4.4 Change of Variable in Integrals: The Jaco- bianksuweb.kennesaw.edu/~plaval/math2203/jacobian.pdf · 4.4 Change of Variable in Integrals: The Jaco- ... of integration. ... We usually

232 CHAPTER 4. MULTIPLE INTEGRALS

4.4 Change of Variable in Integrals: The Jaco-bian

In this section, we generalize to multiple integrals the substitution techniqueused with definite integrals. For functions of two or more variables, there is asimilar process we can use. It is a little bit more involved though. In addition,in higher dimensions, a change of variable can also be used to simplify the regionof integration. We begin with an important definition.

4.4.1 The Jacobian

The Jacobian, named after the German mathematician Carl Gustav Jacobi(1804-1851), plays an important role in higher dimensional mathematics. Inthis section, we see how it is used when changing variables to simplify a regionor an integrand. We begin with its definition.

Definition 352 (2-D case) Suppose that x and y are two independent vari-ables which can be expressed in term of two other independent variables u andv by the formula x = g (u, v) and y = h (u, v). The Jacobian of x and y with

respect to u and v, denoted∂ (x, y)

∂ (u, v)or J (u, v), is

J (u, v) =∂ (x, y)

∂ (u, v)=

∣∣∣∣∣∣∣∂x

∂u

∂x

∂v∂y

∂u

∂y

∂v

∣∣∣∣∣∣∣ =∂x

∂u

∂y

∂v− ∂y

∂u

∂x

∂v

Definition 353 (3-D case) Suppose that x, y and z are three independentvariables which can be expressed in term of three other independent variablesu, v and w by the formula x = g (u, v, w), y = h (u, v, w) and z = l (u, v, w).

The Jacobian of x, y and z with respect to u, v and w, denoted∂ (x, y, z)

∂ (u, v, w)or

J (u, v), is

J (u, v) =∂ (x, y, z)

∂ (u, v, w)=

∣∣∣∣∣∣∣∣∣∣

∂x

∂u

∂x

∂v

∂x

∂w∂y

∂u

∂y

∂v

∂y

∂w∂z

∂u

∂z

∂v

∂z

∂w

∣∣∣∣∣∣∣∣∣∣Example 354 Recall, when switching from Cartesian to polar coordinates, wehave x = r cos θ and y = r sin θ. The Jacobian of x and y with respect to r and

4.4. CHANGE OF VARIABLE IN INTEGRALS: THE JACOBIAN 233

θ is

∂ (x, y)

∂ (r, θ)=

∣∣∣∣∣∣∣∂x

∂r

∂x

∂θ∂y

∂r

∂y

∂θ

∣∣∣∣∣∣∣=

∣∣∣∣ cos θ −r sin θsin θ r cos θ

∣∣∣∣= r cos2 θ + r sin2 θ

∂ (x, y)

∂ (r, θ)= r

Example 355 Recall, when switching from Cartesian to Spherical coordinates,we have x = ρ sinφ cos θ, y = ρ sinφ sin θ, and z = ρ cosφ. The Jacobian of x,y, and z with respect to ρ, θ and φ is

∂ (x, y, z)

∂ (ρ, θ, φ)=

∣∣∣∣∣∣∣∣∣∣∣

∂x

∂ρ

∂x

∂θ

∂x

∂φ∂y

∂ρ

∂y

∂θ

∂y

∂φ∂z

∂ρ

∂z

∂θ

∂z

∂φ

∣∣∣∣∣∣∣∣∣∣∣=

∣∣∣∣∣∣sinφ cos θ −ρ sinφ sin θ ρ cosφ cos θsinφ sin θ ρ sinφ cos θ ρ cosφ sin θ

cosφ 0 −ρ sinφ

∣∣∣∣∣∣The easiest way to compute this determinant is to expand it using the third rowsince one of its entries is 0. We obtain

∂ (x, y, z)

