Multivariable Calculus 515 16.9: Change of Variables in ... · in Multiple Integralsin Multiple Integrals Monday, November 24 andMonday, November 24 and Tuesday, November 25, 2008Tuesday,

Multivariable Calculus 515Multivariable Calculus 515

16.9: Change of Variables 16.9: Change of Variables

in Multiple Integralsin Multiple Integrals

Monday, November 24 andMonday, November 24 andTuesday, November 25, 2008Tuesday, November 25, 2008

IntroductionIntroduction

Consider integrating some function, such asover the parallelogram in the xy-

plane with vertices This would require several separate integrals because of the shape of the domain. In this lesson, we explore a less complicated way to integrate over strangely-shaped regions.

( ), 4 8 ,f x y x y= +( ) ( ) ( ) ( )1,3 , 1, 3 , 3, 1 , and 1,5 .− − −

Exercise 1Exercise 1

Find equations in terms of x and y for the lines that comprise the boundary of the region of integration.


Find equations in terms of x and y for the lines that comprise the boundary of the region of integration.

The line from (–1,3) to (1,–3) is (A)The line from (1,–3) to (3,–1) is (B)The line from (3,–1) to (1,5) is (C)The line from (1,5) to (–1,3) is (D)

3 .y x=−4.y x= −

3 8.y x=− +4.y x= +


Use the substitutions to find

equations in terms of u and v for the lines that comprise the boundary of the region of integration.

( )

( )

141 34

x u v

y v u

⎧⎪ = +⎪⎪⎨⎪ = −⎪⎪⎩


Use the substitutions to find

equations in terms of u and v for the lines that comprise the boundary of the region of integration.

(A): (B):(C):(D):

( )

( )

141 34

x u v

y v u

⎧⎪ = +⎪⎪⎨⎪ = −⎪⎪⎩

0v=4u=8v=

4u=−

The Jacobian Determinant (2D)The Jacobian Determinant (2D)

For a transformation T2 from ² to ² that takes (x,y) to (u,v) whose inverse is given by

and we define the Jacobian determinantas

( ),x x u v=( ), ,y y u v=

( )( )

,,

x u x vx yy u y vu v

∂ ∂ ∂ ∂∂=∂ ∂ ∂ ∂∂

The Jacobian Determinant (3D)The Jacobian Determinant (3D)

For a transformation T3 from ³ to ³ that takes (x,y,z) to (u,v,w) whose inverse is given by

( )( )

, ,, ,

x u x v x wx y z y u y v y wu v w z u z v z w

∂ ∂ ∂ ∂ ∂ ∂∂ =∂ ∂ ∂ ∂ ∂ ∂∂ ∂ ∂ ∂ ∂ ∂ ∂

( ) ( ) ( ), , , , , , and , , , theJacobian isx x u v w y y u v w z z u v w= = =


With T2 given by find the Jacobian

determinant for the transformation.

( )

( )

14

14

,3

x u v

y v u

⎧ = +⎪⎪⎨⎪ = −⎪⎩


With T2 given by find the Jacobian

determinant for the transformation.

( )( ) ( )( )

1 4 1 4

3 4 1 4

1 4 1 4 3 4 1 4

1 16 3 161 4

x u x vy u y v

∂ ∂ ∂ ∂=−∂ ∂ ∂ ∂

= − −

= +=

( )

( )

14

14

,3

x u v

y v u

⎧ = +⎪⎪⎨⎪ = −⎪⎩

ApplicationApplication

To integrate a new function given by making the substitutions from a transformation T2 with a nonzero Jacobian into the original we evaluate

over the new region found by transforming the boundaries of the xy-region of integration to that in the uv-plane.

