Spherical Coordinates MATH 311, Calculus III J. Robert Buchanan Department of Mathematics Fall 2011
Spherical Coordinates Another means of locating points in three-dimensional space is known as the spherical coordinate system. z 0,0,z y x,y,z Φ Θ Ρ x,y,0 x
Coordinate Definitions If the point P has Cartesian coordinates (x, y, z), the points spherical coordinates (ρ, φ, θ) are as follows: ρ: the distance from the origin O to P. ρ = x 2 + y 2 + z 2 0 φ: the angle between vector x, y, z and the positive z-axis. ( ) 0 φ = cos 1 z π x 2 + y 2 + z 2 θ: the angle between vector x, y, 0 and the positive x-axis. ( ) θ = cos 1 x x 2 + y 2
Converting from Spherical to Cartesian Coordinates Given the spherical coordinates (ρ, φ, θ), x = ρ sin φ cos θ y = ρ sin φ sin θ z = ρ cos φ.
Converting from Spherical to Cylindrical Coordinates Given the spherical coordinates (ρ, φ, θ), r = ρ sin φ θ = θ z = ρ cos φ.
Example (1 of 6) If point P has spherical coordinates (ρ, φ, θ) = (2, π/4, π/3), find the coordinates of P in 1 Cartesian coordinates. 2 Cylindrical coordinates.
Example (2 of 6) 1 Cartesian coordinates: x = ρ sin φ cos θ = 2 sin π 4 cos π 3 = 1 2 y = ρ sin φ sin θ = 2 sin π 4 sin π 3 = 3 2 z = ρ cos φ = 2 cos π 4 = 2 2 Cylindrical coordinates: r = ρ sin φ = 2 sin π 4 = 2 θ = θ = π 3 z = ρ cos φ = 2 cos π 4 = 2
Example (3 of 6) If point P has Cartesian coordinates (x, y, z) = (0, 2 3, 2), find the coordinates of P in 1 Spherical coordinates. 2 Cylindrical coordinates.
Example (4 of 6) 1 Spherical coordinates: ρ = x 2 + y 2 + z 2 = 0 2 + (2 3) 2 + ( 2) 2 = 4 ( ) ( φ = cos 1 z = cos 1 2 ) = 2π x 2 + y 2 + z 2 4 3 ( ) θ = cos 1 x = cos 1 (0) = π x 2 + y 2 2 2 Cylindrical coordinates: r = x 2 + y 2 = 0 2 + (2 3) 2 = 2 3 ( ) θ = cos 1 x = cos 1 (0) = π x 2 + y 2 2 z = z = 2
Example (5 of 6) Find the equation in spherical coordinates of the hyperboloid of 2 sheets: x 2 y 2 z 2 = 1.
Example (5 of 6) Find the equation in spherical coordinates of the hyperboloid of 2 sheets: x 2 y 2 z 2 = 1. 1 = x 2 y 2 z 2 = ρ 2 sin 2 φ cos 2 θ ρ 2 sin 2 φ sin 2 θ ρ 2 cos 2 ( φ ) = ρ 2 sin 2 φ cos 2 θ sin 2 φ sin 2 θ cos 2 φ ( [ ] ) = ρ 2 sin 2 φ cos 2 θ sin 2 θ cos 2 φ ( ) = ρ 2 sin 2 φ cos 2θ cos 2 φ
Example (6 of 6) Find the equation in Cartesian coordinates of the surface whose equation in spherical coordinates is ρ = sin φ sin θ.
Example (6 of 6) Find the equation in Cartesian coordinates of the surface whose equation in spherical coordinates is ρ = sin φ sin θ. ρ = sin φ sin θ ρ 2 = ρ sin φ sin θ x 2 + y 2 + z 2 = y ( x 2 + y 1 ) 2 + z 2 = 1 2 4
Spherical Coordinate Equations (1 of 3) Sphere: ρ = c > 0, a constant y z x x 2 + y 2 + z 2 = c 2
Spherical Coordinate Equations (2 of 3) Plane: θ = θ 0, with 0 θ 0 2π z y x
Spherical Coordinate Equations (3 of 3) Cone: φ = φ 0, with 0 < φ 0 < π z y x
Volume Element in Spherical Coordinates 2.0 1.5 1.0 0.5 2.0 0.0 1.5 1.0 0.5 0.0 0.0 0.5 1.0 1.5 2.0 V (ρ sin φ θ)(ρ φ)( ρ) = ρ 2 sin φ ρ φ θ dv = ρ 2 sin φ dρ dφ dθ
Iterated Integrals in Spherical Coordinates The triple integral of f (ρ, φ, θ) over the solid region Q = {(ρ, φ, θ) g 1 (φ, θ) ρ g 2 (φ, θ), h 1 (θ) φ h 2 (θ), α θ β} is Q f (ρ, φ, θ) dv = β h2 (θ) g2 (φ,θ) α h 1 (θ) g 1 (φ,θ) f (ρ, φ, θ)ρ 2 sin φ dρ dφ dθ.
