Chapter 7: Triple Integration
Section 7.3: Regions with Curved Boundaries
Example 7.3.8
If R is that portion of the first-octant lying under the cylinder x2+z2=1, but in front of the plane y=x, iterate the triple integral ∫∫∫R1 dv in the order dz dx dy.
Solution
Mathematical Solution
The region R is shown in Figures 7.3.8(a, b), one a view from the left; and the other, from the right. Figure 7.3.8(c) is a schematic of the region R. The intersection of the plane y=x and the cylinder x2+z2=1 is drawn in green in this figure. The projection of the green curve onto the xz-plane is drawn in black, while its projection onto the yz-plane is drawn in red.
The green curve of intersection, the simultaneous solution of the equations y=x and x2+z2=1, is given parametrically by equations of the form x=x,y=x,z=±1−x2. The projection of the green curve onto the yz-plane is found by eliminating x from the equations y=x and x2+z2=1. Since y=x, the projection y2+z2=1 easily follows. The black curve is the boundary of the cylinder, but it is also the projection of the green curve of intersection onto the xz-plane.
Figure 7.3.8(a) Region R
Figure 7.3.8(b) Region R
Figure 7.3.8(c) Region R
The region R has a physical meaning. The cylinder x2+z2=1 can be interpreted as a piece of quarter-round molding commonly used as wood-working trim and edging. When such trim is installed in a right-angled corner, at least one end must be mitered, or cut at a 45° bevel angle. (The saw blade would be in the plane y=x.) A piece of quarter-round with a square end, so beveled, wastes a chip in the shape of region R. The triple integral in this example computes the volume of such a cut-off scrap.
The relevant iteration of the triple integral is ∫01∫y1∫01−x21 dz dx dy=13. Figures 7.3.8(d - f) contain animations illustrating how the limits of integration might be deduced.
use plots,plottools in module() local a,s1,s2,s3,p1,p2,p3,p4,p5,p6,p7,k; s1 := [seq([k/30,0,sqrt(1-(k/30)^2)],k=0..30)]: s2 := [seq([k/30,k/30,sqrt(1-(k/30)^2)],k=0..30)]: s3 := [seq([0,k/30,sqrt(1-(k/30)^2)],k=0..30)]: p1 := spacecurve({s1,[1,y,0,y=0..1]}, color=black, thickness=3): p2:=spacecurve([x,x,0,x=0..1.2],color=green,thickness=2): p3:=spacecurve(s2,color=green,thickness=2): p4 := spacecurve({[x,1,0,x=0..1],s3}, color=red, thickness=2): p5 := spacecurve({[[1.2,0,0], [0,0,0], [0,1.2,0]], [[0,0,0], [0,0,1.2]]}, color=blue, thickness=2): p6:=display([p1,p2,p3,p4,p5], labels=[x,y,z], scaling = constrained, axes=none); a:=.1; p7:=animate(display,[parallelepiped([a,0,0],[0,a,0],[0,0,a],[.5,.4,b])],b=0..sqrt(3)/2,view=[0..1.2,0..1.2,0..1.2],labels=[x,y,z],orientation=[10,50,0],paraminfo=false,scaling=constrained,background=p6,tickmarks=[3,3,5]); print(p7); end module: end use:
Figure 7.3.8(d) dz
