Chapter 5: Double Integration
Section 5.3: Regions with Curved Boundaries
Example 5.3.8
Integrate fx,y=1+2 x2+3 y2 over the region bounded by the ellipse 4 x2+9 y2=36 and the line y=2−2 x/3.
Solution
Mathematical Solution
To iterate in the order dy dx, describe the bounding curves as in Figure 5.3.8(a) where yT=29−x2/3 is the upper limit of the inner integral, and yB=2−2 x/3 is the lower limit.
To iterate in the order dx dy, describe the bounding curves as in Figure 5.3.8(b) where xR=34−y2/2 is the upper limit of the inner integral, and xL=3−3 y/2 is the lower limit.
Figure 5.3.8(a) Iterating in the order dy dx
Figure 5.3.8(b) Iterating in the order dx dy
The resulting iterated integrals are therefore
∫03∫2−23x239−x21+2 x2+3 y2 ⅆy ⅆx = 514 π−18 = ∫02∫3−32y324−y21+2 x2+3 y2 ⅆx ⅆy
Maple Solution - Interactive
Initialize
Tools≻Load Package: Student Multivariate Calculus
Loading Student:-MultivariateCalculus
Context Panel: Assign Name
f=1+2 x2+3 y2→assign
YT=29−x2/3→assign
YB=2−2 x/3→assign
XR=34−y2/2→assign
XL=3−3 y/2→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 = 2⁢x2+3⁢y2+1→MultiInt∫03∫2−23⁢x23⁢−x2+92⁢x2+3⁢y2+1ⅆyⅆx=−18+514⁢π
Tables 5.3.8(a, b) illustrate the visualization task template wherein a double integral can be iterated in either order. In the drop-down for area-element dA, select an ordering for the iteration. Then fill in the remaining fields as shown.
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 5.3.8(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.
Table 5.3.8(b) shows the iteration in the opposite order.
Table 5.3.8(b) Visualizing R and the resulting volume for iteration in the order dx dy
When integrating in the order dx dy, the first (or inner) integral is taken in the horizontal direction, as indicated by the arrow in the left-hand graph.
The detailed analytic results below are obtained via the palettes and Context Panel.
Iterate dy dx and dx dy
Calculus palette: Template for definite iterated double integral
Context Panel: Evaluate and Display Inline
∫03∫2−2 x/3239−x2f ⅆy ⅆx = −18+514⁢π
∫02∫3−3 y/2324−y2f ⅆx ⅆy = −18+514⁢π
Display the iterated integrals
Calculus palette: Template for definite iterated double integral Context Panel: 2-D Math≻Convert To≻Inert Form Press the Enter key
∫03∫2−2 x/3239−x2f ⅆy ⅆx
∫03∫2−23⁢x23⁢−x2+92⁢x2+3⁢y2+1ⅆyⅆx
=
−18+514⁢π
∫02∫3−3 y/2324−y2f ⅆx ⅆy
∫02∫3−32⁢y32⁢−y2+42⁢x2+3⁢y2+1ⅆxⅆy
Maple Solution - Coded
Install the Student MultivariateCalculus package.
withStudent:-MultivariateCalculus:
Define the integrand.
f≔1+2 x2+3 y2:
Iterate in the order dy dx
Use the Int command to obtain the inert integral and the int command for immediate evaluation
Int⁡f,y=2−23x⁢..23 9−x2⁢,x=0..3=int⁡f,y=2−23x⁢..23 9−x2⁢,x=0..3
∫03∫2−23⁢x23⁢−x2+92⁢x2+3⁢y2+1ⅆyⅆx=−18+514⁢π
Iterate in the order dx dy
Intf,x=3−32y..324−y2,y=0..2=intf,x=3−32y..324−y2,y=0..2
∫02∫3−32⁢y32⁢−y2+42⁢x2+3⁢y2+1ⅆxⅆy=−18+514⁢π
Use the MultiInt command from the Student MultivariateCalculus package
MultiIntf,y=2−23x⁢..23 9−x2⁢,x=0..3,output=integral
MultiIntf,y=2−23x⁢..23 9−x2⁢,x=0..3
MultiIntf,x=3−32y..324−y2,y=0..2,output=integral
MultiIntf,x=3−32y..324−y2,y=0..2
Obtain stepwise evaluations via the MultiInt command
MultiIntf,y=2−23x⁢..23 9−x2⁢,x=0..3,output=steps:
∫03∫2−23⁢x23⁢−x2+92⁢x2+3⁢y2+1ⅆyⅆx
=514π−18
MultiIntf,x=3−32y..324−y2,y=0..2,output=steps:
∫02∫3−32⁢y32⁢−y2+42⁢x2+3⁢y2+1ⅆxⅆy
The stepwise evaluations of the iterated integrals are inserted as images because the actual displays do not fit comfortably in the available space.
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