( ),f u v

( ), ,f x y

( ) ( )( )

,,

,uv

R

x yf u v dA

u v∂∂∫∫

Informal JustificationInformal Justification

This is a higher-dimensional analogue of the technique of integration by u-substitution. Consider substituting to evaluate an integral we find rewrite the limits using the transformation, and since we substitute We have effectively divided by the “size of the transformation”: if we had we would find since du is twice dx, we divide by two in the integral that involves

( )u u x=( ) :b

af x dx∫ ( ),f u

,dudxdu dx=

for .dudu dx

dx2 ,u x=

2 ;du dx=( ).f u


The “size of a transformation” of several variables is given by the Jacobian determinant, a concept which will be explored further in the semester of linear algebra.


For a simple example, consider the transformation

which doubles each coordinate, thus multi-

plying the area of the unit dA by four. We thus must divide by four to undo this. The transform-

ation is equivalent to and the Jacobian is

So when we multiply by that, we divide by four to undo the area increase, as we would expect.

2

2

u xv y

⎧ =⎪⎪⎨⎪ =⎪⎩

2,

2

x uy v

⎧ =⎪⎪⎨⎪ =⎪⎩1 4.


Substitute into the integrand

to find Then evaluate the new integral, including the term of the Jacobian

determinant, to find with Rxy as

described originally.

( )

( )

14

14 3

x u v

y v u

⎧ = +⎪⎪⎨⎪ = −⎪⎩( ), 4 8f x y x y= + ( ), .f u v

( ),xy

R

f x y dA∫∫

The integrand becomes and we have already found the limits and the Jacobian.


Make the given substitution, then evaluate the new integral, including the term of the Jacobian

determinant, to find with Rxy as

described originally.

( ),xy

R

f x y dA∫∫5 3 ,u v− +

Therefore the integral is

which evaluates to 192.

( )84

04

5 3 1 4u v dvdu−

− +∫∫


Use a substitution to evaluate Rxy is

the square with vertices (0,0), (1,1), (2,0), and (1,–1).

;xy

R

xydA∫∫




;xy

R

xydA∫∫

Calling the vertices of the boundary A, B, C, and D, respectively, the lines that enclose the boundary are

:: 2: 2:

AB y xBC y xCD y xAD y x

== −= −=−




;xy

R

xydA∫∫

This suggests that we might wish to use the

substitution which gives

The domain then becomes

Substituting into

,u y xv y x

⎧ = −⎪⎪⎨⎪ = +⎪⎩

1 12 2

1 12 2

.x u vy u v

⎧ =− +⎪⎪⎨⎪ = +⎪⎩[ ] [ ]2,0 and 0,2 .u v∈− ∈

( ) ( ) ² ², gives , .4

u vf x y xy f u v − += =




;xy

R

xydA∫∫

The Jacobian is so the

integral is which

evaluates to 0.

( )( )

1 2 1 2, 1,1 2 1 2, 2

x yu v

−∂ = =−∂

20

02

² ²1 2

4u v dvdu

−

− +−∫∫




;xy

R

xydA∫∫

We might have anticipated this becauseis an odd function of y, and the region described is symmetric about the x-axis.

( ),f x y xy=


Use the transformation to show that the

Jacobian determinant for the transformation from Cartesian to polar coordinates is r.

cos

sin

x ry r

θθ

⎧ =⎪⎪⎨⎪ =⎪⎩


Use the transformation to show that the

Jacobian determinant for the transformation from Cartesian to polar coordinates is r.

The Jacobian is ( )( )

,,

x r xx yy r yr

θθθ

∂ ∂ ∂ ∂∂ =∂ ∂ ∂ ∂∂

( )cos sin

cos² sin² .sin cos

rr r

rθ θ

θ θθ θ

−= = + =

cos

sin

x ry r

θθ

⎧ =⎪⎪⎨⎪ =⎪⎩


Use the transformation to show

that the Jacobian for the transformation from Cartesian to spherical coordinates is

sin cos

sin sin

cos

xyz

ρ ϕ θρ ϕ θρ ϕ

⎧ =⎪⎪⎪ =⎨⎪⎪ =⎪⎩

²sin .ρ ϕ

Exercise 7Exercise 7Use the given transformation to show that the Jacobian for the transformation from Cartesian to spherical coordinates is