Examples (1 of 9) Evaluate the triple integral below by converting to spherical coordinates. e (x 2 +y 2 +z 2 ) 3/2 dv where Q = {(x, y, z) x 2 + y 2 + z 2 1}. Q
Examples (2 of 9) Q e (x 2 +y 2 +z 2 ) 3/2 dv = = 2π = 2π 2π π 1 0 0 0 π 1 0 π 0 1 e (ρ2 ) 3/2 ρ 2 sin φ dρ dφ dθ ρ 2 e ρ3 sin φ dρ dφ 1 0 3 eρ3 0 π sin φ dφ = 2π 1 (e 1) sin φ dφ 0 3 = 2(e 1)π ( cos φ) π 0 3 = 4(e 1)π 3
Examples (3 of 9) Find the volume of the solid that lies above the cone z 2 = x 2 + y 2 and below the sphere x 2 + y 2 + z 2 = z. y 0.0 0.5 1.0 0.5 z 0.5 0.0 0.5 0.0 x 0.5
Example (4 of 9) In spherical coordinates the equations of the cone and the sphere are Cone: φ = π 4 Sphere: ρ = cos φ.
Example (5 of 9) V = = 2π = 2π = 2π 3 = 2π 3 1 dv = Q 0 π/4 cos φ 0 π/4 0 1 0 3 ρ3 0 π/4 0 1/ 2 = π 1 6 u4 1/ = π 2 8 1 2π π/4 cos φ 0 0 ρ 2 sin φ dρ dφ cos φ sin φ dφ cos 3 φ sin φ dφ u 3 du = 2π 3 1 1/ 2 (1)ρ 2 sin φ dρ dφ dθ u 3 du
Examples (6 of 9) Find the mass and center of mass of the solid hemisphere of radius a if the density at any point in the solid is proportional to its distance from the base. z y J. Robert Buchanan x Spherical Coordinates
Examples (7 of 9) The distance of a point in the hemisphere from the base is the z-coordinate of the point. In spherical coordinates z = ρ cos φ. Without loss of generality, we may choose the proportionality constant to be 1. m = M yz = M xz = M xy = Q Q Q Q z dv x z dv y z dv z 2 dv
Examples (8 of 9) m = M xy = = 2π 2π π/2 a = πa4 2 = 2π 0 0 0 π/2 a 0 0 π/2 0 2π π/2 a = 2πa5 5 0 0 0 π/2 a = 2πa5 5 0 0 π/2 0 0 (ρ cos φ)ρ 2 sin φ dρ dφ dθ ρ 3 cos φ sin φ dρ dφ cos φ sin φ dφ = πa4 2 1 (ρ cos φ) 2 ρ 2 sin φ dρ dφ dθ ρ 4 cos 2 φ sin φ dρ dφ cos 2 φ sin φ dφ u 2 du = 2πa5 5 1 1 0 0 u du = πa4 4 u 2 du = 2πa5 15
Examples (9 of 9) M yz = = M xz = = = 0 = 0 2π π/2 a 0 0 0 a π/2 2π 0 0 0 2π π/2 a 0 0 0 a π/2 2π 0 0 0 (ρ sin φ cos θ)ρ 2 sin φ dρ dφ dθ ρ 3 sin 2 φ cos θ dθ dφ dρ (ρ sin φ sin θ)ρ 2 sin φ dρ dφ dθ ρ 3 sin 2 φ sin θ dθ dφ dρ Thus (x, y, z) = ( Myz m, M xz m, M ) ( xy = 0, 0, 8a ). m 15
Homework Read Section 13.7. Exercises: 1 57 odd