This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Solutions to some of the problems in Chapters 15 and 16 of Cain & Herod. 1 Section 15.1 15.1, Problem 1 We use x = u and y = v as parameters. The surface can then be parametrized by r ( u, v ) = u, v, √ u + 2 v 2 , u, v ≥ . 15.1, Problem 2 If we let x = 2 x , y = y and z = 2 √ 2 z , the equation becomes ( x ) 2 + ( y ) 2 + ( z ) 2 = 4 2 , which is a sphere centered at the origin with radius 4 in the x , y , z coordinates. We can use spherical coordinates to parametrize the sphere: x = 4 cos θ sin φ, y = 4 sin θ sin φ, z = 4 cos φ, ≤ θ ≤ 2 π, ≤ φ ≤ π. Reverting back to the x, y, z coordinates, we have thus x = 2 cos θ sin φ, y = 4 sin θ sin φ, z = √ 2 cos φ, ≤ θ < 2 π, ≤ φ ≤ π. or, letting u = θ and v = φ , r ( u, v ) = 2 cos u sin v, 4 sin u sin v, √ 2 cos v , ≤ u ≤ 2 π, ≤ v ≤ π. 15.1, Problem 3 If x 2 + y 2 = 1, we can write, using polar coordinates that x = cos θ and y = sin θ for some θ with 0 ≤ θ < 2 π . Using u = θ and v = z as parameters, the cylinder can thus be parametrized by r ( u, v ) = cos u, sin u, v , ≤ u ≤ 2 π,∞ < v < ∞ . 15.1, Problem 4 If the surface is parametrized by r ( u, v ) = u cos v, u sin v, u , ≤ v ≤ 2 π, 1 ≤ u ≤ 1 , 1 we have x 2 + y 2 = u 2 cos 2 v + u 2 sin 2 v = u 2 = z 2 . The surface is thus the part of the hyperboloid x 2 + y 2 = z 2 between the planes z = 1 and z = 1. 15.1, Problem 5 If the surface is parametrized by r ( u, v ) = u cos v, u sin v, u 2 , ≤ v ≤ 2 π, 1 ≤ u ≤ 2 , we have x 2 + y 2 = u 2 cos 2 v + u 2 sin 2 v = u 2 = z . The surface is thus the part of the paraboloid z = x 2 + y 2 between the planes z = 1 and z = 2. 15.1, Problem 6 The equation for the sphere of radius 3 centered at the point (1 , 2 , 3) is ( x 1) 2 + ( y 2) 2 + ( z 3) 2 = 3 2 . If we let x = x 1, y = y 2 and z = z 3, we get the equation ( x ) 2 + ( y ) 2 + ( z ) 2 = 3 2 , which is a sphere centered at the origin with radius 3 in the x , y , z coordinates. We can use spherical coordinates to parametrize the sphere: x = 3 cos θ sin φ, y = 3 sin θ sin φ, z = 3 cos φ, ≤ θ ≤ 2 π, ≤ φ ≤ π. Reverting back to the x, y, z coordinates, we have thus x = 1 + 3 cos θ sin φ, y = 2 + 3 sin θ sin φ, z = 3 + 3 cos φ, ≤ θ ≤ 2 π, ≤ φ ≤ π. or, letting u = θ and v = φ , r ( u, v ) = 1 + 3 cos u sin v, 2 + 3 sin u sin v, 3 + 3 cos v , ≤ u ≤ 2 π, ≤ v ≤ π. 15.1, Problem 7 We use the parametrization x = a cos θ sin φ, y = a sin θ sin φ, z = a cos φ, ≤ θ ≤ 2 π, ≤ φ ≤ π. or r ( φ, θ ) = a cos θ sin φ, a sin θ sin φ, a cos φ , ≤ θ ≤ 2 π, ≤ φ ≤ π. We have r φ = a cos θ cos φ, a sin θ cos φ, a sin φ , and r θ = a sin θ sin φ, a cos θ sin φ, ....
View
Full
Document
This note was uploaded on 01/29/2012 for the course ENG 2zz3 taught by Professor Proff during the Spring '11 term at McMaster University.
 Spring '11
 proff
 Meter

Click to edit the document details