Chapter 7: Triple Integration
Section 7.4: Integration in Cylindrical Coordinates
Example 7.4.13
Use cylindrical coordinates to integrate the function f=1 over R, the region above the plane z=0, below the plane z=x, inside the cylinder whose cross section is x2+y2=1, and lying in the half-space x≥0.
Solution
Mathematical Solution
Figure 7.4.13(a) shows the region R. In cylindrical coordinates, the upper surface of R is
Zr,θ=r cosθ
while the lower surface is zr,θ= 0.
The lateral surface is a cylinder whose cross section is the circle r=1. However, the restriction x≥0 requires −π/2≤θ≤π/2.
The iteration in the order dz dr dθ is found to be
∫−π/2π/2∫01∫ 0r cosθr dz dr dθ=2/3
Figure 7.4.13(a) The region R
Maple Solution - Interactive
In cylindrical coordinates, the upper surface of R is Zr,θ=r cosθ, while the lower surface is zr,θ= 0. The lateral surface is the cylinder whose cross section is the circle r=1, but the restriction to x≥0 requires π/2≤θ≤π/2.
Table 7.4.13(a) provides a solution by a visualization task template. After selecting the order of iteration, the integrand and the fields for the limits of integration are given. The resulting value of the integral and a graph of the region of integration are generated by pressing the appropriate buttons.
Tools≻Tasks≻Browse:
Calculus - Multivariate≻Integration≻Visualizing Regions of Integration≻Cylindrical
Evaluate ∭RΨr,θ,z dv and Graph R
Volume Element dv
r dz dr dθ
r dz dθ dr
r dr dθ dz
r dr dz dθ
r dθ dr dz
r dθ dz dr
, where Ψ=
F=
G=
b=
f=
g=
a=
Table 7.4.13(a) Task template for iterating a triple integral in cylindrical coordinates
Constrained scaling has been applied to the graph via its Context Panel.
Table 7.4.13(b) provides a solution from first principles. Note that the Jacobian, r, must be inserted into the integrand.
Calculus palette: Iterated triple-integral template
Context Panel: Evaluate and Display Inline
∫−π/2π/2∫01∫ 0r cosθr ⅆz ⅆr ⅆθ = 23
Table 7.4.13(b) Solution from first principles
Table 7.4.13(c) provides a solution by a task template that uses the MultiInt command from the Student MultivariateCalculus package, and iterates in the order dz dr dθ. The command takes a coordinate option, and hence, inserts the appropriate Jacobian automatically.
Tools≻Tasks≻Browse: Calculus - Multivariate≻Integration≻Multiple Integration≻Cylindrical
Iterated Triple Integral in Cylindrical Coordinates
Integrand:
1
Region: z1r,θ≤z≤z2r,θ,r1θ≤r≤r2θ,a≤θ≤b
z1r,θ
0
z2r,θ
r cosθ
2r
r1θ
r2θ
a
−π/2
−12π
b
π/2
12π
Inert Integral: dz dr dθ
(Note automatic insertion of Jacobian.)
StudentMultivariateCalculusMultiInt,z=..,r=..,θ=..,coordinates=cylindricalr,θ,z,output=integral
∫−12π12π∫01∫02rrⅆzⅆrⅆθ
Value:
StudentMultivariateCalculusMultiInt,z=..,r=..,θ=..,coordinates=cylindricalr,θ,z
23π
Stepwise Evaluation:
StudentMultivariateCalculusMultiInt,z=..,r=..,θ=..,coordinates=cylindricalr,θ,z,output=steps
∫−π2π2∫01∫02rrⅆzⅆrⅆθ=∫−π2π2∫01rzz=0..2r|rzz=0..2rⅆrⅆθ=∫−π2π2∫012r2ⅆrⅆθ=∫−π2π22r33r=0..1|2r33r=0..1ⅆθ=∫−π2π223ⅆθ=2θ3θ=−π2..π2|2θ3θ=−π2..π2
Table 7.4.13(c) Solution by task template that implements the MultiInt command
Initialize
Tools≻Load Package: Student Multivariate Calculus
Loading Student:-MultivariateCalculus
Access the MultiInt command via the Context Panel
Type the integrand, 1.
Context Panel: Student Multivariate Calculus≻Integrate≻Iterated Fill in the fields of the two dialogs shown below.
Context Panel: Evaluate Integral
1→MultiInt∫−12π12π∫01∫0rcosθrⅆzⅆrⅆθ=23
Maple Solution - Coded
Install the Student MultivariateCalculus package.
withStudent:-MultivariateCalculus:
Top-level solution using the Int and int commands
Intr,z= 0..r cosθ,r=0..1,θ=−π2..π2=intr,z= 0..r cosθ,r=0..1,θ=−π2..π2
∫−12π12π∫01∫0rcosθrⅆzⅆrⅆθ=23
Solution via the MultiInt command from the Student MultivariateCalculus package
MultiInt1,z= 0..r cosθ,r=0..1,θ=−π2..π2,coordinates=cylindrical,output=integral=MultiInt1,z= 0..r cosθ,r=0..1,θ=−π2..π2,coordinates=cylindrical
MultiInt1,z= 0..r cosθ,r=0..1,θ=−π2..π2,coordinates=cylindrical,output=steps
23
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