∂ (ρ, θ, φ)= cosφ

∣∣∣∣ −ρ sinφ sin θ ρ cosφ cos θρ sinφ cos θ ρ cosφ sin θ

∣∣∣∣− ρ sinφ

∣∣∣∣ sinφ cos θ −ρ sinφ sin θsinφ sin θ ρ sinφ cos θ

∣∣∣∣= cosφ

(−ρ2 sinφ cosφ sin2 θ − ρ2 sinφ cosφ cos2 θ

)−ρ sinφ

(ρ sin2 φ cos2 θ + ρ sin2 φ sin2 θ

)= −ρ2 sinφ cos2 φ− ρ2 sin3 φ

= −ρ2 sinφ

4.4.2 Change of Variable

You will recall in one-dimensional calculus, when given an integral of the form∫g′ (u) f (g (u)) du, we performed the change of variable x = g (u) which gave

us dx = g′ (u) du and thus∫g′ (u) f (g (u)) du =

∫f (x) dx


We can write this slightly differently as follows. Since x = g (u),dx

du= g′ (u)

hence, we have ∫f (x) dx =

∫f (g (u))

dx

dudu

The Jacobian is what generalizesdx

duin the above formula. We begin with

the change of variable theorem for double integrals. We then look at severalexamples to see how one can benefit from a change of variable. These benefitsinclude using a change of variable to simplify an integrand, using a change ofvariable to simplify a region. As an application, we will look at double integralsin polar coordinates. Note that most of the explanations given below will be forregions in the xy-plane or for functions of two variables.

General Case

Let us first introduce some notation. Let R denote a region in the xy-planeand S a region in the uv-plane. A change of variable is usually described as atransformation T from the uv-plane to the xy-plane given by T (u, v) = (x, y)where x and y are given by

x = g (u, v)

y = h (u, v)

We usually assume that the first order partials of g and h are continuous. Whenit is the case we say that T is a C1 transformation. Such a transformationwill map a region S in the uv-plane into another region R into the xy-plane (seefigure 4.4.2). In most cases, we are given R and are looking for a region S anda transformation T from S to R in which S is simpler than R. We will onlyconsider simple cases here.When we talk about the Jacobian of the transformation T , we mean the

Jacobian of the change of variable

x = g (u, v)

y = h (u, v)

We begin with the change of variable theorem for integrals, given withoutproof.

Theorem 356 (Change of Variable in Double Integrals) Let R and S beregions of the xy-planes and uv-planes respectively. Let T : S → R be a C1

transformation such that T (u, v) = (x, y) where

x = g (u, v)

y = h (u, v)


such that each point in R is the image of a unique point in S. If f is continuous

on R and∂ (x, y)

∂ (u, v)6= 0 on R then

∫∫R

f (x, y) dxdy =

∫∫S

f (g (u, v) , h (u, v))

∣∣∣∣∂ (x, y)

∂ (u, v)

∣∣∣∣ dudvRemark 357 It is important to understand that R is given. We must find anappropriate change of variable in which the integral on the right is simpler. Asstated, the theorem is deceiving. It makes it look like the integral on the left iseasier than the integral on the right which appears to contain much more. Butthis is not the case in practice.

Remark 358 As in the one dimensional case, it is important to understandthat when one performs a change of variable, not only the integrand changes,but also the region of integration. As mentioned already, there two possiblereasons for performing a change of variable:

1. To obtain a simpler integrand.

2. To obtain an easier region over which to integrate. The ideal region is arectangle with sides parallel to the coordinate axes. In this case, we canuse Fubini’s theorem.

Remark 359 Deciding what region works best for an integral takes practice.Finding the transformation T which maps one region into another is also veryinvolved and could be the purpose of an entire course. Here, we will only considersimple examples. Problems will fall in two categories. The change of variablewill be suggested or it will not.


Figure 4.21: Region bounded by y = 0, y = 4, y = 2x and y = 2x− 1

Example 360 Evaluate∫ 4

0

∫ y2+1y2

2x− y2

dxdy by applying the transformation

u =2x− y

2and v =

y

2.

The region corresponding to this integral is R ={

(x, y) : 0 ≤ y ≤ 4, y2 ≤ x ≤y2 + 1

}.

