Chapter 6: Applications of Double Integration
Section 6.2: Volume
Example 6.2.1
If F=x y and R is the finite region bounded by the graph of y=x 1−x and the x-axis, calculate the volume of the region bounded above by the surface z=Fx,y and below by the plane z=0. See Example 6.1.1.
Solution
Mathematical Solution
The region whose volume is to be computed is shown in Figure 6.2.1(a). The simplest iteration of the double integral that gives the required volume takes the integrand as F and uses the order dy dx:
∫01∫0x 1−xx y dy dx = 1120
If the order of integration is taken as dx dy, then the iterated integral would be the more difficult
∫01/4∫1 −1−4 y/21+1−4 y/2x y dx dy = 1120
Figure 6.2.1(a) The volume to be computed
Maple Solution - Interactive
Initialize
Tools≻Load Package: Student Multivariate Calculus
Loading Student:-MultivariateCalculus
Context Panel: Assign Name
f=x 1−x→assign
F=x y→assign
Solve y=fx for x=xy
Write the equation y=f and press the Enter key.
Context Panel: Solve≻Obtain Solutions for≻x
Context Panel: Assign to a Name≻X (Thus, XL=X2 and XR=X1.)
y=f
y=x⁢1−x
→solutions for x
12+12⁢1−4⁢y,12−12⁢1−4⁢y
→assign to a name
X
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 = x⁢y→MultiInt∫01∫0x⁢1−xx⁢yⅆyⅆx=1120
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
∫01∫0fF ⅆy ⅆx = 1120
Iterate in the order dx dy via the template in the Calculus palette
∫01/4∫X2X1F ⅆx ⅆy = 1120
The task template in Table 6.2.1(a) iterates in the order dy dx. The upward-pointing arrow in the left-hand graph indicates that the inner (first) integration is in the y-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 6.2.1(a) Iteration in the order dy dx via visualization task-template
The task template in Table 6.2.1(b) 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.
Table 6.2.1(b) Iteration in the order dx dy via visualization task-template
Maple Solution - Coded
Install the Student MultivariateCalculus package.
withStudent:-MultivariateCalculus:
Define F as the integrand.
F≔x y:
Apply the solve command to the equation y=x 1−x to obtain x=xy.
X≔solvey=x 1−x,x
Top-level, using the Int and int commands
IntF,y=0..x 1−x,x=0..1=intF,y=0..x 1−x,x=0..1
∫01∫0x⁢1−xx⁢yⅆyⅆx=1120
Use the MultiInt command from the Student MultivariateCalculus package
MultiIntF,y=0..x 1−x,x=0..1 = 1120
MultiIntF,y=0..x 1−x,x=0..1,output=integral
∫01∫0x⁢1−xx⁢yⅆyⅆx
MultiIntF,y=0..x 1−x,x=0..1,output=steps
1120
Use the MultiInt command with a pre-defined domain option
MultiIntF,x,y=Region0..1,0.. x 1−x = 1120
MultiIntF,x,y=Region0..1,0.. x 1−x,output=integral
MultiIntF,y,x=Region0..1/4,X2..X1 = 1120
MultiIntF,y,x=Region0..1/4,X2..X1,output=integral
∫014∫12−12⁢1−4⁢y12+12⁢1−4⁢yx⁢yⅆxⅆy
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