Chapter 6: Applications of Double Integration
Section 6.2: Volume
Example 6.2.3
If F=2 x+3 y+1 and R is the region bounded by the graphs of fx=sinx and gx=sin2 x on 0≤x≤π, calculate the volume of the region bounded above by the surface z=Fx,y and below by the plane z=0. See Example 6.1.3.
Solution
Mathematical Solution
The region whose volume is to be computed is shown in Figure 6.2.3(a). The simplest iteration of the double integral that gives the area of R takes the integrand as F and uses the order dy dx:
∫0π/3∫sinxsin2 xF ⅆy ⅆx+∫π/3π∫sin2 xsinxF ⅆy ⅆx =52+4⁢π−1516⁢3≐ 13.44
It takes two iterated double integrals to obtain the requisite area because the curves "cross" at x=π/3.
use plots in module() local F,p1,p2,p3; F:=2*x+3*y+1; p1:=plot3d(F,x=0..Pi/3,y=sin(x)..sin(2*x),filled=true,lightmodel=none); p2:=plot3d(F,x=Pi/3..Pi,y=sin(2*x)..sin(x),filled=true,lightmodel=none); p3:=display(p1,p2,orientation=[-140,60,0],labels=[x,y,z],tickmarks=[4,3, 4]); print(p3); end module: end use:
Figure 6.2.3(a) The volume to be computed
Maple Solution - Interactive
Initialize
Context Panel: Assign Name
F=2 x+3 y+1→assign
Obtain the intersections of the curves bounding R
Write a sequence of the two equations defining the bounding curves.
Context Panel: Solve≻Solve
sinx=sin2 x→solvex=0,x=π,x=13⁢π,x=−13⁢π
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
Context Panel: Approximate≻10 (digits)
∫0π/3∫sinxsin2 xF ⅆy ⅆx+∫π/3π∫sin2 xsinxF ⅆy ⅆx = 52+4⁢π−1516⁢3→at 10 digits13.44257299
Maple Solution - Coded
Install the Student MultivariateCalculus package.
withStudent:-MultivariateCalculus:
Define the integrand F.
F≔2 x+3 y+1:
Top-level, using the Int and int commands
IntF,y=sinx..sin2 x,x=0..π/3+IntF,y=sin2 x..sinx,x=π/3..π=intF,y=sinx..sin2 x,x=0..π/3+intF,y=sin2 x..sinx,x=π/3..π
∫013⁢π∫sin⁡xsin⁡2⁢x2⁢x+3⁢y+1ⅆyⅆx+∫13⁢ππ∫sin⁡2⁢xsin⁡x2⁢x+3⁢y+1ⅆyⅆx=52−1516⁢3+4⁢π
Use the MultiInt command from the Student MultivariateCalculus package
MultiIntF,y=sinx..sin2 x,x=0..π/3+ MultiIntF,y=sin2 x..sinx,x=π/3..π
52+4⁢π−1516⁢3
MultiIntF,y=sinx..sin2 x,x=0..π/3,output=integral+ MultiIntF,y=sin2 x..sinx,x=π/3..π,output=integral
∫013⁢π∫sin⁡xsin⁡2⁢x2⁢x+3⁢y+1ⅆyⅆx+∫13⁢ππ∫sin⁡2⁢xsin⁡x2⁢x+3⁢y+1ⅆyⅆx
Use the MultiInt command with a pre-defined domain option
MultiIntF,x,y=Region0..π/3,sinx..sin2 x+ MultiIntF,x,y=Regionπ/3..π,sin2 x..sinx
MultiIntF,x,y=Region0..π/3,sinx..sin2 x,output=integral+ MultiIntF,x,y=Regionπ/3..π,sin2 x..sinx,output=integral
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