This is a type II region. It is shown in figure 4.21. In the equations defining uand v, we need to solve for x and y since we need to compute the Jacobian ofthe transformation giving us xand y in terms of u and v. Simple algebra givesx = u+ v and y = 2v. Therefore

∂ (x, y)

∂ (u, v)=

∣∣∣∣ 1 10 2

∣∣∣∣= 2

We also need to find what the new region, we call S will be. It is enough to findits boundaries. We illustrate how to do this in the table below:

xy equations for the boundary of R uv equations for the boundary of S Simplified uv equationsx = y

2 u+ v = 2v2 = v u = 0

x = y2 + 1 u+ v = 2v

2 + 1 = v + 1 u = 1y = 0 2v = 0 v = 0y = 4 2v = 4 v = 2

Hence S = {(u, v) : 0 ≤ u ≤ 1, 0 ≤ v ≤ 2}. Thisregion is shown in figure 4.22.


Figure 4.22: Region bouded by u = 0, u = 1, v = 0 and v = 2.

We are now ready to apply the change of variable formula.

∫ 4

0

∫ y2+1

y2

2x− y2

dxdy =

∫ 2

0

∫ 1

0

u

∣∣∣∣∂ (x, y)

∂ (u, v)

∣∣∣∣ dudv=

∫ 2

0

∫ 1

0

2ududv

=

∫ 2

0

[u2]∣∣1

0dv

=

∫ 2

0

dv

= 2

Example 361 Evaluate∫ 1

0

∫ 1−x0

√x+ y (y − 2x)

2dydx using a transformation.

First, let us find the region R over which we are integrating. It is R = {(x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1− x}.It is shown in figure 4.23. The integrand suggests the change of variable u = x+yand v = y − 2x. First, we need to solve for x and y. Simple algebra suggests


Figure 4.23: Region bounded by x = 0, y = 0, x+ y = 1.

that x =u− v

3and y =

2u+ v

3. Next, we compute

∂ (x, y)

∂ (u, v).

∂ (x, y)

∂ (u, v)=

∣∣∣∣∣∣∣1

3−1

32

3

1

3

∣∣∣∣∣∣∣=

1

3

Finally, we need to find the corresponding uv region, we call it S. As in theprevious example, we use a table.

xy equations for the boundary of R uv equations for the boundary of S Simplified uv equations

x = 0u− v

3= 0 v = u

y = 02u+ v

3= 0 v = −2u

x+ y = 1u− v

3+

2u+ v

3= 1 u = 1

So, we see that S is the region bounded by u = 0, v = u and v = −2u. It is shownin figure 4.24.We can write it as a type I region. S = {(x, y) : 0 ≤ u ≤ 1,−2u ≤ v ≤ u}.


Figure 4.24: Region bounded by u = 0, u = v and v = −2u.

We are now ready to do the change of variable.∫ 1

0

∫ 1−x

0

√x+ y (y − 2x)

2dydx =

∫ 1

0

∫ u

−2u

√uv2

∣∣∣∣∂ (x, y)

∂ (u, v)

∣∣∣∣ dvdu=

∫ 1

0

[1

3

√uv3

3

]∣∣∣∣u−2u

du

=

∫ 1

0

(u72

9+

8u72

9

)du

=

∫ 1

0

u72 du

=2

9u92

∣∣∣∣10

=2

9

Example 362 Evaluate∫∫R

(x+ y)2dxdy where R is the region bounded by

x+ y = 0, x+ y = 1, 2x− y = 0, y = 3 by first performing a change of variablewhich maps a region S into R where S is a rectangle in the uv-plane.


This region R is shown in figure 362. You will note that R is a parallelogram,that is opposite sides are parallel. This can be seen on figure 362 as well as inthe equations of R. The variable part of these equations fall in two categories:x+ y and 2x− y. So, we only need to set u and v to these and solve for x. Welet u = x+ y and v = 2x− y and solve for x and y.

