Download presentation
Presentation is loading. Please wait.
1
The Triple Integral
2
Remark The triple integral of f over the region { (x,y,z) : x є [ a, b ], α(x) ≤ y ≤ β(x), g(x,y) ≤ z ≤ h(x,y) } is equal to: a b α(x) β(x) g(x,y) h(x,y) f(x,y,z) dz dy dx = a b [ α(x) β(x) [ g(x,y) h(x,y) f(x,y,z) dz ] dy ] dx
3
Example 1 Evaluat the triple integral I of the given function f over the given region D. f(x,y,z) = 8xyz D = { (x,y,z) : x є [ 0, 3 ], y є [ 0, 2], zє [ 0, 1 }
4
Solution I = 0 3 0 2 0 1 8xyz dz dy dx = 0 3 0 2 [ ( 4xyz 2 )│ z = 1 - ( 4xyz 2 )│ z = 0 ] dy dx = 0 3 0 2 4xy dy dx = 0 3 [ ( 2xy 2 )│y= 2 - ( 2x y 2 )│y= 0 ] dx = 0 3 8x dx = ( 4 x 2 )│x= 3 - ( 4 x 2 )│x= 0 = 36
5
Example 1 Evaluat the triple integral I of the given function f over the given region D. f(x,y,z) = z D = { (x,y,z) : x є [ 0, 2 ], 0 ≤ y ≤ 3x, 0 ≤ z ≤ xy }
6
Solution I = 0 2 0 3x 0 xy z dz dy dx = 0 2 0 3x [ (z 2 / 2)│z = xy - (z 2 / 2 )│z = 0 ] dy dx = (1/2) 0 2 0 3x x 2 y 2 dy dx = (1/6) 0 2 [ (x 2 y 3 )│y= 3x - (x 2 y 3 )│y= 0 ] dx = (1/6) 0 2 27 x 5 dy dx = ?
7
Triple Integral in Cylindrical Coordinates Let g( r, θ ) ≤ h( r,θ), K(θ) ≤ M(θ) ; α ≤ θ ≤ β Let D be the region ( in the cylindrical coordinates) defined by: D = { (r,θ,z) : α ≤θ ≤ β, K(θ) ≤ r ≤ M(θ), g(r,θ) ≤ z ≤ h(r,θ) } If f is continuous in x = cosθ, y = rsinθ and z Then the triple integral of f over D is : ∭ D f( rcosθ, rsinθ, z ) r dz dr dθ = α β K(θ) M(θ) g(r,θ) h(r,θ) f( rcosθ, rsinθ, z ) r dz dr dθ
8
Example Let f(x,y,z) = z and D = { (r,θ,z) : 0 ≤ θ ≤ 2π, 0 ≤ r ≤ 2, 0 ≤ z ≤ √ 16 - r 2 } Evaluate the triple integral of f over D
9
Solution I = 0 2π 0 2 0 √16 – r2 z r dz dr dθ = (1/2) 0 2π 0 2 [ (z 2 r )│z =√16 – r 2 - (z 2 r )│z = 0 ] dr dθ = (1/2) 0 2π 0 2 ( 16 - r 2 ) r dr dθ = (1/2) 0 2π 0 2 ( 16r - r 3 ) r dr dθ = (1/2) 0 2π [ ( 8 r 2 - (1/4)r 4 )│r = 2 - (8 r 2 - (1/4)r 4 )│r = 0 ] dθ = 0 2π 14 dθ = 28π
10
Triple Integral in Spherical Coordinates Let D be the region ( in the spherical coordinates) defined by: D = { ( ρ, θ, φ ) : α ≤θ ≤ β, γ ≤ φ ≤ ω, a ≤ ρ≤ b } If f is continuous in: x = ρ cosθ sinφ, y = ρ sinθ sinφ and z = ρ cosφ Then the triple integral of f over D is : ∭ D f( ρ cosθ sinφ, ρ sinθ sinφ, ρ cosφ ) ρ 2 sinφ dρ dφ dθ = α β γ ω a b f( ρ cosθ sinφ, ρ sinθ sinφ, ρ cosφ) ρ 2 sinφ dρ dφ dθ
11
Example Let f be a constant function with the value 5 on R 3 Find the triple integral of f on the following region D = { (ρ,θ,φ) : α ≤θ ≤ π, 0≤ φ ≤π/3, 0≤ ρ≤3 }
12
Solution I = 0 π 0 π/3 0 3 5 ρ 2 sinφ dρ dφ dθ = (5/3) 0 π 0 π/3 [ ( ρ 3 sinφ) │ρ = 3 - ( ρ 3 sinφ) │ρ = 0 ]dφ dθ = (5/3) 0 π 0 π/3 27 sinφ dφ dθ = 45 0 π 0 π/3 [ ( - cosφ )│ φ = π/3 - ( - cosφ )│ φ = 0 ] dθ = - 45 0 π [ (1/2) - 1 ] dθ = - 45 0 π ( - ½ ) dθ = (45/2 ) [ ( θ )│ θ = π - ( θ )│ θ = 0 = 45π/2
13
Example 2 Let f(x,y,z) = (x 2 + y 2 + z 2 ) ½ And let D be the region in the first octant in R 3 and bounded by the graphs: 4 – z = 9x 2 + y 2, y = 4x, z = 0 and y = 0. Graph the region D and set up the triple integral of the function f over D.
14
Solution x 2 = 1/9 [ 4 – y 2 ] → x = (⅓) √4 – y 2 z = 4 – 9x 2 – y 2 I = ∫∫∫D f(x,y,z) dz dx dy = 0 ∫ 8/5 y/4 ∫ (1/3(√4 – y2) 0 ∫ 4 - 9 x2 – y2 (x 2 + y 2 +z 2 ) 1/2 dz dy dx
Similar presentations
© 2024 SlidePlayer.com. Inc.
All rights reserved.