Final Exam Solution

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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

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.

Ask a homework question - tutors are online