Chapter 7: Triple Integration
Section 7.4: Integration in Cylindrical Coordinates
Example 7.4.1
Use cylindrical coordinates to integrate the function f=1 over R, the region bounded above by the upper hemisphere in a sphere of radius 1, and below by the plane through the "equator".
Solution
Mathematical Solution
Figure 7.4.1(b) shows the hemisphere enclosing the region R. The schematic in Figure 7.4.1(a) represents a vertical cross section taken through the z-axis. The connection between the height z and the radius r is established via the right triangle whose hypotenuse (red) is 1, whose horizontal leg is r, and whose vertical leg (green) is z=zr=1−r2.
Figure 7.4.1(a) Vertical cross-section
Figure 7.4.1(b) Upper hemisphere
From Figure 7.4.1(a), the iteration in the order dz dr dθ is found to be
∫02 π∫01∫01−r2r dz dr dθ=23π
Maple Solution - Interactive
Table 7.4.1(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 7.4.1(a) Task template for iterating a triple integral in cylindrical coordinates
Constrained scaling is applied to the graph via its Context Panel.
Table 7.4.1(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
∫02 π∫01∫01−r2r ⅆz ⅆr ⅆθ = 23π
Table 7.4.1(b) Solution from first principles
Table 7.4.1(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
1−r2
−r2+1
r1θ
r2θ
a
b
2 π
2π
Inert Integral: dz dr dθ
(Note automatic insertion of Jacobian.)
StudentMultivariateCalculusMultiInt,z=..,r=..,θ=..,coordinates=cylindricalr,θ,z,output=integral
∫02π∫01∫0−r2+1rⅆzⅆrⅆθ
Value:
StudentMultivariateCalculusMultiInt,z=..,r=..,θ=..,coordinates=cylindricalr,θ,z
23π
Stepwise Evaluation:
StudentMultivariateCalculusMultiInt,z=..,r=..,θ=..,coordinates=cylindricalr,θ,z,output=steps
∫02π∫01∫0−r2+1rⅆzⅆrⅆθ=∫02π∫01rzz=0..−r2+1|rzz=0..−r2+1ⅆrⅆθ=∫02π∫01r−r2+1ⅆrⅆθ=∫02πr−1r+1−r2+13r=0..1|r−1r+1−r2+13r=0..1ⅆθ=∫02π13ⅆθ=θ3θ=0..2π|θ3θ=0..2π
Table 7.4.1(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∫02π∫01∫0−r2+1rⅆzⅆrⅆθ=23π
Maple Solution - Coded
Top-level solution using the Int and int commands
Intr,z=0..1−r2,r=0..1,θ=0..2 π=intr,z=0..1−r2,r=0..1,θ=0..2 π
∫02π∫01∫0−r2+1rⅆzⅆrⅆθ=23π
Solution via the MultiInt command from the Student MultivariateCalculus package
Install the Student MultivariateCalculus package.
withStudent:-MultivariateCalculus:
MultiInt1,z=0..1−r2,r=0..1,θ=0..2 π,coordinates=cylindrical,output=integral
MultiInt1,z=0..1−r2,r=0..1,θ=0..2 π,coordinates=cylindrical = 23π
MultiInt1,z=0..1−r2,r=0..1,θ=0..2 π,coordinates=cylindrical,output=steps
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