Chapter 6: Applications of Double Integration
Section 6.2: Volume
Example 6.2.9
Calculate the volume bounded by the ellipsoid 3 x2+5 y2+7 z2=1.
Solution
Mathematical Solution
The volume to be computed is shown in Figure 6.2.9(a). To obtain expressions for the upper and lower bounding surfaces in the form z=zx,y, solve 3 x2+5 y2+7 z2=1 for
z±= ± 7−21 x2−35⁢y2/7
Iterate in the order dy dx, in which case bounds for the plane region R are y=±⁢5−15 x2/5, found by solving 3 x2+5 y2=1 for y=yx.
The lateral bounds for R are x=±1/3.
The requisite volume is then found via the integral
Figure 6.2.9(a) The volume to be computed
∫x−x+∫y−y+z+−z− dy dx = 4 π315⁢105 ≐ 0.41
Maple Solution - Interactive
Solve 3 x2+5 y2+7 z2=1 for z=zx,y
Context Panel: Assign to a Name≻q
3 x2+5 y2+7 z2=1→assign to a nameq
Type the name q and press the Enter key.
Context Panel: Solve≻Obtain Solutions for≻z
Context Panel: Assign to a Name≻Z
q
3⁢x2+5⁢y2+7⁢z2=1
→solutions for z
17⁢−21⁢x2−35⁢y2+7,−17⁢−21⁢x2−35⁢y2+7
→assign to a name
Z
The simplest approach is to employ the task template in Table 6.2.9(a), after noting that the plane region R is the ellipse 3 x2+5 y2=1.
Tools≻Tasks≻Browse:
Calculus - Vector≻Integration≻Multiple Integration≻2-D≻Over an Ellipse
Integrate fx,y over an Ellipse
fx,y=
Equation of ellipse:
From θ= to θ=
Table 6.2.9(a) Task template for integration over an ellipse
Maple elects to implement the integration in polar coordinates, representing the ellipse as
r
=115⁢tan2θ+115+13tan2θ
=sec2θ3+5 tan2θ
=13 cos2θ+5 sin2θ
=131−sin2θ+5 sin2θ
=13+2 sin2θ
which is what would be obtained by the following "direct" conversion to polar coordinates.
Expression palette: Evaluation template
Context Panel: Solve≻Obtain Solutions for≻r
3 x2+5 y2=1x=a|f(x)x=r cosθ,y=r sinθ
3⁢r2⁢cos⁡θ2+5⁢r2⁢sin⁡θ2=1
→solutions for r
12⁢sin⁡θ2+3,−12⁢sin⁡θ2+3
A solution from first principles is given in Table 6.2.9(b).
Solve for y=yx
Expression palette: Evaluation template (Set z=0 in the equation defining the ellipsoid.) Context Panel: Evaluate and Display Inline
Context Panel: Solve≻Obtain Solutions for≻y
Context Panel: Assign to a Name≻Y
qx=a|f(x)z=0 = 3⁢x2+5⁢y2=1→solutions for y15⁢−15⁢x2+5,−15⁢−15⁢x2+5→assign to a nameY
Write an appropriate iterated integral and evaluate
Calculus palette: Iterated double-integral template
Context Panel: Evaluate and Display Inline
Context Panel: Combine≻power
Context Panel: Approximate≻10 (digits)
∫−1313∫Y2Y1Z1−Z2 ⅆy ⅆx = 4315⁢7⁢5⁢π⁢3= combine 4315⁢π⁢105→at 10 digits0.4087840668
Table 6.2.9(b) Solution from first principles
An alternative solution using the MultiInt command accessed through the Context Panel appears in Table 6.2.9(c).
Tools≻Load Package: Student Multivariate Calculus
Loading Student:-MultivariateCalculus
Access the MultiInt command via the Context Panel
Write Z1−Z2 as the integrand and press the Enter key.
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
Z1−Z2
27⁢−21⁢x2−35⁢y2+7
→MultiInt
∫−13⁢313⁢3∫−15⁢−15⁢x2+515⁢−15⁢x2+527⁢−21⁢x2−35⁢y2+7ⅆyⅆx
=
4315⁢7⁢5⁢π⁢3
= combine
4315⁢π⁢105
→at 10 digits
0.4087840668
Table 6.2.9(c) Solution via the MultiInt command accessed through the Context Panel
Maple Solution - Coded
Initialize
Install the Student MultivariateCalculus package.
withStudent:-MultivariateCalculus:
Define the bounding surfaces zx,y=Z± and the bounds of the region R as yx=Y±, and X=±1/3. Note the use of the solve and eval commands.
q≔3 x2+5 y2+7 z2=1:Z≔solveq,z:Y≔solveevalq,z=0,y:X≔solveevalq,y=0,z=0,x:
Top-level, using the Int and int commands
IntZ1−Z2,y=Y2..Y1,x=X2..X1=intZ1−Z2,y=Y2..Y1,x=X2..X1
∫−13⁢313⁢3∫−15⁢−15⁢x2+515⁢−15⁢x2+527⁢−21⁢x2−35⁢y2+7ⅆyⅆx=4315⁢7⁢5⁢π⁢3
Use the MultiInt command from the Student MultivariateCalculus package
MultiIntZ1−Z2,y=Y2..Y1,x=X2..X1 = 4315⁢7⁢5⁢π⁢3
MultiIntZ1−Z2,y=Y2..Y1,x=X2..X1,output=integral
Use the MultiInt command with a pre-defined domain option
MultiIntZ1−Z2,x,y=RegionX2..X1,Y2..Y1 = 4315⁢7⁢5⁢π⁢3
MultiIntZ1−Z2,x,y=RegionX2..X1,Y2..Y1,output=integral
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