Chapter 5: Double Integration
Section 5.6: Changing Variables in a Double Integral
Example 5.6.5
Let R be the region bounded by the curves C1:16⁢x2⁢y+16⁢y3+8⁢x−88 y=0, C2:144⁢x2⁢y+144⁢y3−24 x−120 y=0, C3:16⁢x2⁢y+16⁢y3−8 x−8 y=0.
Integrate fx,y=x2+y2/y2 over R, noting that it takes two iterations to cover R. Hint: Solve each bounding curve for x=xy and integrate in the order dx dy.
Make the change of variables u=x2+y2, v=x/y and evaluate the integral of f over the image of R under this change of variables.
Solution
Mathematical Solution
Figure 5.6.5(a) shows the region R; Figure 5.6.5(b) shows R′, the image of R under the given change of variables.
Figure 5.6.5(a) Region R
Figure 5.6.5(b) Region R′
Table 5.6.5(a) shows the correspondence between the "corner" points in regions R and R′. Table 5.6.5(b) lists the equations for the mappings between regions R and R′.
Corners in R
Corners in R′
a:1/2,1/2
1,1
b:7/5,1/5
2,7
c:53/26,3/26
3,5
Table 5.6.5(a) Corners in regions R and R′
Mapping R′→R
Mapping R→R′
x=vu1+v2
u=x2+y2
y=u1+v2
v=x/y
Table 5.6.5(b) Mappings R↔R′
Integration over region R is best done by expressing each bounding curve for R in the form x=xy and choosing the order dx dy. Table 5.6.5(c) lists the expressions for the resulting xky,k=1,2,3. The result is
∫32612∫x2x3f dx dy+∫15326∫x2x1f dx dy = 2
x1y=⁢−1+1−16 y4+88⁢y24 y
x2y=⁢1+1−144 y4+120⁢y212 y
x3y=⁢1+1−16 y4+8⁢y24 y
Table 5.6.5(c) Ck solved for x=xy
Direct calculation of the Jacobian ∂x,y∂u,v is tedious. Alternatively, the matrix for the Jacobian ∂u,v∂x,y is 2 x2 y−1/y−x/y2 so this Jacobian is −21+x2/y2. The requisite Jacobian is the reciprocal of this expressed in terms of u and v, that is, ∂x,y∂u,v=−121+v2.
Table 5.6.5(d) lists equations for Ck/ , the image of each Ck in region R′.
Since fxu,v,yu,v=1+v2, the requisite integral is then
C1→v=11−2 u
C2→v=6 u−5
C3→v=2 u−1
Table 5.6.5(d) Images of Ck in R′
∫∫R′fxu,v,yu,v∂x,y∂u,v dv du
= 12∫12∫2 u−16 u−5dv du+12∫23∫2 u−111−2 udv du
=2
Maple Solution - Interactive
Initialize
Tools≻Load Package: Student Multivariate Calculus
Loading Student:-MultivariateCalculus
Context Panel: Assign Name
L=16⁢x2⁢y+16⁢y3+8⁢x−88 y=0,144⁢x2⁢y+144⁢y3−24 x−120 y=0,16 x2⁢y+16⁢y3−8 x−8 y=0→assign
Context Panel: Assign function
fx,y=x2+y2/y2→assign as functionf
U=x2+y2→assign
Context Panel: Assign name
V=x/y→assign
Implement the integration over the region R. Begin by obtaining the coordinates of the "corners" of region R.
Write a sequence of a pair of equations for intersecting edges and press the Enter key.
Context Panel: Solve≻Solve (explicit)
a
L1,L3
16⁢x2⁢y+16⁢y3+8⁢x−88⁢y=0,16⁢x2⁢y+16⁢y3−8⁢x−8⁢y=0
→solve
x=0,y=0,x=526⁢78,y=126⁢78,x=−526⁢78,y=−126⁢78
b
L1,L2
16⁢x2⁢y+16⁢y3+8⁢x−88⁢y=0,144⁢x2⁢y+144⁢y3−24⁢x−120⁢y=0
x=0,y=0,x=75,y=15,x=−75,y=−15
c
Express each edge of region R in the form x=xy.
