Integrals in clindrical, spherical coordinates (Sect. 5.7 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 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
Integrals in clindrical, spherical coordinates (Sect. 5.7 Integration in spherical coordinates. Review: Clindrical coordinates. Spherical coordinates in space. Triple integral in spherical coordinates. Spherical coordinates in R Definition The spherical coordinates of a point P R 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 +.
Spherical coordinates in R Use spherical coordinates to epress region between the sphere + + = and the cone = +. Solution: ( = ρ sin(φ cos(θ, = ρ sin(φ sin(θ, = ρ cos(φ. = = + The top surface is the sphere ρ =. The bottom surface is the cone: ρ cos(φ = ρ sin (φ cos(φ = sin(φ, + = / Hence: R = / so the cone is φ = π 4. { (ρ, φ, θ : θ, π, φ, π }, ρ,. 4 Integrals in clindrical, spherical coordinates (Sect. 5.7 Integration in spherical coordinates. Review: Clindrical coordinates. Spherical coordinates in space. Triple integral in spherical coordinates.
Theorem If the function f : R R 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, V = π cos(π + cos( R ; hence: V = 4 πr.
Use spherical coordinates to find the volume below the sphere + + = and above the cone = +. { Solution: R = (ρ, φ, θ : θ, π, φ, π }, ρ,. 4 The calculation is simple, the region is a simple section of a sphere. V = π π/4 π π/4 V = dθ V = π V = π cos(φ + ρ sin(φ dρ dφ dθ, sin(φ dφ π/4 ( ρ, ρ dρ, V = π (. Triple integral in spherical coordinates Find the integral of f (,, = e ( + + / in the region R = {,,, + + } using spherical coordinates. { Solution: R = θ, π, φ, π }, ρ,. Hence, R f dv = π/ π/ π/ π/ dθ sin(φ dφ Use substitution: u = ρ, hence du = ρ dρ, so π cos(φ π e u du e ρ ρ sin(φ dρ dφ dθ, R e ρ ρ dρ. f dv = π 6 (e.
Change to spherical coordinates and compute the integral 4 4 + + 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 + + 4. Triple integral in spherical coordinates Change to spherical coordinates and compute the integral 4 4 + + 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.
Change to spherical coordinates and compute the integral Solution: 4 π π/ π π/ sin(θ dθ ( cos(θ π 4 + + d d d. ρ sin(φ sin(θ ( ρ sin(φ dρ dφ dθ. π/ ( π ( sin(φ sin (φ dφ ρ 4 dρ, ( ( ρ 5 cos(φ dφ 5 π/ 5 5, 4 π 5. Triple integral in spherical coordinates Compute the integral π π/ sec(φ ρ sin(φ dρ dφ dθ. Solution: Recall: sec(φ = / cos(φ. π π π/ π/ (ρ ( sec(φ sin(φ dφ, cos sin(φ dφ (φ In the second term substitute: u = cos(φ, du = sin(φ dφ. π ( cos(φ π/ + / du u.
Compute the integral π π/ sec(φ ρ sin(φ dρ dφ dθ. Solution: π ( cos(φ π/ + / du u. π ( + u du = π 4 / ( u /, π 4 + ( u / = π 4 + ( 8 = π We conclude: 5π. Triple integral in spherical coordinates (Sect. 5.7 Use spherical coordinates to find the volume of the region outside the sphere ρ = cos(φ and inside the half sphere ρ = with φ, π/. Solution: First sketch the integration region. ρ = cos(φ is a sphere, since ρ = ρ cos(φ + + = rho = + + ( =. ρ = is a sphere radius and φ, π/ sas we onl consider the upper half of the sphere. rho = cos (
(Sect. 5.7 Use spherical coordinates to find the volume of the region outside the sphere ρ = cos(φ and inside the sphere ρ = with φ, π/. Solution: rho = rho = cos ( V = V = π V = 6π ( cos(φ π π/ V = π π/ π/ π/ cos(φ ( ρ ρ sin(φ dρ dφ dθ. cos(φ sin(φ dφ 8 sin(φ 8 cos (φ sin(φ dφ. π/ cos (φ sin(φ dφ. Triple integral in spherical coordinates (Sect. 5.7 Use spherical coordinates to find the volume of the region outside the sphere ρ = cos(φ and inside the sphere ρ = with φ, π/. Solution: V = 6π ( cos(φ π/ π/ cos (φ sin(φ dφ. Introduce the substitution: u = cos(φ, du = sin(φ dφ. V = 6π + u du = 6π + ( u 4 4 = 6π (. 4 V = 6π 4 V = 4π.