Chapter 7: Triple Integration
Section 7.3: Regions with Curved Boundaries
Example 7.3.19
Implement an appropriate iteration of the triple integral ∫∫∫R1 dv, where R is the region bounded above by z=6−x2/2−2 y2 and below by z=x2+4 y2.
Solution
Mathematical Solution
Figure 7.3.19(a) contains an image of R, the region of integration.
Iterating in the order dz dy dx leads to
∫−22∫−4−x224−x22∫x2+4 y26−x22−2 y21 ⅆz ⅆy ⅆx = 6π
Figure 7.3.19(a) The region R
Maple Solution - Interactive
Initialize
Tools≻Load Package: Student Multivariate Calculus
Loading Student:-MultivariateCalculus
Access the MultiInt command via the Context Panel
a=x2+4 y2→assign
b=6−x2/2−2 y2→assign
c=4−x2/2→assign
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∫−22∫−12−x2+412−x2+4∫x2+4y26−12x2−2y21ⅆzⅆyⅆx=6π
Table 7.3.19(a) contains a solution provided by a visualization task template. After the order of iteration is selected, fill in the fields that correspond to the limits of integration. If the graph of the region swept by these limits is correct, then the integral is correctly formulated and evaluated.
Tools≻Tasks≻Browse:
Calculus - Multivariate≻Integration≻Visualizing Regions of Integration≻Cartesian 3-D
Evaluate ∭RΨx,y,z dv and Graph R
Volume Element dv
Select dvdz dy dxdz dx dydx dy dzdx dz dydy dx dzdy dz dx
, where Ψ=
F=
G=
b=
f=
g=
a=
Table 7.3.19(a) Solution by visualization task template
This template employs the MultiInt command from the Student MultivariateCalculus package, but the graphic are coded from first principles.
Table 7.3.19(b) contains a solution implemented with the iterated triple-integral template found in the Calculus palette.
Calculus palette: Iterated triple-integral template
Context Panel: Evaluate and Display Inline
Table 7.3.19(b) Solution via iterated triple-integral template in the Calculus palette
Table 7.3.19(c) contains a solution provided by a task template that employs the modified int command in the Student VectorCalculus package. Unfortunately, this template iterates in the single order dz dy dx.
Calculus - Vector≻Integration≻Multiple Integration≻3-D≻Over a General 3-D Region
Integrate fx,y,z over a General Region
∫x1x2∫y1xy2x∫z1x,yz2x,yf ⅆz ⅆy ⅆx fx,y,z=
z2=
y2=
x2=
z1=
y1=
x1=
dz dy dx
Table 7.3.19(c) Solution by a VectorCalculus task template
Table 7.3.19(d) contains a solution by a task template that embodies the MultiInt command from the Student MultivariateCalculus package.
Calculus - Multivariate≻Integration≻Multiple Integration≻Cartesian 3-D
Iterated Triple Integrals in Cartesian Coordinates
Integrand:
1
Region: z1x,y≤z≤z2x,y,y1x≤y≤y2x,a≤x≤b
z1x,y
x2+4 y2
x2+4y2
z2x,y
6− x2/2−2 y2
6−12x2−2y2
y1x
−4−x2/2
−12−x2+4
y2x
4−x2/2
12−x2+4
a
−2
b
2
Inert Integral: dz dy dx
StudentMultivariateCalculusMultiInt,z=..,y=..,x=..,output=integral
∫−22∫−12−x2+412−x2+4∫x2+4y26−12x2−2y21ⅆzⅆyⅆx
Value:
StudentMultivariateCalculusMultiInt,z=..,y=..,x=..
6π
Stepwise Evaluation:
StudentMultivariateCalculusMultiInt,z=..,y=..,x=..,output=steps
∫−22∫−−x2+42−x2+42∫x2+4y26−x22−2y21ⅆzⅆyⅆx=∫−22∫−−x2+42−x2+42zz=x2+4y2..6−x22−2y2|zz=x2+4y2..6−x22−2y2ⅆyⅆx=∫−22∫−−x2+42−x2+426−3x22−6y2ⅆyⅆx=∫−226y−32x2y−2y3y=−−x2+42..−x2+42|6y−32x2y−2y3y=−−x2+42..−x2+42ⅆx=∫−226−x2+4−3x2−x2+42−−x2+4322ⅆx=x−x2+4324+3x−x2+42+6arcsinx2x=−2..2|x−x2+4324+3x−x2+42+6arcsinx2x=−2..2
Table 7.3.19(d) Task template with the MultiInt command, Student MultivariateCalculus package
Maple Solution - Coded
Top-level: Int and int commands
Int1,z=x2+4 y2..6−x2/2−2 y2,y=−4−x2/2..4−x2/2,x=−2..2=int1,z=x2+4 y2..6−x2/2−2 y2,y=−4−x2/2..4−x2/2,x=−2..2
∫−22∫−12−x2+412−x2+4∫x2+4y26−12x2−2y21ⅆzⅆyⅆx=6π
The MultiInt command in the Student MultivariateCalculus package
Install the Student MultivariateCalculus package.
withStudent:-MultivariateCalculus:
MultiInt1,z=x2+4 y2..6−x2/2−2 y2,y=−4−x2/2..4−x2/2,x=−2..2,output=integral
MultiInt1,z=x2+4 y2..6−x2/2−2 y2,y=−4−x2/2..4−x2/2,x=−2..2 = 6π
MultiInt1,z=x2+4 y2..6−x2/2−2 y2,y=−4−x2/2..4−x2/2,x=−2..2,output=steps
MultiInt1,x,y,z=Region−2..2,−4−x22..4−x22,x2+4 y2..6−x22−2 y2,output=integral
MultiInt1,x,y,z=Region−2..2,−4−x22..4−x22,x2+4 y2..6−x22−2 y2 = 6π
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