {for r = 0 to 2} = 0. ----------------- 2) curl div(f(x,y,z)) = curl

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