Chapter 8: Applications of Triple Integration
Section 8.1: Volume
Example 8.1.8
Use an iterated triple integral to obtain the volume of R, the region common to the two cylinders x2+y2=1 and x2+z2=1.
Solution
Mathematical Solution
Figure 8.1.8(a) shows the two intersecting cylinders; Figure 8.1.8(b), the actual region R; and Figure 8.1.8(c), a cut-away view in the first octant.
use plottools, plots in EX818:=module() local p1,p2; export p3; p1:=cylinder([0,0,-2],1,4); p2:=rotate(p1,0,Pi/2,0); p3:=display(p1,p2,scaling=constrained,labels=[x,y,z],axes=frame,tickmarks=[5,2,5],lightmodel=light4,orientation=[-75,65,0]); print(p3); end module: end use:
Figure 8.1.8(a) Cylinders
Figure 8.1.8(b) Region R
Figure 8.1.8(c) Cut-away
The volume of R can be found by iterating in the order dz dy dx, with the horizontal cylinder x2+z2=1 solved for z=±1−x2 and the vertical cylinder x2+y2=1 solved for y=±1−x2. As a result, the requisite volume is then
∫−11∫−1−x21−x2∫−1−x21−x21 dz dy dx = 163
Maple Solution - Interactive
Table 8.1.8(a) provides, via a visualization task template, a solution in Cartesian coordinates.
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 8.1.8(a) Solution in Cartesian coordinates via a visualization task template
Table 8.1.8(b) provides, via a visualization task template, a solution in cylindrical coordinates.
Tools≻Tasks≻Browse: Calculus - Multivariate≻Integration≻Visualizing Regions of Integration≻Cylindrical
Evaluate ∭RΨr,θ,z dv and Graph R
r dz dr dθ
r dz dθ dr
r dr dθ dz
r dr dz dθ
r dθ dr dz
r dθ dz dr
Table 8.1.8(b) Solution in cylindrical coordinates via a visualization task template
Table 8.1.8(c) provides, via a task template that implements the MultiInt command from the Student MultivariateCalculus package, a solution in Cartesian coordinates.
Tools≻Tasks≻Browse: 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
−1−x2
−−x2+1
z2x,y
1−x2
−x2+1
y1x
y2x
a
−1
b
Inert Integral: dz dy dx
StudentMultivariateCalculusMultiInt,z=..,y=..,x=..,output=integral
∫−11∫−−x2+1−x2+1∫−−x2+1−x2+11ⅆzⅆyⅆx
Value:
StudentMultivariateCalculusMultiInt,z=..,y=..,x=..
163
Stepwise Evaluation:
StudentMultivariateCalculusMultiInt,z=..,y=..,x=..,output=steps
∫−11∫−−x2+1−x2+1∫−−x2+1−x2+11ⅆzⅆyⅆx=∫−11∫−−x2+1−x2+1zz=−−x2+1..−x2+1|zz=−−x2+1..−x2+1ⅆyⅆx=∫−11∫−−x2+1−x2+12−x2+1ⅆyⅆx=∫−112−x2+1yy=−−x2+1..−x2+1|2−x2+1yy=−−x2+1..−x2+1ⅆx=∫−11−4x2+4ⅆx=−43x3+4xx=−1..1|−43x3+4xx=−1..1
Table 8.1.8(c) Task template implementation of MultiInt solution in Cartesian coordinates
Initialize
Tools≻Load Package: Student Multivariate Calculus
Loading Student:-MultivariateCalculus
Access the MultiInt command via the Context Panel
c=1−x2→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∫−11∫−−x2+1−x2+1∫−−x2+1−x2+11ⅆzⅆyⅆx=163
Table 8.1.8(d) provides, via a task template that implements the MultiInt command from the Student MultivariateCalculus package, a solution in cylindrical coordinates.
Tools≻Tasks≻Browse: Calculus - Multivariate≻Integration≻Multiple Integration≻Cylindrical
Iterated Triple Integral in Cylindrical Coordinates
Region: z1r,θ≤z≤z2r,θ,r1θ≤r≤r2θ,a≤θ≤b
z1r,θ
−1−r2cos2θ
−1−r2cosθ2
z2r,θ
1−r2cos2θ
1−r2cosθ2
r1θ
0
r2θ
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∫−1−r2cosθ21−r2cosθ2rⅆzⅆrⅆθ
StudentMultivariateCalculusMultiInt,z=..,r=..,θ=..,coordinates=cylindricalr,θ,z
StudentMultivariateCalculusMultiInt,z=..,r=..,θ=..,coordinates=cylindricalr,θ,z,output=steps
∫02π∫01∫−1−r2cosθ21−r2cosθ2rⅆzⅆrⅆθ=∫02π∫01rzz=−1−r2cosθ2..1−r2cosθ2|rzz=−1−r2cosθ2..1−r2cosθ2ⅆrⅆθ=∫02π∫012r1−r2cosθ2ⅆrⅆθ=∫02π−21−r2cosθ2323cosθ2r=0..1|−21−r2cosθ2323cosθ2r=0..1ⅆθ=∫02π−21−cosθ232−13cosθ2ⅆθ=2sinθ3cosθ+2cosθ2−1cosθ2+13cosθsinθ1−cosθ2θ=0..2π|2sinθ3cosθ+2cosθ2−1cosθ2+13cosθsinθ1−cosθ2θ=0..2π
Table 8.1.8(d) Task template implementation of MultiInt solution in cylindrical coordinates
C=1−r2cos2θ→assign
1→MultiInt∫02π∫01∫−1−r2cosθ21−r2cosθ2rⅆzⅆrⅆθ=163
Table 8.1.8(e) provides solutions from first principles: on the left, a solution in Cartesian coordinates; on the right, in cylindrical coordinates.
∫−11∫−1−x21−x2∫−1−x21−x21 ⅆz ⅆy ⅆx = 163
∫02 π∫01∫−1−r2cos2θ1−r2cos2θr ⅆz ⅆr ⅆθ = 163
Table 8.1.8(e) From first principles, solutions in both Cartesian and cylindrical coordinates
Maple Solution - Coded
Table 8.1.8(f) provides, from first principles using the top-level Int and int commands, solutions in Cartesian and cylindrical coordinates.
Int1,z=−1−x2..1−x2,y=−1−x2..1−x2,x=−1..1=int1,z=−1−x2..1−x2,y=−1−x2..1−x2,x=−1..1
∫−11∫−−x2+1−x2+1∫−−x2+1−x2+11ⅆzⅆyⅆx=163
Intr,z=−1−r2cos2θ..1−r2cos2θ,r=0..1,θ=0..2 π=intr,z=−1−r2cos2θ..1−r2cos2θ,r=0..1,θ=0..2 π
∫02π∫01∫−1−r2cosθ21−r2cosθ2rⅆzⅆrⅆθ=163
Table 8.1.8(f) From first principles, solutions in Cartesian and cylindrical coordinates.
Table 8.1.8(g) demonstrates the syntax applying the MultiInt command in both Cartesian and cylindrical coordinates.
Install the Student
