Chapter 6: Applications of Double Integration
Section 6.2: Volume
Example 6.2.6
If F=x−32 y−62/30 and R is the interior of the triangle whose vertices are 1,3,7,4,5,9, calculate the volume of the region bounded above by the surface z=Fx,y and below by the plane z=0.
See Example 6.1.6.
Solution
Mathematical Solution
The region whose volume is to be computed is shown in Figure 6.2.6(a). The equations for the edges of R are
YAB=x+17/6
YBC=43−5 x/2
YAC=3x+1/2
The chosen order of iteration results in the sum
∫15∫YABYACF dy dx+∫57∫YABYBCF dy dx = 2272675
use plots in module() local YAB,YBC,YAC,p1,p2,p3,F; F:=(x-3)^2*(y-6)^2/30; YAB:=(x+17)/6; YBC:=(43-5*x)/2; YAC:=3*(x+1)/2; p1:=shadebetween(0,F,x=1..5,y=YAB..YAC,changefill=[transparency=0]); p2:=shadebetween(0,F,x=5..7,y=YAB..YBC,changefill=[transparency=0]); p3:=display(p1,p2,axes=frame,labels=[x,y,z],tickmarks=[3,3,3],lightmodel=none,orientation=[150,70,0]); print(p3); end module: end use:
Figure 6.1.6(a) The region R
Either order of iteration requires two double integrals.
Maple Solution - Interactive
The equations of the edges of the triangle defining the region R are found in Table 6.2.6(a).
Initialize
Context Panel: Assign Name
F=x−32 y−62/30→assign
Tools≻Load Package: Student Precalculus
Loading Student:-Precalculus
Obtain equations of the edges
Write a sequence of two lists, each list representing an endpoint of an edge.
Context Panel: Student Precalculus≻Lines And Segments≻Line≻Equation
Context Panel: Right-hand Side
Context Panel: Assign to a Name≻Y[AB] (or Y[BC] or Y[AC], as appropriate)
1,3,7,4→equation of liney=16⁢x+176→right hand side16⁢x+176→assign to a nameYAB
7,4,5,9→equation of liney=−52⁢x+432→right hand side−52⁢x+432→assign to a nameYBC
1,3,5,9→equation of liney=32⁢x+32→right hand side32⁢x+32→assign to a nameYAC
Table 6.2.6(a) Obtaining the equations of the edges of the triangle defining region R
An iteration by means of which the volume over R can be found is given in Table 6.2.6(b).
Iterate in the order dy dx via the template in the Calculus palette
Calculus palette: Iterated double-integral template
Context Panel: Evaluate and Display Inline
∫15∫YABYACF ⅆy ⅆx+∫57∫YABYBCF ⅆy ⅆx = 2272675
Table 6.2.6(b) Iterated double-integrals for finding the area of region R
Table 6.2.6(c) contains a task-template solution.
Tools≻Tasks≻Browse:
Calculus - Vector≻Integration≻Multiple Integration≻2-D≻Over a Triangle
Integrate fx,y over a Triangle
fx,y=
Vertices
x1=
x2=
x3=
y1=
y2=
y3=
dy dx
Table 6.2.6(c) Task-template for integrating over a triangle
The astute observer will note that each iterated integral produced by the task template is preceded by a minus sign. This is because Maple does not rearrange the limits of integration so that the upper edge in the triangle becomes the upper limit of integration. However, the equations for the edges are the same as listed in Table 6.2.6(a), the advantage being that the only input to the task template is the set of three vertices. Maple determines the equations of the edges and sets up the iterated integral(s).
Maple Solution - Coded
Install the Student MultivariateCalculus package.
withStudent:-MultivariateCalculus:
Define the integrand F.
F≔x−32 y−62/30:
Obtain the equations of the edges of the triangle forming region R
Use the Line and GetRepresentation commands.
Use isolate to solve for y and rhs to select the right-hand side.
YAB≔rhsisolateGetRepresentationLine1,3,7,4,form=equation,y:YBC≔rhsisolateGetRepresentationLine7,4,5,9,form=equation,y:YAC≔rhsisolateGetRepresentationLine1,3,5,9,form=equation,y:
YAB = 176+16⁢x
YBC = −52⁢x+432
YAC = 32⁢x+32
Top-level, using the Int and int commands
IntF,y=YAB..YAC,x=1..5+IntF,y=YAB..YBC,x=5..7=intF,y=YAB..YAC,x=1..5+intF,y=YAB..YBC,x=5..7
∫15∫16⁢x+17632⁢x+32130⁢x−32⁢y−62ⅆyⅆx+∫57∫16⁢x+176−52⁢x+432130⁢x−32⁢y−62ⅆyⅆx=2272675
Use the MultiInt command from the Student MultivariateCalculus package
MultiIntF,y=YAB..YAC,x=1..5+MultiIntF,y=YAB..YBC,x=5..7 = 2272675
MultiIntF,y=YAB..YAC,x=1..5,output=integral+ MultiIntF,y=YAB..YBC,x=5..7,output=integral
∫15∫176+16⁢x32⁢x+32130⁢x−32⁢y−62ⅆyⅆx+∫57∫176+16⁢x−52⁢x+432130⁢x−32⁢y−62ⅆyⅆx
Use the MultiInt command with a pre-defined domain option
MultiIntF,x,y=Triangle1,3,7,4,5,9 = 2272675
MultiIntF,x,y=Triangle1,3,7,4,5,9,output=integral
∫15∫32⁢x+32176+16⁢x−130⁢x−32⁢y−62ⅆyⅆx+∫57∫−52⁢x+432176+16⁢x−130⁢x−32⁢y−62ⅆyⅆx
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