Change of VariablesIn

Multiple Integrals

When we convert a double integral from rectangularto polar coordinates, recall the changes that mustbe made to x, y and dA.

( , ) cos( , ) sin

x x r ry y r rdA r drd

θ θθ θθ

= == ==

r

θ

x

yT R

r

θ

x

y

In the polar coordinate system, an element of areais generally a rectangle corresponding to a range ofvalues for r and a range for θ.

a r b≤ ≤

c dθ≤ ≤

T R

However, in the xy coordinate system, this rectangle usually takes on a different shape, and the formula for an element of area changes.

r

θ

x

y

A r r θΔ ≈ ⋅Δ ⋅Δ

T R

r

θ

x

y

This leads to the following formula for the doubleintegral in polar coordinates.

( , ) ( cos , sin )R T

f x y dA f r r r drdθ θ θ=∫ ∫ ∫ ∫

T R

r

θ

x

y

We’ll now develop a general method for findingchange of variable formulas such as this one.

T R

( , ) ( cos , sin )R T

f x y dA f r r r drdθ θ θ=∫ ∫ ∫ ∫

Suppose we have a rectangle in an st-coordinatesystem and a pair of functions that converts (s,t)coordinates into (x,y) coordinates.

s

t

x

y

( , )( , )

x x s ty y s t==

T

R

s

t

x

y

Suppose also that these functions are differentiableand that the transformation from the st-coordinatesto xy-coordinates is one-to-one.

( , )( , )

x x s ty y s t==

T

R

s

t

x

y

Then because of differentiability, local linearitywill be present and a small rectangle in the st system will be mapped onto approximately a parallelogram in the xy-coordinate system.

( , )( , )

x x s ty y s t==T

R

x

If we add some coordinates, then it looks like this.

R

s

t

( ),s t ( ),s s t+ Δ

( ),s t t+ Δ

y

( ) ( )( ), , ,x s t y s t

( ) ( )( ), , ,x s s t y s s t+ Δ + Δ

( ) ( )( ), , ,x s t t y s t t+ Δ + Δ

a

b

( , )( , )

x x s ty y s t==

T

s

t

x( ),s t ( ),s s t+ Δ

( ),s t t+ Δ

y

( ) ( )( ), , ,x s t y s t

( ) ( )( ), , ,x s s t y s s t+ Δ + Δ

( ) ( )( ), , ,x s t t y s t t+ Δ + Δ

a

b

An element of area in our xy-coordinate system isrepresented by a parallelogram.

( , )( , )

x x s ty y s t==

T

R

x

y

( ) ( )( ), , ,x s t y s t

( ) ( )( ), , ,x s s t y s s t+ Δ + Δ

( ) ( )( ), , ,x s t t y s t t+ Δ + Δ

a

b

The area of this parallelogram is given by the normof a cross product.

Area a b= ×

R

x

y

( ) ( )( ), , ,x s t y s t

( ) ( )( ), , ,x s s t y s s t+ Δ + Δ

( ) ( )( ), , ,x s t t y s t t+ Δ + Δ

a

b

Notice that, ( ) ( )( ) ( ) ( )( )

( ) ( )( ) ( ) ( )( )

ˆ ˆ, , , ,

ˆ ˆ

ˆ ˆ, , , ,

ˆ ˆ

a x s s t x s t i y s s t y s t j

x ys i s js s

b x s t t x s t i y s t t y s t j

x yt i t jt t

= + Δ − + + Δ −

∂ ∂≈ Δ + Δ∂ ∂

= + Δ − + + Δ −

∂ ∂≈ Δ + Δ∂ ∂

R

x

y

( ) ( )( ), , ,x s t y s t

( ) ( )( ), , ,x s s t y s s t+ Δ + Δ

( ) ( )( ), , ,x s t t y s t t+ Δ + Δ

a

b

Hence,

ˆˆ ˆ

ˆ0

0

i j kx y x y y xa b s s s t s t ks s s t s tx yt tt t

∂ ∂ ∂ ∂ ∂ ∂⎛ ⎞× ≈ Δ Δ = Δ ⋅ Δ − Δ ⋅ Δ⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠∂ ∂

Δ Δ∂ ∂

R

x

y

( ) ( )( ), , ,x s t y s t

( ) ( )( ), , ,x s s t y s s t+ Δ + Δ

( ) ( )( ), , ,x s t t y s t t+ Δ + Δ

a

b

And,

x y y x x y y xa b s t s t s ts t s t s t s t∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

× ≈ Δ ⋅ Δ − Δ ⋅ Δ = − Δ Δ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

R

x

y

( ) ( )( ), , ,x s t y s t

( ) ( )( ), , ,x s s t y s s t+ Δ + Δ

( ) ( )( ), , ,x s t t y s t t+ Δ + Δ

a

b

The expression inside the last absolute value signis called the Jacobian, and it is usually written as follows:

( , )( , )

x xx y x y y xs t

y ys t s t s ts t

∂ ∂∂ ∂ ∂ ∂ ∂∂ ∂= = −

∂ ∂∂ ∂ ∂ ∂ ∂∂ ∂

R

x

y

( ) ( )( ), , ,x s t y s t

( ) ( )( ), , ,x s s t y s s t+ Δ + Δ

( ) ( )( ), , ,x s t t y s t t+ Δ + Δ

a

b

Therefore,

( )( )

,,

x yx y y xArea A a b s t s ts t s t s t

∂∂ ∂ ∂ ∂= Δ = × ≈ − Δ Δ = Δ Δ

∂ ∂ ∂ ∂ ∂

R

x

y

( ) ( )( ), , ,x s t y s t

( ) ( )( ), , ,x s s t y s s t+ Δ + Δ

( ) ( )( ), , ,x s t t y s t t+ Δ + Δ

a

b

And as a result of local linearity, this approximationimproves as the changes in s and t become smaller.

