Chapter 7: Triple Integration
Section 7.3: Regions with Curved Boundaries
Example 7.3.4
If R is that portion of the first-octant lying under the cylinder x2+z=4 for y∈0,3, iterate the triple integral ∫∫∫R1 dv in the order dy dz dx.
Solution
Mathematical Solution
The required iteration is
∫02∫04−x2∫031 dy dz dx=16
The animations in Figures 7.3.4(b - d) indicate how the element of volume (a small parallelepiped) must move so as to sweep out the region R.
According to the animation in Figure 7.3.4(b), the element of volume must first move in the y-direction, from y=0 to y=3.
Figure 7.3.4(a) The region R
The result of this inner integration is to produce the "post" animated in Figure 7.3.4(c). This "post" must move in the z-direction, from z=0 to the surface z=4−x2 , resulting in the "slab" that is animated in Figure 7.3.4(d). This "slab" now must move in the x-direction, from x=0 to x=2.
use plots, plottools in module() local a,p1,p2; a:=.2; p1:=plot3d(4-x^2,x=0..2,y=0..3); p2:=animate(display,[parallelepiped([a,0,0],[0,a,0],[0,0,a],[1,b,1.5])],b=0..2.8,view=[0..2,0..3,0..4],labels=[x,y,z],orientation=[-130,75,0],paraminfo=false,scaling=constrained,background=p1,tickmarks=[3,3,5]); print(p2); end module: end use:
Figure 7.3.4(b) dy
use plots, plottools in module() local a,p1,p2; a:=.2; p1:=plot3d(4-x^2,x=0..2,y=0..3); p2:=animate(display,[parallelepiped([a,0,0],[0,3,0],[0,0,a],[1,0,b])],b=0..2.4,view=[0..2,0..3,0..4],labels=[x,y,z],orientation=[-130,75,0],paraminfo=false,scaling=constrained,background=p1,tickmarks=[3,3,5]); print(p2); end module: end use:
Figure 7.3.4(c) dz
use plots, plottools in module() local a,p1,p2; a:=.2; p1:=plot3d(4-x^2,x=0..2,y=0..3); p2:=animate(display,[parallelepiped([a,0,0],[0,3,0],[0,0,4-b^2],[b,0,0])],b=0..2,view=[0..2,0..3,0..4],labels=[x,y,z],orientation=[-130,75,0],paraminfo=false,scaling=constrained,background=p1,tickmarks=[3,3,5]); print(p2); end module: end use:
Figure 7.3.4(d) dx
Maple Solution - Interactive
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∫0−x2+4∫031ⅆyⅆzⅆx=16
Table 7.3.4(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 7.3.4(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 7.3.4(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
∫02∫04−x2∫031 ⅆy ⅆz ⅆx = 16
Table 7.3.4(b) Solution via iterated triple-integral template in the Calculus palette
Maple Solution - Coded
Top-level: Int and int commands
Int1,y=0..3,z=0..4−x2,x=0..2=int1,y=0..3,z=0..4−x2,x=0..2
∫02∫0−x2+4∫031ⅆyⅆzⅆx=16
The MultiInt command in the Student MultivariateCalculus package
Install the Student MultivariateCalculus package.
withStudent:-MultivariateCalculus:
MultiInt1,y=0..3,z=0..4−x2,x=0..2,output=integral
∫02∫0−x2+4∫031ⅆyⅆzⅆx
MultiInt1,y=0..3,z=0..4−x2,x=0..2 = 16
MultiInt1,y=0..3,z=0..4−x2,x=0..2,output=steps
16
The MultiInt command with a pre-defined domain option
MultiInt1,x,z,y=Region0..2,0..4−x2,0..3,output=integral
MultiInt1,x,z,y=Region0..2,0..4−x2,0..3 = 16
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