Chapter 5: Double Integration
Section 5.3: Regions with Curved Boundaries
Example 5.3.4
Integrate fx,y=1+2 x2+3 y2 over the region R={x,y | 0≤x≤1,x3/2≤y≤2−2−x2}.
Solution
Mathematical Solution
To iterate in the order dy dx, describe the bounding curves as in Figure 5.3.4(a) where yB=x3/2 is the lower limit of the inner integral, and yT=2−2− x2 is the upper limit.
Figure 5.3.4(b) suggests that iteration in the order dx dy requires three different doubly iterated integrals, a chore not to be pursued here.
Figure 5.3.4(a) Iterating in the order dy dx
Figure 5.3.4(b) Iterating in the order dx dy
The resulting iterated integral is therefore
∫01∫x3/22−2− x21+2 x2+3 y2 ⅆy ⅆx = 3247240−318π ≐ 1.355
Maple Solution - Interactive
Initialize
Tools≻Load Package: Student Multivariate Calculus
Loading Student:-MultivariateCalculus
Context Panel: Assign Name
f=1+2 x2+3 y2→assign
YB=x3/2→assign
YT=2−2−x2→assign
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+1→MultiInt∫01∫12x32−−x2+22x2+3y2+1ⅆyⅆx=3247240−318π
Table 5.3.4(a) illustrates the visualization task template keyed to iterate in the order dy dx. The display in the table shows the end-state of the task template. To re-initialize it so that, for example, the buttons under "Value of Integral" work, re-select the Area Element. (If one of these buttons is pressed first, the Area Element will reset to "Select dA.")
Tools≻Tasks≻Browse:
Calculus - Multivariate≻Integration≻Visualizing Regions of Integration≻
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 5.3.4(a) Visualizing R and the resulting volume for iteration in the order dy dx
The vertical arrow in the left-hand graph indicates that the iteration is in the order dy dx, whereby the first (or inner) integration is in the vertical direction, from the lowermost boundary curve to the uppermost. Because the integrand is positive, the double integral calculates the volume below the surface z=f but above the plane z=0. The solid whose volume is thereby calculated is seen in the right-hand graph.
The detailed analytic results below are obtained via the palettes and Context Panel.
Iterate in the order dy dx
Calculus palette: Template for definite iterated double integral
Context Panel: Evaluate and Display Inline
∫01∫x3/22−2−x2f ⅆy ⅆx = 3247240−318π
Display the iterated integrals
Context Panel: 2-D Math≻Convert To≻Inert Form
Press the Enter key
∫01∫x3/22−2−x2f ⅆy ⅆx
∫01∫12x32−−x2+22x2+3y2+1ⅆyⅆx
=
3247240−318π
Maple Solution - Coded
Install the Student MultivariateCalculus package.
withStudent:-MultivariateCalculus:
Define the integrand.
f≔1+2 x2+3 y2:
Use the Int command to obtain the inert integral and the int command for immediate evaluation
Intf,y=x3/2..2−2−x2,x=0..1=intf,y=x3/2..2−2−x2,x=0..1
∫01∫12x32−−x2+22x2+3y2+1ⅆyⅆx=3247240−318π
Use the MultiInt command from the Student MultivariateCalculus package
MultiIntf,y=x3/2..2−2−x2,x=0..1,output=integral
MultiIntf,y=x3/2..2−2−x2,x=0..1
Obtain stepwise evaluations via the MultiInt command
MultiIntf,y=x3/2..2−2−x2,x=0..1,output=steps:
∫01∫x3/22−2−x22x2+3y2+1ⅆyⅆx
=3247240−318π
The stepwise evaluation of the iterated integral is inserted as an image because the actual display does not fit comfortably in the available space.
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