( )( )

,,

x yx y y xArea A a b s t s ts t s t s t

∂∂ ∂ ∂ ∂= Δ = × ≈ − Δ Δ = Δ Δ

∂ ∂ ∂ ∂ ∂

R

x

y

( ) ( )( ), , ,x s t y s t

( ) ( )( ), , ,x s s t y s s t+ Δ + Δ

( ) ( )( ), , ,x s t t y s t t+ Δ + Δ

a

b

And finally,

( )( )

( )( )

0

, 0

( , ) lim ( , )

,lim ( ( , ), ( , ))

,

,( ( , ), ( , ))

,

R A

s t

T

f x y dA f x y A

x yf x s t y s t s t

s t

x yf x s t y s t dsdt

s t

Δ →

Δ Δ →

= ⋅Δ

∂= ⋅ Δ Δ

∂

∂= ⋅

∂

∑∫∫

∑

∫∫

R

x

y

( ) ( )( ), , ,x s t y s t

( ) ( )( ), , ,x s s t y s s t+ Δ + Δ

( ) ( )( ), , ,x s t t y s t t+ Δ + Δ

a

b

Now let’s verify that this formula works for polarcoordinates.

( )( )

( )( )

0

, 0

( , ) lim ( , )

, ,lim ( , ) ( , )

, ,

R A

Ts t

f x y dA f x y A

x y x yf x y s t f x y dsdt

s t s t

Δ →

Δ Δ →

= ⋅Δ

∂ ∂= ⋅ Δ Δ = ⋅

∂ ∂

∑∫∫

∑ ∫∫

R

( )( )

,( , ) ( ( , ), ( , ))

,R T

x yf x y dA f x s t y s t drd

rθ

θ∂

= ⋅∂∫ ∫ ∫ ∫

cossin

x ry r

θθ

==

( )2 2 2 2

cos sin( , )sin cos( , )

cos sin cos sin

x xrx y r

y y rrr

r r r r

θ θθθ θθ

θ

θ θ θ θ

∂ ∂−∂ ∂ ∂= =

∂ ∂∂∂ ∂

= + = + =

( , )( , )x y r rr θ

∂= = =

∂

( )( )

,( , ) ( cos , sin )

,

( cos , sin )

R T

T

x yf x y dA f r r drd

r

f r r r drd

θ θ θθ

θ θ θ

∂= ⋅

∂

= ⋅

∫ ∫ ∫ ∫

∫ ∫

Therefore,

Let’s try one more example! Consider the equationbelow for an ellipse.

2 2

2 2 1x ya b

+ =

The following substitution will change it into anequation for a unit circle.

2 2 2 22 2

2 2 2 2( ) ( )1 1 1x y a s b t s t

a b a b⋅ ⋅

+ = ⇒ + = ⇒ + =

x a sy b t= ⋅= ⋅

Now find the absolute value of the Jacobian.

x a sy b t= ⋅= ⋅

0( , )0( , )

x xax y s t ab

y y bs ts t

∂ ∂∂ ∂ ∂= = =

∂ ∂∂∂ ∂

( , )( , )x y abs t

∂=

∂

We can now easily find the area of the ellipse.

ellipse unit circle

unit circle

Areaof ellipse dA abdsdt

ab dsdt ab abπ π

= =

⎛ ⎞= = ⋅ =⎜ ⎟⎝ ⎠

∫∫ ∫∫

∫∫

If we have a function of three variables, then ourJacobian looks like this.

( , , )( , , )

x x xs t u

x y z y y ys t u s t u

z z zs t u

∂ ∂ ∂∂ ∂ ∂

∂ ∂ ∂ ∂=

∂ ∂ ∂ ∂∂ ∂ ∂∂ ∂ ∂

Find the volume of the ellipsoid.

2 2 2

2 2 2 1x y za b c

+ + =

The coordinate change below transforms the ellipsoid into a unit sphere.

0 0( , , ) 0 0( , , )

0 0

x x xs t u a

x y z y y y b abcs t u s t u

cz z zs t u

∂ ∂ ∂∂ ∂ ∂

∂ ∂ ∂ ∂= = =

∂ ∂ ∂ ∂∂ ∂ ∂∂ ∂ ∂

2 2 2

2 2 2 1x y za b c

+ + =x a sy b tz c u

= ⋅= ⋅= ⋅

( , , )( , , )

4 43 3

R T

T T

x y zVolume dV dsdtdus t u

abcdsdtdu abc dsdtdu

abc abcπ π

∂= =

∂

= =

= ⋅ =

∫∫∫ ∫∫∫

∫∫∫ ∫∫∫

2 2 2

2 2 2 1x y za b c

+ + =

Any Questions?