Surface Area and Surface Integrals - Lecture Slides | MAT 272, Exams of Analytical Geometry and Calculus

Download Surface Area and Surface Integrals - Lecture Slides | MAT 272 and more Exams Analytical Geometry and Calculus in PDF only on Docsity! Surface area and surface integrals Eric Kostelich DEPT. OF MATHEMATICS AND STATISTICS April 3, 2009 Reading for this week Sections 7.1–7.4 Next exam postponed until Friday, April 10 Homework is posted and is due next Friday MAT 272 April 3 E. Kostelich MATHEMATICS AND STATISTICS 2 / 15 Important notes Assume that S is smooth except possibly at a finite number of points S can be a “quilt”, i.e., the union of a finite number of smaller surfaces In this case we require smoothness except at the corners of the patches in the quilt MAT 272 April 3 E. Kostelich MATHEMATICS AND STATISTICS 5 / 15 Justification for the area formula In R3, the area of the parallelogram spanned by vectors a and b is ‖a×b‖ Proof: Some algebra shows that ‖a×b‖2 = ‖a‖2‖b‖2− (a ·b)2 = ‖a‖2‖b‖2(1− cos2 θ) = ‖a‖2‖b‖2 sin2 θ MAT 272 April 3 E. Kostelich MATHEMATICS AND STATISTICS 6 / 15 Justification, 2 A little box ∆u∆v at (ui,vj) gets mapped by Φ to a little parallelogram of area ‖Tu∆u×Tv∆v‖ where the derivatives are evaluated at (ui,vj) MAT 272 April 3 E. Kostelich MATHEMATICS AND STATISTICS 7 / 15 Caveats The mapping Φ must be “mostly” 1-to-1 Example: The sphere of radius ρ may be parametrized as x = ρ sinφ cosθ , y = ρ sinφ sinθ , z = ρ cosφ But if we take 0 ≤ θ ≤ 2π and 0 ≤ φ ≤ 2π then we cover the sphere twice! (Need 0 ≤ φ ≤ π instead) Φ is not 1-to-1 when φ = 0 or θ = 2π but overlaps are a point or curve with no area MAT 272 April 3 E. Kostelich MATHEMATICS AND STATISTICS 10 / 15 The surface area of the sphere Given the parametrization x = ρ sinφ cosθ , y = ρ sinφ sinθ , z = ρ cosφ we have Tθ = (−ρ sinφ sinθ ,ρ sinφ cosθ ,0) Tφ = (ρ cosφ cosθ ,ρ cosφ sinθ ,−ρ sinφ) Tθ ×Tφ = −ρ2 sinφ(sinφ cosθ ,sinφ sinθ ,cosφ) ‖Tθ ×Tφ‖ = ρ2 sinφ Then SA = ∫ 2π 0 ∫ π 0 ρ 2 sinφ dφ dθ = 4πρ2 MAT 272 April 3 E. Kostelich MATHEMATICS AND STATISTICS 11 / 15 Surface integrals Given a parametrization r(t), a ≤ t ≤ b of a curve C, we define the path integral∫ C f ds = ∫ b a f (r(t)) ‖r′(t)‖ dt Given a parametrization Φ(u,v), (u,v) ∈ [a,b]× [c,d] of a surface S, we define the surface integral∫∫ S f dS = ∫ d c ∫ b a f (Φ(u,v)) ‖Tu×Tv‖ du dv MAT 272 April 3 E. Kostelich MATHEMATICS AND STATISTICS 12 / 15