{u = x+ yv = 2x− y

Solving for y in both equations gives y = u− x and y = 2x− v thus

u− x = 2x− v ⇐⇒ u+ v = 3x

⇐⇒ x =u+ v

3

Using one of the expressions we got for y gives

y = u− x

= u− u+ v

3

=2u− v

3

We need to find the corresponding uv region we will call S . As before, we do itwith a table.

xy equations for the boundary of R uv equations for the boundary of S Simplified uv equationsx+ y = 0 u = 0 u = 0x+ y = 1 u = 1 u = 12x− y = 0 v = 0 v = 02x− y = 3 v = 3 v = 3

So S = {(x, y) : 0 ≤ u ≤ 1and0 ≤ v ≤ 3}. It is a rectangle shown in figure 362.The transformation T (or change of variable) which takes points (u, v) in S topoints (x, y) in R is T (u, v) = (x, y) where

x =u

3+v

3

y =2u

3− v

3


The Jacobian for this transformation is

∂ (x, y)

∂ (u, v)=

∣∣∣∣∣∣∣∂x

∂u

∂x

∂v∂y

∂u

∂y

∂v

∣∣∣∣∣∣∣=

∣∣∣∣∣∣∣1

3

1

32

3

−1

3

∣∣∣∣∣∣∣=−1

9− 2

9

= −1

3

Using the change of variable theorem ,we have

∫∫R

(x+ y)2dxdy =

∫∫S

f (g (u, v) , h (u, v))

∣∣∣∣∂ (x, y)

∂ (u, v)

∣∣∣∣ dudv=

∫∫S

u2

3dudv


Since S is a rectangular region, we can use Fubini’s theorem to get∫∫R

(x+ y)2dxdy =

∫∫S

u2

3dudv

=1

3

∫∫S

u2dudv

=1

3

∫ 3

0

dv

∫ 1

0

u2du

=1

3(3)

(1

3

)=

1

3


(x+ y)2

sin(

2x− y)dA if R is the region bounded

by the square with vertices (0, 1), (1, 2), (2, 1) and (1, 0)The region R is shown in figure 363. Note that the sides of R lie on the linesx− y = 1, x− y = −1, x+ y = 1 and x+ y = 3. As above, we let u = x+ y andv = x− y. It is easy to determine the bounds of u and v for the S region. Fromthe boundaries of R, we see that S = {(u, v) : 1 ≤ u ≤ 3 and − 1 ≤ v ≤ 1}. We


need to express x and y in terms of u and v so we can compute the Jacobian.Since {

u = x+ yv = x− y

solving for y gives us y = u− x and y = x− v thus

u− x = x− v ⇐⇒ u+ v = 2x

⇐⇒ x =u

2+v

2

replacing in one of the expressions for y gives

y = u− x= u− u

2− v

2

=u

2− v

2

Thus

∂ (x, y)

∂ (u, v)=

∣∣∣∣∣∣∣∂x

∂u

∂x

∂v∂y

∂u

∂y

∂v

∣∣∣∣∣∣∣=

∣∣∣∣∣∣∣1

2

1

21

2−1

2

∣∣∣∣∣∣∣=−1

4− 1

4

=−1

2

Using the change of variables theorem, we see that∫∫R

(x+ y)2

sin2 (x− y) dA =

∫∫S

f (g (u, v) , h (u, v))

∣∣∣∣∂ (x, y)

∂ (u, v)

∣∣∣∣ dudv=

∫∫S

u2 sin2 v

(1

2

)dudv

Since S is a rectangular region, by Fubini’s theorem, we get∫∫R

(x+ y)2

sin2 (x− y) dA =

∫∫S

u2 sin2 v

(1

2

)dudv

=1

2

∫ 1

−1

∫ 3

1

u2 sin2 v

(1

2

)dudv

=1

2

∫ 1

−1

sin2 vdv

∫ 3

1

u2du


We can do each integral separately.∫ 3

1

u2du =u3

3

∣∣∣∣31

= 9− 1

3

=26

3

and, using the fact that sin2x =1− cos 2x

2∫ 1

−1

sin2 vdv =1

2

∫ 1

−1

(1− cos 2v) dv

=1

2

(v − 1

2sin 2v

)∣∣∣∣1−1

=1

2

((1− 1

2sin 2

)−(−1− 1

2sin (−2)

