Orientation is an issue here switching the orientation will introduce a negative sign to the integral. dS as in §7.6, we are integrating a vector field F : R3 → R3 and the definition is ∫∫ S F.Orientation is not a concern for this type of integral since it is only the length, and not the direction, of the normal vector which plays a role. When we compute ∫∫ S f dS as in §7.5, we are dealing with a real-valued function f : R3 → R and the defintion is∫∫ S f dS = ∫∫ D f(Φ(u, v))||Tu × Tv|| dudv. Once we have that we can compute the normal vector Tu × Tv. (4) In both cases we need a parametrization Φ : D → R3 for which Φ(D) = S. When you’re done, go to the surfaces applet and plot it! (You’ll need to use that |u− v| = √ (u− v)2.) (3) Since ||Tu × Tv|| = √ 3, you don’t need to break it into two regions. You can graph the figure by thinking about what happens on the line u = v and then thinking about what happens to each boundary segment of D. (2) You can get the derivative everywhere except where u = v. The easy way to get the unit normal is to write it as 1 3 (x− 4, y− 5, z +5), but you could also compute −1 3 Tθ × Tφ = 1 3 Tφ × Tθ. Frank, Spring 2005 (1) Let θ ∈ and let φ ∈. Download Final Exam Review Sheet Hints - Multivariable Calculus | MATH 222 and more Calculus Study notes in PDF only on Docsity!Final Review Sheet Hints-Math 222 Prof.
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