Integrals in cylindrical, spherical coordinates (Sect. 15.7)
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1 Integrals in clindrical, spherical coordinates (Sect Integration in spherical coordinates. Review: Clindrical coordinates. Spherical coordinates in space. Triple integral in spherical coordinates. Clindrical coordinates in space. Definition The clindrical coordinates of a point P R 3 is the ordered triple (r, θ, defined b the picture. Remark: Clindrical coordinates are just polar coordinates on the plane = together with the vertical coordinate. Theorem (Cartesian-clindrical transformations The Cartesian coordinates of a point P = (r, θ, are given b = r cos(θ, = r sin(θ, and =. The clindrical coordinates of a point P = (,, in the first and fourth quadrant are r = +, θ = arctan(/, and =. r P
2 Integrals in clindrical, spherical coordinates (Sect Integration in spherical coordinates. Review: Clindrical coordinates. Spherical coordinates in space. Triple integral in spherical coordinates. Spherical coordinates in R 3 Definition The spherical coordinates of a point P R 3 is the ordered triple (ρ, φ, θ defined b the picture. rho Theorem (Cartesian-spherical transformations The Cartesian coordinates of P = (ρ, φ, θ in the first quadrant are given b = ρ sin(φ cos(θ, = ρ sin(φ sin(θ, and = ρ cos(φ. The spherical coordinates of P = (,, in the first quadrant are ρ = ( ( + +, θ = arctan, and φ = arctan +.
3 Spherical coordinates in R 3 Use spherical coordinates to epress region between the sphere + + = 1 and the cone = +. Solution: ( = ρ sin(φ cos(θ, = ρ sin(φ sin(θ, = ρ cos(φ. = 1 = + The top surface is the sphere ρ = 1. The bottom surface is the cone: ρ cos(φ = ρ sin (φ cos(φ = sin(φ, + = 1/ Hence: R = 1/ so the cone is φ = π 4. { (ρ, φ, θ : θ, π], φ, π ] }, ρ, 1]. 4 Integrals in clindrical, spherical coordinates (Sect Integration in spherical coordinates. Review: Clindrical coordinates. Spherical coordinates in space. Triple integral in spherical coordinates.
4 Theorem If the function f : R R 3 R is continuous, then the triple integral of function f in the region R can be epressed in spherical coordinates as follows, f dv = f (ρ, φ, θ ρ sin(φ dρ dφ dθ. R R Remark: Spherical coordinates are useful when the integration region R is described in a simple wa using spherical coordinates. Notice the etra factor ρ sin(φ on the right-hand side. Triple integral in spherical coordinates Find the volume of a sphere of radius R. Solution: Sphere: S = {θ, π], φ, π], ρ, R]}. V = π π R π ] π V = dθ V = π cos(φ ρ sin(φ dρ dφ dθ, ] R ] sin(φ dφ ρ dρ, π ] R 3 3, V = π cos(π + cos( ] R 3 3 ; hence: V = 4 3 πr3.
5 Use spherical coordinates to find the volume below the sphere + + = 1 and above the cone = +. { Solution: R = (ρ, φ, θ : θ, π], φ, π ] }, ρ, 1]. 4 The calculation is simple, the region is a simple section of a sphere. V = π π/4 1 π ] π/4 V = dθ V = π V = π cos(φ ] ρ sin(φ dρ dφ dθ, ] 1 sin(φ dφ π/4 ]( ρ 3 3 1, ] ρ dρ, V = π 3 (. Triple integral in spherical coordinates Find the integral of f (,, = e ( + + 3/ in the region R = {,,, + + 1} using spherical coordinates. { Solution: R = θ, π ], φ, π ] }, ρ, 1]. Hence, R f dv = π/ π/ 1 π/ ] π/ ] 1 dθ sin(φ dφ Use substitution: u = ρ 3, hence du = 3ρ dρ, so π cos(φ π ] 1 e u 3 du e ρ3 ρ sin(φ dρ dφ dθ, R ] e ρ3 ρ dρ. f dv = π 6 (e 1.
6 Change to spherical coordinates and compute the integral d d d. Solution: ( = ρ sin(φ cos(θ, = ρ sin(φ sin(θ, = ρ cos(φ. Limits in : ; Limits in : 4, so the positive side of the disk + 4. Limits in : 4, so a positive quarter of the ball Triple integral in spherical coordinates Change to spherical coordinates and compute the integral d d d. Solution: ( = ρ sin(φ cos(θ, = ρ sin(φ sin(θ, = ρ cos(φ. Limits in θ: θ.π]; Limits in φ: φ, π/]; Limits in ρ: ρ, ]. The function to integrate is: f = ρ sin(φ sin(θ. π π/ ρ sin(φ sin(θ ( ρ sin(φ dρ dφ dθ.
7 Change to spherical coordinates and compute the integral Solution: 4 π π/ π ] π/ sin(θ dθ ( cos(θ π d d d. ρ sin(φ sin(θ ( ρ sin(φ dρ dφ dθ. π/ 1 ( π 1 ( sin(φ ] sin ] (φ dφ ρ 4 dρ, 1( ]( ρ 5 1 cos(φ dφ 5 π/ ] 5 5, 4 π 5. Triple integral in spherical coordinates Compute the integral π π/3 sec(φ 3ρ sin(φ dρ dφ dθ. Solution: Recall: sec(φ = 1/ cos(φ. π π π/3 π/3 (ρ 3 ( 3 sec(φ sin(φ dφ, 1 cos 3 sin(φ dφ (φ In the second term substitute: u = cos(φ, du = sin(φ dφ. π 3( cos(φ π/3 + 1/ 1 du u 3 ].
8 Compute the integral π π/3 sec(φ 3ρ sin(φ dρ dφ dθ. Solution: π 3( cos(φ π/3 + 1/ 1 du ] u 3. π 3( 1 1 ] + 1 u 3 du = π 4 1/ ( u 1 1/ ], π ( u 1 1/ ] = π ( ] 8 = π ] 3 We conclude: 5π.
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