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Marsden Vector Calculus 6e: Section 1.4 - Exercise 3 Page 1 of 4 Exercise 3 (a) The following points are given in cylindrical coordinates; express each in rectangular coordinates and spherical coordinates: (1, 45➦, 1), (2, π/2, −4), (0, 45➦, 10), (3, π/6, 4), (1, π/6, 0), and (2, 3π/4, −2). (Only the first point is solved in the Study Guide.) (b) Change each of the following points from rectangular coordinates to spherical coordinates √ √ and to cylindrical coordinates: (2, 1, −2), (0, 3, 4), ( 2, 1, 1), (−2 3, −2, 3). (Only the first point is solved in the Study Guide.) Solution Part (a) Cartesian coordinates (x,y,z) and spherical coordinates (ρ,θ,ϕ), with ϕ being the polar angle, can be written in terms of cylindrical coordinates (r,θ,z) as x=rcosθ ρ2 = r2 +z2 y = rsinθ θ = θ z = z ρcosϕ=z. (r = 1,θ = 45➦,z = 1) x=1cos45➦ √ √ ! y = 1sin45➦ → x= 2,y= 2,z=1 2 2 z = 1 p2 2 ρ = 1 +1 θ = 45➦ √ → ρ= 2,θ=45➦,ϕ=45➦ 1 −1 ϕ=cos √ 2 2 1 +1 (r = 2,θ = π/2,z = −4) π x=2cos 2 y = 2sin π → (x=0,y=2,z=−4) 2 z = −4 p2 2 ρ = 2 +(−4) π θ = 2 √ π ! → ρ= 20,θ= ,ϕ≈153➦ 2 −1 −4 ϕ=cos p 2 2 2 +(−4) www.stemjock.com Marsden Vector Calculus 6e: Section 1.4 - Exercise 3 Page 2 of 4 (r = 0,θ = 45➦,z = 10) x=0cos45➦ y = 0sin45➦ → (x=0,y=0,z=10) z = 10 p2 2 ρ = 0 +10 θ = 45➦ → (ρ=10,θ=45➦,ϕ=0) 10 −1 ϕ=cos √ 2 2 0 +10 (r = 3,θ = π/6,z = 4) π x=3cos √ ! 6 y = 3sin π → x=3 3,y= 3,z=4 6 2 2 z = 4 p2 2 ρ = 3 +4 π θ = 6 → ρ=5,θ=π,ϕ≈36.9➦ 6 4 −1 ϕ=cos √ 2 2 3 +4 (r = 1,θ = π/6,z = 0) π x=1cos √ ! 6 y = 1sin π → x= 3,y=1,z=0 6 2 2 z = 0 p2 2 ρ = 1 +0 π θ = 6 → ρ=1,θ=π,ϕ=π 6 2 0 −1 ϕ=cos √ 2 2 1 +0 www.stemjock.com Marsden Vector Calculus 6e: Section 1.4 - Exercise 3 Page 3 of 4 (r = 2,θ = 3π/4,z = −2) 3π x=2cos 4 3π √ √ y = 2sin → x=− 2,y= 2,z=−2 4 z = −2 p2 2 ρ = 2 +(−2) 3π θ = 4 √ 3π 3π ! → ρ = 8,θ = , ϕ = 4 4 −2 −1 ϕ=cos p 2 2 2 +(−2) Part (b) Cylindrical coordinates (r,θ,z) and spherical coordinates (ρ,θ,ϕ), with ϕ being the polar angle, can be written in terms of Cartesian coordinates (x,y,z) as 2 2 2 2 2 2 2 r =x +y ρ =x +y +z tanθ = y tanθ = y x x z = z ρcosϕ=z. (x = 2,y = 1,z = −2) p2 2 r = 2 +1 −1 1 √ θ = tan 2 → r = 5,θ ≈ 26.6➦,z = −2 z = −2 p2 2 2 ρ = 2 +1 +(−2) 1 −1 θ = tan 2 → (ρ=3,θ≈26.6➦,ϕ≈132➦) ! −1 −2 ϕ=cos p 2 2 2 2 +1 +(−2) www.stemjock.com Marsden Vector Calculus 6e: Section 1.4 - Exercise 3 Page 4 of 4 (x = 0,y = 3,z = 4) p2 2 r = 0 +3 −1 3 π θ = tan 0 → r = 3,θ = 2,z = 4 z = 4 p2 2 2 ρ = 0 +3 +4 3 −1 π θ = tan 0 → ρ=5,θ= ,ϕ≈36.9➦ 2 4 −1 ϕ=cos √ 2 2 2 0 +3 +4 (x = √2,y = 1,z = 1) q√ 2 2 r = ( 2) +1 −1 1 √ θ = tan √ → r= 3,θ≈35.3➦,z=1 2 z = 1 q√ 2 2 2 ρ = ( 2) +1 +1 1 −1 θ = tan √ π 2 → ρ=2,θ≈35.3➦,ϕ= 3 1 −1 ϕ=cos q √ ( 2)2 +12 +12 √ q (x = −2 3,y = −2,z = 3) √ 2 2 r = (−2 3) +(−2) −1 1 → r = 4,θ = 7π,z = 3 θ = π +tan √ 6 3 z = 3 q √ 2 2 2 ρ = (−2 3) +(−2) +3 1 −1 θ = π +tan √ 7π 3 → ρ = 5,θ = , ϕ ≈ 53.1➦ 6 3 −1 ϕ=cos q √ 2 2 2 (−2 3) +(−2) +3 www.stemjock.com
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