))= 1− 1

2sin 2

Therefore ∫∫R

(x+ y)2

sin2 (x− y) dA =1

2

(26

3

)(1− 1

2sin 2

)≈ 2. 363 2

Application of Change of Variable to Polar Coordinates

You will recall that the change of variable from Cartesian to polar coordinatesis x = r cos θ and y = r sin θ. Also, from example 354, the Jacobian of this

transformation is∂ (x, y)

∂ (r, θ)= r. Therefore, from the change of variable theorem,

we have ∫∫R

f (x, y) dxdy =

∫∫S

f (r cos θ, r sin θ)

∣∣∣∣∂ (x, y)

∂ (r, θ)

∣∣∣∣ drdθ=

∫∫S

f (r cos θ, r sin θ) rdrdθ

where S is the region in the rθ-plane corresponding to R in the xy-plane.




ex2+y2dxdy where R is the upper half portion of

the unit circle.The region R can be written as R =

{(x, y) : −1 ≤ x ≤ 1 and 0 ≤ y ≤

√1− x2

}.

The corresponding region S is S = {(r, θ) : 0 ≤ r ≤ 1 and 0 ≤ θ ≤ π}. There-fore

∫∫R

ex2+y2dxdy =

∫∫S

rer2

drdθ

∫ π

0

∫ 1

0

rer2

drdθ

We can evaluate the inner integral using the substitution u = r2 hence du = 2rdrthus

∫ 1

0

rer2

dr =1

2

∫ 1

0

eudu

=1

2(e− 1)

You will note that in the process, we updated the limits of integration. It doesnot sow because they end up being the same. But make sure not to forget to doit. The original limits were 0 and 1, they were limits for r. When we substituteand use u instead, we must find the limits on u. Since u = r2 when r = 0,u = 02 = 0 and when r = 1, u = 12 = 1. We can now finish the integral.

∫∫R

ex2+y2dxdy =

∫ π

0

1

2(e− 1) dθ

=π

2(e− 1)


(x2 + y

)dA where R is the annular region between

the two circles x2 + y2 = 1 and x2 + y2 = 5. The corresponding region S is

S ={

(r, θ) | 1 ≤ r ≤√

5 and 0 ≤ θ ≤ 2π}


Therefore ∫∫R

(x2 + y

)dA =

∫ 2π

0

∫ √5

1

(r2 cos2 θ + r sin θ

)rdrdθ

=

∫ 2π

0

∫ √5

1

(r3 cos2 θ + r2 sin θ

)drdθ

=

∫ 2π

0

(r4

4cos2 θ +

r3

3sin θ

)∣∣∣∣√

5

1

dθ

=

∫ 2π

0

(6 cos2 θ +

5√

5− 1

3sin θ

)dθ

Using the fact that cos2 θ =1 + cos 2x

2, we have

∫∫R

(x2 + y

)dA =

∫ 2π

0

(3 + 3 cos 2θ +

5√

5− 1

3sin θ

)dθ

= 6π

4.4.3 Assignment

1. This section contains more material than what you need to know for thefinal. I want you to know the following:

(a) The Jacobian. Given a transformation, be able to find the Jacobiancorresponding to it.

(b) Being able to evaluate integrals using polar coordinates.

2. Solve the system{

u = x− yv = 2x+ y

for x and y then find∂ (x, y)

∂ (u, v).

3. Solve the system{u = x+ 2yv = x− y for x and y then find

∂ (x, y)

∂ (u, v).

4. Rewrite the given integrals from Cartesian to polar coordinates, then eval-uate the integral.

(a)∫ 1

−1

∫√1−x20

dydx

(b)∫ 1

0

∫√1−x20

(x2 + y2

)dydx

(c)∫ a−a∫√a−x2−√a−x2 dydx

(d)∫ 6

0

∫ y0xdxdy

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4.4 Change of Variable in Integrals: The Jaco- bianksuweb.kennesaw.edu/~plaval/math2203/jacobian.pdf · 4.4 Change of Variable in Integrals: The Jaco- ... of integration. ... We usually