use plots,plottools in module() local a,s1,s2,s3,p1,p2,p3,p4,p5,p6,p7,k; s1 := [seq([k/30,0,sqrt(1-(k/30)^2)],k=0..30)]: s2 := [seq([k/30,k/30,sqrt(1-(k/30)^2)],k=0..30)]: s3 := [seq([0,k/30,sqrt(1-(k/30)^2)],k=0..30)]: p1 := spacecurve({s1,[1,y,0,y=0..1]}, color=black, thickness=3): p2:=spacecurve([x,x,0,x=0..1.2],color=green,thickness=2): p3:=spacecurve(s2,color=green,thickness=2): p4 := spacecurve({[x,1,0,x=0..1],s3}, color=red, thickness=2): p5 := spacecurve({[[1.2,0,0], [0,0,0], [0,1.2,0]], [[0,0,0], [0,0,1.2]]}, color=blue, thickness=2): p6:=display([p1,p2,p3,p4,p5], labels=[x,y,z], scaling = constrained, axes=none); a:=.1; p7:=animate(display,[parallelepiped([a,0,0],[0,a,0],[0,0,sqrt(1-b^2)],[b-.1,.4,0])],b=.55..1,view=[0..1.2,0..1.2,0..1.2],labels=[x,y,z],orientation=[10,50,0],paraminfo=false,scaling=constrained,background=p6,tickmarks=[3,3,5]); print(p7); end module: end use:
Figure 7.3.8(e) dx
use plots,plottools in module() local a,s1,s2,s3,p1,p2,p3,p4,p5,p6,p7,k; s1 := [seq([k/30,0,sqrt(1-(k/30)^2)],k=0..30)]: s2 := [seq([k/30,k/30,sqrt(1-(k/30)^2)],k=0..30)]: s3 := [seq([0,k/30,sqrt(1-(k/30)^2)],k=0..30)]: p1 := spacecurve({s1,[1,y,0,y=0..1]}, color=black, thickness=3): p2:=spacecurve([x,x,0,x=0..1.2],color=green,thickness=2): p3:=spacecurve(s2,color=green,thickness=2): p4 := spacecurve({[x,1,0,x=0..1],s3}, color=red, thickness=2): p5 := spacecurve({[[1.2,0,0], [0,0,0], [0,1.2,0]], [[0,0,0], [0,0,1.2]]}, color=blue, thickness=2): p6:=display([p1,p2,p3,p4,p5], labels=[x,y,z], scaling = constrained, axes=none); a:=.1; p7:=animate(plot3d,[[x,b,z],x=b..1,z=0..sqrt(1-x^2)],b=0..1,view=[0..1.2,0..1.2,0..1.2],labels=[x,y,z],orientation=[10,50,0],paraminfo=false,scaling=constrained,background=p6,tickmarks=[3,3,5]); print(p7); end module: end use:
Figure 7.3.8(f) dy
In Figure 7.3.8(d), the volume element first moved in the z-direction, from z=0 to the surface given in the form z=1−x2. This forms the "post" animated in Figure 7.3.8(e). The "post" next moves in the x-direction, from x=y to x=1. This forms the "slab" animated in Figure 7.3.8(f). The "slab" moves in the y-direction, from y=0 to y=1.
Once the inner integration takes place, and the equivalent of the "post" has been formed, the base of the "post" must traverse the projection of R onto the plane of the remaining two variables. Here, those two variables are y and x, so the region that the bottom of the post must traverse is the triangle formed by the x-axis, the line y=x, and the line x=1.
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∫01∫y1∫0−x2+11ⅆzⅆxⅆy=13
Table 7.3.8(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 7.3.8(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 7.3.8(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
∫01∫y1∫01−x21 ⅆz ⅆx ⅆy = 13
Table 7.3.8(b) Solution via iterated triple-integral template in the Calculus palette
Maple Solution - Coded
Top-level: Int and int commands
Int1,z=0..1−x2,x=y..1, y=0..1=int1,z=0..1−x2,x=y..1, y=0..1
∫01∫y1∫0−x2+11ⅆzⅆxⅆy=13
The MultiInt command in the Student MultivariateCalculus package
Install the Student MultivariateCalculus package.
withStudent:-MultivariateCalculus:
MultiInt1,z=0..1−x2,x=y..1, y=0..1,output=integral
∫01∫y1∫0−x2+11ⅆzⅆxⅆy
MultiInt1,z=0..1−x2,x=y..1, y=0..1 = 13
MultiInt1,z=0..1−x2,x=y..1, y=0..1,output=steps
13
The MultiInt command with a pre-defined domain option
MultiInt1,y,x,z=Region0..1,y..1,0..1−x2,output=integral
MultiInt1,y,x,z=Region0..1,y..1,0..1−x2 = 13
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