Phys105A_W09_HW6sols

# Phys105A_W09_HW6sols - Physics 105A Homework 6 Solutions...

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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 π yds = 2 π 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 = κ is a constant. Solving for x , x = κ y 2- κ 2 . Using separation of varialbes dx = κ y 2- κ 2 dy. The integral on the right can be done using the u- substitution y = κ cosh u. Since dy = κ sinh u and cosh 2 u- 1 = sinh 2 u, dx = κ du. Hence x- x = κ u Solving for y, y = κ cosh x- x κ 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 θ 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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Phys105A_W09_HW6sols - Physics 105A Homework 6 Solutions...

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