Chapter 6: Applications of Double Integration
Section 6.2: Volume
Example 6.2.2
If F=2 x2+3 y2 and R is the finite region bounded by the graphs of x=y2 and y=3−2 x, calculate the volume of the region bounded above by the surface z=Fx,y and below by the plane z=0.
See Example 6.1.2.
Solution
Mathematical Solution
use plots in module() local p1,p2,p3; p1:=shadebetween(-sqrt(x),sqrt(x),x=0..1,color=red,changefill=[color=blue,transparency=.7],thickness=3): p2:=shadebetween(-sqrt(x),3-2*x,x=1..9/4,color=black,changefill=[color=green,transparency=.7],thickness=3): p3:=display(p1,p2,labels=[x,y],tickmarks=[3,[-1.5,-1,0,1]]); print(p3); end module: end use:
Figure 6.2.2(a) The region R
Figure 6.2.2(b) The volume to be computed
The region whose volume is to be computed is shown in Figure 6.2.2(b). The simplest iteration of the double integral that gives the required volume takes the integrand as F and uses the order dx dy:
∫−3/21∫y23−y/2F dx dy = 168751792
If the order of integration is taken as dx dy, then, according to Figure 6.2.2(a), the iterated integral is more difficult:
∫01∫−xxF dy dx+∫19/4∫−x3−2 xF dy dx = 168751792
Maple Solution - Interactive
Initialize
Tools≻Load Package: Student Multivariate Calculus
Loading Student:-MultivariateCalculus
Context Panel: Assign Name
F=2 x2+3 y2→assign
XL=y2→assign
XR=3−y/2→assign
Obtain the intersections of the curves bounding R
Write a sequence of the two equations defining the bounding curves.
Context Panel: Solve≻Solve
x=y2,y=3−2 x→solvex=1,y=1,x=94,y=−32
Access the MultiInt command via the Context Panel
Write F, the name of the integrand. Context Panel: Evaluate and Display Inline
Context Panel: Student Multivariate Calculus≻Integrate≻Iterated Fill in both panes (see Figures 5.3.(1, 2)) and select "integral" for the Output
Context Panel: Evaluate Integral
F = 2x2+3y2→MultiInt∫−321∫y232−12y2x2+3y2ⅆxⅆy=168751792
Iterate in the order dx dy via the template in the Calculus palette
Calculus palette: Iterated double-integral template
Context Panel: Evaluate and Display Inline
∫−3/21∫y23−y/2F ⅆx ⅆy = 168751792
Iterate in the order dy dx via the template in the Calculus palette
∫01∫−xxF ⅆy ⅆx+∫19/4∫−x3−2 xF ⅆy ⅆx = 168751792
The task template in Table 6.2.2(a) iterates in the order dx dy. The right-pointing arrow in the left-hand graph indicates that the inner (first) integration is in the x-direction. The right-hand image contains a graph of the volume computed by the integral. Since the height of the region in the graph is the constant 1, the volume and the area are the same number.
Tools≻Tasks≻Browse:
Calculus - Multivariate≻Integration≻Visualizing Regions of Integration≻Cartesian 2-D
Evaluate ∬RΨx,y dA and Graph R
Area Element dA
Select dAdy dxdx dy
, Ψ=
Value of Integral
G=
b=
g=
a=
Bounding Curves
"Volume"
Table 6.2.2(a) Iteration in the order dx dy via visualization task-template
Maple Solution - Coded
Install the Student MultivariateCalculus package.
withStudent:-MultivariateCalculus:
Define the integrand F.
F≔2 x2+3 y2:
Top-level, using the Int and int commands
IntF,x=y2..3−y/2,y=−3/2..1=intF,x=y2..3−y/2,y=−3/2..1
∫−321∫y232−12y2x2+3y2ⅆxⅆy=168751792
Use the MultiInt command from the Student MultivariateCalculus package
MultiIntF,x=y2..3−y/2,y=−3/2..1 = 168751792
MultiIntF,x=y2..3−y/2,y=−3/2..1,output=integral
∫−321∫y232−12y2x2+3y2ⅆxⅆy
MultiIntF,x=y2..3−y/2,y=−3/2..1,output=steps
168751792
Use the MultiInt command with a pre-defined domain option
MultiIntF,y,x=Region−3/2..1,y2..3−y/2 = 168751792
MultiIntF,y,x=Region−3/2..1,y2..3−y/2,output=integral
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