Chapter 7: Triple Integration
Section 7.4: Integration in Cylindrical Coordinates
Example 7.4.10
Use cylindrical coordinates to integrate the function f=1 over R, that part of the interior of the sphere x2+y2+z2=4 that lies inside the cylinder whose cross section is r=2 sinθ.
Solution
Mathematical Solution
Figure 7.4.10(a) shows the region R. In cylindrical coordinates, the upper surface of R is
Zr,θ=4−r2
while the lower surface is zr,θ= −Zr,θ.
The lateral surface is the cylinder whose cross section is the circle r=2 sinθ.
The iteration in the order dz dr dθ is found to be
∫02 π∫02 sinθ∫ −ZZr dz dr dθ=323 π−43≐19.3
Figure 7.4.10(a) The region R
Maple Solution - Interactive
In cylindrical coordinates, the upper surface of R is given by
Context Panel: Assign Name
Z=4−r2→assign
while the lower surface is zr,θ= −Z. The lateral surface is a cylinder whose cross section is the circle r=2 sinθ.
Table 7.4.10(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 7.4.10(a) Task template for iterating a triple integral in cylindrical coordinates
Constrained scaling is applied to the graph via its Context Panel.
Table 7.4.10(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∫02 sinθ∫ −ZZr ⅆz ⅆr ⅆθ = 323π−1289
Table 7.4.10(b) Solution from first principles
Table 7.4.10(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
−Z
−−r2+4
z2r,θ
Z
−r2+4
r1θ
0
r2θ
2 sinθ
2sinθ
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∫02sinθ∫−−r2+4−r2+4rⅆzⅆrⅆθ
Value:
StudentMultivariateCalculusMultiInt,z=..,r=..,θ=..,coordinates=cylindricalr,θ,z
323π−1289
Stepwise Evaluation:
StudentMultivariateCalculusMultiInt,z=..,r=..,θ=..,coordinates=cylindricalr,θ,z,output=steps
∫02π∫02sinθ∫−−r2+4−r2+4rⅆzⅆrⅆθ=∫02π∫02sinθrzz=−−r2+4..−r2+4|rzz=−−r2+4..−r2+4ⅆrⅆθ=∫02π∫02sinθ2r−r2+4ⅆrⅆθ=∫02π2r−2r+2−r2+43r=0..2sinθ|2r−2r+2−r2+43r=0..2sinθⅆθ=∫02π163+16−sinθ2+1sinθ23−16−sinθ2+13ⅆθ=16θ3−16sinθ2−1sinθ39cosθ−sinθ2+1+16sinθ2−1sinθ3cosθ−sinθ2+1θ=0..2π|16θ3−16sinθ2−1sinθ39cosθ−sinθ2+1+16sinθ2−1sinθ3cosθ−sinθ2+1θ=0..2π
Table 7.4.10(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∫02sinθ∫−−r2+4−r2+4rⅆzⅆrⅆθ=323π−1289
Maple Solution - Coded
Install the Student MultivariateCalculus package.
withStudent:-MultivariateCalculus:
Define the upper surface Zr,θ.
Z≔4−r2:
Top-level solution using the Int and int commands
Intr,z= −Z..Z,r=0..2 sinθ,θ=0..2 π=intr,z= −Z..Z,r=0..2 sinθ,θ=0..2 π
∫02π∫02sinθ∫−−r2+4−r2+4rⅆzⅆrⅆθ=323π−1289
Solution via the MultiInt command from the Student MultivariateCalculus package
MultiInt1,z= −Z..Z,r=0..2 sinθ,θ=0..2 π,coordinates=cylindrical,output=integral=MultiInt1,z= −Z..Z,r=0..2 sinθ,θ=0..2 π,coordinates=cylindrical
MultiInt1,z= −Z..Z,r=0..2 sinθ,θ=0..2 π,coordinates=cylindrical,output=steps
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