Final Exam Solution

# At the point 2 0 0 we have 2 0 0 1 cosh u 2 sinh

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Unformatted text preview: that satisﬁes the following: (x − 1)2 − y2 = 1, 4 2, x − 3 2 y 3 , 2 0 z 2. Let n (x, y, z ) be the unit normal of S with n (2, 0, 0) = (1, 0, 0). ˆ ˆ (a) (4 points) Give any parametrization Φ (u, v ) of S . Include the ranges of u and v . Solution: Φ (u, v ) = (1 + cosh u, 2 sinh u, v ) . Obviously v ∈ [0, 2]. To solve for the range of u, let eu = w 2 sinh u = 3 3 1 3 3 ⇒ eu − e−u = ⇒ w− = ⇒ w2 − w − 1 = 0 2 2 w 2 2 The positive solution to this quadratic equation is w = 2 so u = ln 2. To solve for the other limit of u, solve 2 sinh u = − 3 3 1 3 3 ⇒ eu − e−u = − ⇒ w − = − ⇒ w 2 + w − 1 = 0 2 2 w 2 2 1 The positive solution to this quadratic equation is w = − 1 so u = ln 2 = − ln 2. 2 So we have u ∈ [− ln 2, ln 2]. (b) (3 points) Find Tu × Tv . Is the parametrization orientation-preserving or reversing? Solution: Tu = (sinh u, 2 cosh u, 0) , Tv = (0, 0, 1) , Tu × Tv = (2 cosh u, − sinh u, 0) . At the point (2, 0, 0) we have (2, 0, 0) = (1 + cosh u, 2 sinh u, v ) which yields u = v = 0. At that point, Tu ×Tv = (2 cosh 0,,− sinh 0,,0) = (1, 0, 0) = n (2, 0, 0). ˆ (2 cosh 0 − sinh 0 0) T ×T u v So the parametrization is orientation preserving. (c) (3 points) Evaluate S F · dS where F (x, y, z ) = (x − 1, y, xyz ). Solution: S F · dS = D 2 F (Φ (u, v )) · Tu × Tv du dv ln 2 = 0 2 − ln 2 ln 2 2 − ln 2 ln 2 2 − ln 2 ln 2 = 0 = 0 = F (Φ (u, v )) · Tu × Tv du dv (cosh u, 2 sinh u, (1 + cosh u) 2 sinh u v ) · (2 cosh u, − sinh u, 0) du dv 2 cosh2 u − si...
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## This note was uploaded on 01/19/2014 for the course MATH 3010 taught by Professor Magpantay during the Winter '13 term at York University.

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