Verify Stokes\'s theorem. S is a plane surface with straight edges. The vertices
ID: 3087234 • Letter: V
Question
Verify Stokes's theorem. S is a plane surface with straight edges. The vertices and orientation of C are given. v=xy^2zj , C: (0,0,1) to (0,1,2) to (1,2,0) to (0,0,1)Explanation / Answer
curl F = (1 - 0, -(0 - 1), 2) = (1, 1, 2). As for the surface, the sphere and cone intersect when x^2 + y^2 = 1/2 ==> z = v2/2. In spherical coordinates (with ? = 1), this corresponds to f = p/4. Thus, we can parameterize the surface as r(u, v) = (cos u sin v, sin u sin v, cos v) for u in [0, 2p] and v in [0, p/4]. r_u x r_v = (-cos u sin^2(v), -sin u sin^2(v), -sin v cos v) To get the orientation correct (so the normal points outward), negate this. n = (cos u sin^2(v), sin u sin^2(v), sin v cos v) Therefore Stokes' Theorem yields ?c F · dr = ?? curl F · dS = ?? curl F · n dA = ?? (1, 1, 2) · (cos u sin^2(v), sin u sin^2(v), sin v cos v) dA = ?(v in [0, p/4]) ?(u in [0, 2p]) (cos u sin^2(v)+ sin u sin^2(v) + 2 sin v cos v) du dv = ?(v in [0, p/4]) (0 + (2p) 2 sin v cos v) dv = 2p sin^2(v) {for v = 0 to p/4} = p. I hope this helps!
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