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Unformatted text preview: Physics 105A Homework # 6 Solutions Richard Eager Problem (1) Taylor 6.3 Attached. (a) Problem (2) Taylor 6.19 The area of the surface of revolution is A = 2 yds = 2 y 2 y 1 y 1 + x 2 dy So we wish to minimize the functional f ( x ( y ) , x ( y ) , y ) . As in Problem 6.10 the Euler-Lagrange equation d dy f x = f x simplifies since f x = 0 . Thus f x = is a constant. Solving for x , x = y 2- 2 . Using separation of varialbes dx = y 2- 2 dy. The integral on the right can be done using the u- substitution y = cosh u. Since dy = sinh u and cosh 2 u- 1 = sinh 2 u, dx = du. Hence x- x = u Solving for y, y = cosh x- x 1 Note on Surface Area dS 2 = ( dxdy ) 2 + ( dxdz ) 2 + ( dydz ) 2 dS = 1 + x y 2 + x z 2 dydz Converting to polar, dS = ( 1 + r 2 ) 1 / 2 rdrd Problem (3) Taylor 6.20 Given a functional f ( y ( x ) , y ( x ) , x ) that does not explicitly depend on x, e.g....
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- Fall '08