MA1104 Week 5
Double Integral over region on the xy-plane
1. Double Integral over Rectangle
In this section, we shall introduce double integral of a two-variable function
f (x, y) over a rectangle R. We start by reviewing how we arrive at the
definite integral of functions of a single variable:
Z b
f (x) dx
a
This is motivated by the problem of finding area under the curve y = f (x)
(assuming f (x) ≥ 0) above the x-axis.
The problem can be solved using the following procedures:
Step 1. Suppose f (x) is defined for a ≤ x ≤ b. We divide the interval [a, b] into
n subintervals of equal size 4x = b−a
.
n
Step 2. We choose sample points x∗i from these subintervals and form the Riemann Sum
n
X
f (x∗i )4x.
i=1
1
Step 3. Take the limit of such sum as n → ∞ to obtain the definite integral of
f from a to b:
Z b
f (x)dx = lim
n→∞
a
n
X
f (x∗i )4x.
i=1
To define double integral, we suppose f (x, y) is a function of two variables
defined on a closed rectangle
R = [a, b] × [c, d]
= {(x, y) ∈ R2 : a ≤ x ≤ b, c ≤ y ≤ d}.
For convenience, we suuppose f (x, y) ≥ 0.
The graph of f is a surface with z = f (x, y) above the region R.
Let S be the solid that lies above R and under the graph of f . How can we
find the volume of S?
This can be solved using the following procedures:
Step 1. Divide the rectangle R into subrectangles.
• dividing the interval [a, b] into m subintervals [xi−1 , xi ] of equal
length 4x = b−a
, and
m
• dividing the interval [c, d] into n subintervals [yj−1 , yj ] of equal
length 4y = d−c
.
n
2
Form subrectangles
Rij = [xi−1 , xi ] × [yj−1 , yj ],
for all 1 ≤ i ≤ m, 1 ≤ j ≤ n.
Each of these subrectangles has area 4A = 4x4y.
Step 2. Choose a sample point (x∗ij , yij∗ ) in each Rij .
Then approximate the part of S lies above Rij by a thin rectangle box
with base Rij and height f (x∗ij , yij∗ ). The volume of this box is given by
f (x∗ij , yij∗ )4A.
3
It follows that by adding the volumes of all these thin boxes, we get an
approximation of the total volume of S:
V ≈
n
m X
X
f (x∗ij , yij∗ )4A.
i=1 j=1
Step 3: This approximation becomes better as m, n → ∞. So we would expect
V = lim
m,n→∞
m X
n
X
f (x∗ij , yij∗ )4A.
i=1 j=1
In general, we define
ZZ
f (x, y) dA = lim
R
m,n→∞
4
m X
n
X
i=1 j=1
f (x∗ij , yij∗ )4A.
Definition 1 (Double Integral over Rectangle). The double integral of f
over the rectangle R is
ZZ
f (x, y) dA = lim
R
m,n→∞
m X
n
X
f (x∗ij , yij∗ )4A
i=1 j=1
provided the limit exists and is the same for any choice of the sample points
(x∗ij , yij∗ ) in Rij , for 1 ≤ i ≤ m, 1 ≤ j ≤ n.
2. Double Integral over General Region
In the last section, we have seen the definition of double integral over rectangle. How do we define double integrals over regions D of more general shape
like the one shown below?
First, we enclose it in a rectangle R as shown:
5
Next, we define a new function:
f (x, y) if (x, y) ∈ D
F (x, y) =
0
if (x, y) ∈ R \ D
Definition 2 (Double Integral over General Region). We define the double
integral of f over general region D by
ZZ
ZZ
f (x, y) dA =
F (x, y) dA.
D
R
The above makes sense since roughly we have
ZZ
XX
F (x, y)4A
F (x, y) dA ≈
R
| {z }
R
XX
XX
F (x, y)4A +
F (x, y)4A
≈
| {z }
| {z }
D
R\D
≈ 0+
XX
f (x, y)4A
| {z }
D
ZZ
≈
f (x, y, ) dA.
D
As in the case of a rectangle, we can interpret a double integral as volume
when f (x, y) ≥ 0 for all (x, y) ∈ D.
6
Theorem 3 (Volume as a Double Integral). If f (x, y) ≥ 0, the the volume
V of the solid that lies above the region D and below the surface z = f (x, y)
is
ZZ
V =
f (x, y) dA.
D
We do ourselves no favour by only thinking double integrals as volumes.
There are many other interpretations depending on the units we give to x, y
as well as f (x, y). Below are two different interpretations:
Double Integral as Mass of a plane region D:
• f (x, y) is the mass density in kg per meter2
• x and y are measured in meters.
ZZ
• The unit for
f (x, y) dA is kg, and it is the total mass of the region
D.
D
Double Integral as Probability of the event D:
• f (x, y) is the joint probability density for the random variable x and y
7
ZZ
• Then
f (x, y) dA is the probability of the event D
D
Below are some basic properties of double integrals that we use sometimes
even without mention. Assuming all the integrals exist, we have
ZZ
(f (x, y) + g(x, y)) dA
D
ZZ
ZZ
=
f (x, y) dA +
g(x, y) dA.
D
D
ZZ
ZZ
cf (x, y) dA = c
D
f (x, y) dA.