Write the name of the equation for an edge and press the Enter key.
Context Panel: Solve≻Obtain Solutions for≻x
Context Panel: Select Element≻1
Context Panel: Assign to a Name≻e[k], k=1,2,3
L1
16⁢x2⁢y+16⁢y3+8⁢x−88⁢y=0
→solutions for x
14⁢−1+−16⁢y4+88⁢y2+1y,−14⁢1+−16⁢y4+88⁢y2+1y
→select entry 1
14⁢−1+−16⁢y4+88⁢y2+1y
→assign to a name
e1
L2
144⁢x2⁢y+144⁢y3−24⁢x−120⁢y=0
112⁢1+−144⁢y4+120⁢y2+1y,−112⁢−1+−144⁢y4+120⁢y2+1y
112⁢1+−144⁢y4+120⁢y2+1y
e2
L3
16⁢x2⁢y+16⁢y3−8⁢x−8⁢y=0
14⁢1+−16⁢y4+8⁢y2+1y,−14⁢−1+−16⁢y4+8⁢y2+1y
14⁢1+−16⁢y4+8⁢y2+1y
e3
Write and evaluate the iterated integral.
Calculus palette: Iterated double-integral template
Context Panel: Evaluate and Display Inline
∫3/261/2∫e2e3fx,y ⅆx ⅆy+∫1/53/26∫e2e1fx,y ⅆx ⅆy = 2
Change coordinates and implement the integration in the new coordinate system.
Obtain the equations S=x=xu,v,y=yu,v for the mapping u,v→x,y
Write the equations for the mapping v⁡x,t⁢will now be displayed as⁢v. Press the Enter key.
Context Panel: Solve≻Solve for Variables≻x,y
Context Panel: All Values
Select Element≻1
Context Panel: Simplify≻Symbolic
Context Panel: Assign to a Name≻S
u=U,v=V
u=x2+y2,v=xy
→solve (specified)
x=v⁢RootOf⁡v2+1⁢_Z2−u,y=RootOf⁡v2+1⁢_Z2−u
→all values
x=v⁢uv2+1,y=uv2+1,x=−v⁢uv2+1,y=−uv2+1
x=v⁢uv2+1,y=uv2+1
→simplify symbolic
S
Obtain the Jacobian matrix and the Jacobian
Expression palette: Evaluation template Evaluate x and then y using the information in set S
Context Panel: Assign to a Name≻X or Y, as appropriate
Form the list X,Y Context Panel: Evaluate and Display Inline
Context Panel: Student Multivariate Calculus≻Differentiate≻Jacobian
Context Panel: Assign to a Name≻J
xx=a|f(x)S = v⁢uv2+1→assign to a nameX
yx=a|f(x)S = uv2+1→assign to a nameY
X,Y = v⁢uv2+1,uv2+1→Jacobian−12⁢v2+1→assign to a nameJ
Obtain the Jacobian matrix and the Jacobian from first principles
Matrix palette: Insert template for 2×2 matrix.
Calculus palette: Partial derivative operator
Context Panel: Standard Operations≻Determinant
∂∂ u X∂∂ v X∂∂ u Y∂∂ v Y = →determinant−12⁢v2+1
Obtain the images of the edges of the region R
Expression palette: Evaluation template Evaluate the equation of each edge of R using the equations in set S. Press the Enter key.
