Chapter 7: Triple Integration
Section 7.5: Integration in Spherical Coordinates
Example 7.6.1
Use spherical coordinates to integrate the function f=1 over R, the region above the plane z=3/4 in a unit sphere with center at the origin.
(This region was graphed in Example 7.5.3.)
Solution
Mathematical Solution
The graph in Figure 7.6.1(a) was obtained in Example 7.5.3.
Iteration in the order dρ dφ dθ leads to
∫02 π∫0cos−134∫34 cosφ1ρ2sinφ ⅆρ ⅆφ ⅆθ = 11192⁢π
Figure 7.6.1(a) The region R
The lower limit of the innermost integral is obtained by solving z=ρ cosφ=3/4 for ρ=3/4 cosφ.
The upper limit of the middle integral is obtained by solving z=ρ cosφ=3/4 and ρ=1 for φ=arccos3/4.
Maple Solution - Interactive
Table 7.6.1(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 7.6.1(a) Solution by visualization task template
In Table 7.6.1(a), constrained scaling has been applied to the graph via its Context Panel.
Table 7.6.1(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
34 cosφ
34⁢cos⁡φ
ρ2φ,θ
φ1θ
0
φ2θ
arccos3/4
arccos⁡34
a
b
2 π
2⁢π
Inert Integral: dρ dφ dθ
StudentMultivariateCalculusMultiInt,ρ=..,φ=..,θ=..,coordinates=sphericalρ,φ,θ,output=integral
∫02⁢π∫0arccos⁡34∫34⁢cos⁡φ1ρ2⁢sin⁡φⅆρⅆφⅆθ
Value:
StudentMultivariateCalculusMultiInt,ρ=..,φ=..,θ=.., coordinates=sphericalρ,φ,θ
11192⁢π
Stepwise Evaluation:
StudentMultivariateCalculusMultiInt,ρ=.., φ=..,θ=..,coordinates=sphericalρ,φ,θ,output=steps
∫02⁢π∫0arccos⁡34∫34⁢cos⁡φ1ρ2⁢sin⁡φⅆρⅆφⅆθ=∫02⁢π∫0arccos⁡34ρ3⁢sin⁡φ3ρ=34⁢cos⁡φ..1|ρ3⁢sin⁡φ3ρ=34⁢cos⁡φ..1ⅆφⅆθ=∫02⁢π∫0arccos⁡34sin⁡φ⁢1−2764⁢cos⁡φ33ⅆφⅆθ=∫02⁢π−cos⁡φ3−9128⁢cos⁡φ2φ=0..arccos⁡34|−cos⁡φ3−9128⁢cos⁡φ2φ=0..arccos⁡34ⅆθ=∫02⁢π11384ⅆθ=11⁢θ384θ=0..2⁢π|11⁢θ384θ=0..2⁢π
Table 7.6.1(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∫0arccos⁡34∫34⁢cos⁡φ1ρ2⁢sin⁡φⅆρⅆφⅆθ=11192⁢π
Table 7.6.1(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∫0cos−134∫34 cosφ1ρ2sinφ ⅆρ ⅆφ ⅆθ = 11192⁢π
Table 7.6.1(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=34 cosφ..1,φ=0..cos−134,θ=0..2 π,coordinates=sphericalρ,φ,θ,output=integral
MultiInt1,ρ=34 cosφ..1,φ=0..cos−134,θ=0..2 π,coordinates=sphericalρ,φ,θ
MultiInt1,ρ=34 cosφ..1,φ=0..cos−134,θ=0..2 π,coordinates=sphericalρ,φ,θ,output=steps
Solution via the top-level Int and int commands
Intρ2sinφ,ρ=34 cosφ..1,φ=0..cos−134,θ=0..2 π=intρ2sinφ,ρ=34 cosφ..1,φ=0..cos−134,θ=0..2 π
∫02⁢π∫0arccos⁡34∫34⁢cos⁡φ1ρ2⁢sin⁡φⅆρⅆφⅆθ=11192⁢π
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