Chapter 7: Triple Integration
Section 7.5: Integration in Spherical Coordinates
Example 7.6.8
Use spherical coordinates to integrate the function f=1 over R, R is the region bounded inside by the surface ρ=1+cosφ and outside by the sphere ρ=2.
(This region was graphed in Example 7.5.10.)
Solution
Mathematical Solution
The graph in Figure 7.6.8(a) was obtained in Example 7.5.10.
Iteration in the order dρ dφ dθ leads to
∫02 π∫0π∫1+cosφ2ρ2sinφ ⅆρ ⅆφ ⅆθ = 8 π
Figure 7.6.8(a) The region R
Maple Solution - Interactive
Table 7.6.8(a) provides a solution by a visualization task template.
Tools≻Tasks≻Browse:
Calculus - Multivariate≻Integration≻Visualizing Regions of Integration≻Spherical
Evaluate ∭RΨρ,φ,θ dv and Graph R
Volume Element dv=ρ2sinφ×
dρ dφ dθ
dρ dθ dφ
dφ dρ dθ
dφ dθ dρ
dθ dφ dρ
dθ dρ dφ
, where Ψ=
F=
G=
b=
f=
g=
a=
Table 7.6.8(a) Solution by visualization task template
In Table 7.6.8(a), transparency has been applied to the graph via its Context Panel.
Table 7.6.8(b) provides a solution by a task template that embodies the MultiInt command from the Student MultivariateCalculus package. It iterates in just the order dρ dφ dθ, although the command itself will iterate in any of the six possible orders.
Tools≻Tasks≻Browse: Calculus - Multivariate≻Integration≻Multiple Integration≻Spherical
Iterated Triple Integral in Spherical Coordinates
(φ = colatitude, measured down from z-axis)
Integrand:
1
Region: ρ1φ,θ≤ρ≤ρ2φ,θ,φ1θ≤φ≤φ2θ,a≤θ≤b
ρ1φ,θ
1+cosφ
1+cos⁡φ
ρ2φ,θ
2
φ1θ
0
φ2θ
π
a
b
2 π
2⁢π
Inert Integral: dρ dφ dθ
StudentMultivariateCalculusMultiInt,ρ=..,φ=..,θ=..,coordinates=sphericalρ,φ,θ,output=integral
∫02⁢π∫0π∫1+cos⁡φ2ρ2⁢sin⁡φⅆρⅆφⅆθ
Value:
StudentMultivariateCalculusMultiInt,ρ=..,φ=..,θ=.., coordinates=sphericalρ,φ,θ
8⁢π
Stepwise Evaluation:
StudentMultivariateCalculusMultiInt,ρ=.., φ=..,θ=..,coordinates=sphericalρ,φ,θ,output=steps
∫02⁢π∫0π∫1+cos⁡φ2ρ2⁢sin⁡φⅆρⅆφⅆθ=∫02⁢π∫0πρ3⁢sin⁡φ3ρ=1+cos⁡φ..2|ρ3⁢sin⁡φ3ρ=1+cos⁡φ..2ⅆφⅆθ=∫02⁢π∫0πsin⁡φ⁢8−1+cos⁡φ33ⅆφⅆθ=∫02⁢π−8⁢cos⁡φ3+1+cos⁡φ412φ=0..π|−8⁢cos⁡φ3+1+cos⁡φ412φ=0..πⅆθ=∫02⁢π4ⅆθ=4⁢θθ=0..2⁢π|4⁢θθ=0..2⁢π
Table 7.6.8(b) Solution by task template that embodies 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∫02⁢π∫0π∫1+cos⁡φ2ρ2⁢sin⁡φⅆρⅆφⅆθ=8⁢π
Table 7.6.8(c) contains a solution from first principles, with the iterated integral being set via the iterated triple-integral template in the Calculus palette.
Calculus palette: Iterated triple-integral template (In the integrand, be sure to include the Jacobian for spherical coordinates.)
Context Panel: Evaluate and Display Inline
∫02 π∫0π∫1+cosφ2ρ2sinφ ⅆρ ⅆφ ⅆθ = 8⁢π
Table 7.6.8(c) Solution from first principles
Maple Solution - Coded
Install the Student MultivariateCalculus package.
withStudent:-MultivariateCalculus:
Apply the MultiInt command from the Student MultivariateCalculus package
MultiInt1,ρ=1+cosφ..2,φ=0..π,θ=0..2 π,coordinates=sphericalρ,φ,θ,output=integral
MultiInt1,ρ=1+cosφ..2,φ=0..π,θ=0..2 π,coordinates=sphericalρ,φ,θ
MultiInt1,ρ=1+cosφ..2,φ=0..π,θ=0..2 π,coordinates=sphericalρ,φ,θ,output=steps
Solution via the top-level Int and int commands
Intρ2sinφ,ρ=1+cosφ..2,φ=0..π,θ=0..2 π=intρ2sinφ,ρ=1+cosφ..2,φ=0..π,θ=0..2 π
∫02⁢π∫0π∫1+cos⁡φ2ρ2⁢sin⁡φⅆρⅆφⅆθ=8⁢π
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