Context Panel: Solve≻Isolate Expression for≻v
L1x=a|f(x)S
16⁢v2⁢u3/2v2+13/2+16⁢u3/2v2+13/2+8⁢v⁢uv2+1−88⁢uv2+1=0
→isolate for v
v=−2⁢u+11
L2x=a|f(x)S
144⁢v2⁢u3/2v2+13/2+144⁢u3/2v2+13/2−24⁢v⁢uv2+1−120⁢uv2+1=0
v=6⁢u−5
L3x=a|f(x)S
16⁢v2⁢u3/2v2+13/2+16⁢u3/2v2+13/2−8⁢v⁢uv2+1−8⁢uv2+1=0
v=2⁢u−1
Obtain the transformed integrand fxu,v,yu,v∂x,y∂u,v
Context Panel: Simplify≻Assuming Real
fX,Y J = 12⁢uv2+1+v2⁢uv2+1⁢v2+1u⁢v2+1→assuming real12
Implement the integration over the region R′
∫12∫2 u−16 u−512 ⅆv ⅆu+∫23∫2 u−111−2 u12 ⅆv ⅆu = 2
Maple Solution - Coded
Install the Student MultivariateCalculus package.
withStudent:-MultivariateCalculus:
Define the integrand f.
f≔x,y→x2+y2/y2:
Define the edges of the region defining R.
L≔16⁢x2⁢y+16⁢y3+8⁢x−88 y=0,144⁢x2⁢y+144⁢y3−24 x−120 y=0,16 x2⁢y+16⁢y3−8 x−8 y=0:
Define the mapping v⁡x,t⁢will now be displayed as⁢v.
U≔x2+y2:V≔x/y:
Integration over region R is best done in the order dx dy, and this is facilitated by expressing each bounding curve in the form x=xy. This is done in Table 5.6.5(e).
X1≔solveL1,x1:
X1 = 14⁢−1+−16⁢y4+88⁢y2+1y
X2≔solveL2,x1:
X2 = 112⁢1+−144⁢y4+120⁢y2+1y
X3≔solveL3,x1:
X3 = 14⁢1+−16⁢y4+8⁢y2+1y
Table 5.6.5(e) Each Ck in R solved for x=xy
The corners of region R are found by pairwise intersecting the bounding curves Ck. This is done in Table 5.6.5(f), and correlates with Figure 5.6.5(a).
Point
Coordinates
solveL2,L3,explicit
x=0,y=0,x=12⁢2,y=12⁢2,x=−12⁢2,y=−12⁢2
solveL1,L2
solveL1,L3,explicit
Table 5.6.5(f) Corners of region R: Select appropriate solution based on Figure 5.6.5(a)
The iteration over R is then
∫32612∫X2X3fx,y ⅆx ⅆy+∫15326∫X2X1fx,y ⅆx ⅆy = 2
Next, change coordinates and implement the integration in the new coordinate system.
Apply the solve command and simplify it under the assumption of positivity. (The first solution maps the first quadrant to the first quadrant, so it is the one to select.)
temp≔solveu=U,v=V,x,y,explicit;S≔simplifytemp1 assuming positive
x=v⁢v2+1⁢uv2+1,y=v2+1⁢uv2+1,x=−v⁢v2+1⁢uv2+1,y=−v2+1⁢uv2+1
Use the eval command to obtain a list of expressions xu,v,yu,v.
Apply the Jacobian command in its two forms, one to get the matrix, and one to get the determinant. Apply the simplify command to the Jacobian matrix.
T≔evalx,y,S;simplifyJacobianT,u,v;J≔JacobianT,u,v,output=determinant
v⁢uv2+1,uv2+1
Obtain the images of the boundaries of the region R
Use the eval command to make the replacements x→xu,v and y→yu,v in the equation of each edge of the region R.
Use the isolate command to obtain equations of the form v=…
isolateevalL1,S,v
v=11−2⁢u
isolateevalL2,S,v
isolateevalL3,S,v
Obtain the transformed integrand fxu,v,yu,v
Apply the simplify command, where T1=xu,v and T2=yu,v.
simplifyfT1,T2 = v2+1
Use the Int command to generate the inert integral and apply the value command to evaluate it.
q≔Int1/2,v=2 u−1..6 u−5,u=1..2+Int1/2,v=2 u−1..11−2 u,u=2..3;valueq
∫12∫2⁢u−16⁢u−512ⅆvⅆu+∫23∫2⁢u−111−2⁢u12ⅆvⅆu
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