The Jacobian is

²sin .ρ ϕ

( )( )

, ,, ,

x x xx y z y y y

z z z

ρ θ ϕρ θ ϕ

ρ θϕ ρ θ ϕ

∂ ∂ ∂ ∂ ∂ ∂∂ =∂ ∂ ∂ ∂ ∂ ∂∂ ∂ ∂ ∂ ∂ ∂ ∂

sin cos sin sin cos cos

sin sin sin cos cos sin

cos 0 sin

ϕ θ ρ ϕ θ ρ ϕ θϕ θ ρ ϕ θ ρ ϕ θϕ ρ ϕ

−=

−

Exercise 7Exercise 7Use the given transformation to show that the Jacobian for the transformation from Cartesian to spherical coordinates is

which is easiest to evaluate by taking minors across the bottom row:

²sin .ρ ϕ

( )( )

( )

cos ²sin cos sin² ²sin cos cos²sin sin² cos² sin² sin²

²sin cos² 1 ²sin³ ²sin ,of which we take the absolute value to get ²sin .

ϕ ρ ϕ ϕ θ ρ ϕ ϕ θρ ϕ ρ ϕ θ ρ ϕ θρ ϕ ϕ ρ ϕ ρ ϕ

ρ ϕ

− −− +=− − =−


Use the transformation to find the volume

of the general ellipsoid

x auy bvz cw

⎧ =⎪⎪⎪ =⎨⎪⎪ =⎪⎩² ² ² 1.² ² ²

x y za b c+ + =

Exercise 8Exercise 8Use the given transformation to find the volume of the general ellipsoid

The Jacobian is The

ellipsoid’s equation transforms to which is a sphere in uvw-space with radius 1 and volume We multiply that by the Jacobian to

give the ellipsoid’s volume

² ² ² ² ² ² 1.x a y b z c+ + =

( )( )

0 0, , 0 0 ., ,

0 0

ax y z b abcu v w c

∂ = =∂

² ² ² 1,u v z+ + =

4π 3.4π .3

V abc=


A transformation from the uv-plane to the xy-plane

is given by

(a) Find the Jacobian for this transformation.

3 2.

4

u x yv x y

⎧ = +⎪⎪⎨⎪ = +⎪⎩


A transformation from the uv-plane to the xy-plane

is given by

(a) Find the Jacobian for this transformation.

Since the Jacobian is

3 2.

4

u x yv x y

⎧ = +⎪⎪⎨⎪ = +⎪⎩

2 15 5

3110 10

,x u vy u v

⎧ = −⎪⎪⎨⎪ =− +⎪⎩

2 15 5

3110 10

1 10.−

=−

Exercise 9Exercise 9With the previously given transformation,(b) Find the image of the triangular region in the first quadrant of the xy-plane bounded by the coordinate axes and the line 1.x y+ =

Exercise 9Exercise 9With the previously given transformation,(b) Find the image of the triangular region in the first quadrant of the xy-plane bounded by the coordinate axes and the line

The region is bounded bythe lines and

1.x y+ =

3, 2 , andv u v u= =10 3 .v u= −

Exercise 9Exercise 9With the previously given transformation,

(c) evaluate where Rxy is

the region in the xy-plane bounded by the lines

( )2 23 14 8xy

R

x xy y dA+ +∫∫3 3 1 12 2 4 41, 3, , and 1.y x y x y x y x=− + =− + =− =− +

Exercise 9Exercise 9With the previously given transformation,

(c) evaluate the given integral over the given region.

The boundaries transform to, respectively, and making the substitutions

into gives so

the integral is

2,u=6, 0, and 4,u v v= = =

( ), 3 ² 14 8 ²f x y x xy y= + + ( ), ,f u v uv=46

02

1 64 5.10uv dvdu=∫∫

Documents

Multivariable Calculus 515 16.9: Change of Variables in ... · in Multiple Integralsin Multiple Integrals Monday, November 24 andMonday, November 24 and Tuesday, November 25, 2008Tuesday,