D
3. Iterated Double Integral
Although a double integral is the limit of double Riemann sums, in practice,
we almost never compute any double integrals by the first principle!
In fact, we compute double integrals by means of iterated integrals, which
we now define.
The iterated double integral of f on the rectangle R = [a, b] × [c, d] in the
order dy dx is defined to be
Z b Z d
f (x, y) dy dx.
a
c
We must compute the inner integral first, followed by the outer integral.
Z d
• The inner integral
f (x, y) dy is integrated with respect to y from c
c
to d by holding x fixed.
Rd
• In general the inner integral c f (x, y) dy is a function of x only, so the
outer integral with respect to x makes sense.
8
Usually, we omit the brackets and just write:
Z b Z d
Z bZ d
f (x, y) dy dx =
f (x, y) dy dx.
a
c
a
c
Similarly, the iterated double integral of f on the rectangle R = [a, b] × [c, d]
in the order dx dy is defined to be
Z d Z b
f (x, y) dx dy.
c
a
Z b
• The inner integral
f (x, y) dx is integrated with respect to x from a
a
to b by holding y fixed.
Rb
• In general the inner integral a f (x, y) dx is a function of y only, so the
outer integral with respect to y makes sense.
Again, we usually omit the brackets and just write:
Z d Z b
Z dZ b
f (x, y) dx dy.
f (x, y) dx dy =
c
c
a
a
Example 1. Evaluate the iterated integral
Z 2Z 3
x2 y dx dy.
1
0
Solution.
9
Z 2Z 3
Z 2 Z 3
2
x y dx dy =
1
0
1
x y dx dy
2
0
Z 2 3 3
xy
dy
=
3 0
1
Z 2
9y dy
=
1
=
2
9y 2
2 1
=
27
.
2
Example 2. Evaluate the iterated integral
Z 3Z 2
x2 y dy dx.
0
1
Solution.
Z 3Z 2
Z 3 Z 2
2
x y dy dx =
0
1
2
x y dy dx
0
1
Z 3 2 2 2
xy
=
dx
2 1
0
Z 3
3 2
x dx
=
0 2
3 3
x
=
2 0
=
27
.
2
10
Notice in both of the preceding examples, we obtained the same answer.
It seems that the order of integration (with respect to x or y first) does not
matter. Indeed, the fact that this is always the case is given by Fubini’s
Theorem
Theorem 4 (Fubini’s Theorem). If f is continuous on the rectangle R =
[a, b] × [c, d], then
ZZ
Z bZ d
f (x, y) dA =
f (x, y) dy dx
R
a
c
Z dZ b
f (x, y) dx dy.
=
c
a
Example 3. Evaluate
ZZ
y sin(xy) dA
R
where R = [1, 2] × [0, π].
Solution. Using Fubini’s Theorem, we can find the double integral using the
iterated integral in the order
• dx dy, or
• dy dx.
Using the order dx dy:
Z πZ 2
ZZ
y sin(xy) dA =
y sin(xy) dx dy
R
0
Z π
=
1
[− cos(xy)]21 dy
0
Z π
=
(− cos 2y + cos y) dy
0
=
π
1
− sin 2y + sin y = 0.
2
0
11
Using the order dy dx:
ZZ
Z 2Z π
y sin(xy) dy dx.
y sin(xy) dA =
1
R
0
It seems a bit tedious to compute the inner integral:
Z π
y sin(xy) dy.
0
Using integration by parts,
Z π
y sin(xy) dy
0
=
π Z π
y cos(xy)
cos(xy)
−
dy
−
−
x
x
0
0
π cos πx
1
+ 2 [sin xy]π0
x
x
π cos πx sin πx
= −
+
.
x
x2
= −
Substituting the inner integral:
Z 2
ZZ
π cos πx sin πx
+
y sin(xy) dA =
−
dx.
x
x2
1
R
Now, integrating the first term by parts, we have
Z
Z sin πx
sin πx
π cos πx dx = −
−
dx
−
x
x
x2
Rearranging:
Z π cos πx sin πx
sin πx
dx = −
+
.
−
2
x
x
x
Hence
12
ZZ
y sin(xy) dA =
R
Z 2Z π
y sin(xy) dy dx =
1
0
2
sin πx
−
x
1
= −
sin 2π
+ sin π = 0.
2
Here are some observations about the last example:
(1) Though both solution gives the same answer, the first solution is much
easier than the second one.
Therefore, when we evaluate double integrals, it is wise to choose the right order of
integration that yields simpler calculations.
Notice that the function f (x, y) = y sin(xy) takes both positive and negative
values on R = [1, 2] × [0, π].
13
(2) We can think of
RR
R
f (x, y) dA as a difference of volumes:
V1 − V2 where
• V1 is the volume above R and below the graph of f .
• V2 is the volume below R and above the graph.
4. Double Integral over Type I & Type II Regions
Definition 5 (Type I Region). A plane region D is said to be of Type I if
it lies between the graphs of two continuous functions of x, that is,
D = {(x, y) : a ≤ x ≤ b, g1 (x) ≤ y ≤ g2 (x)}
where g1 (x) and g2 (x) are continuous on [a, b].
Some examples of Type I region:
14
Here is one example which is NOT Type I:
Definition 6 (Type II Region). A plane region D is said to be of Type II if
it lies between the graphs of two continuous functions of y, that is,
D = {(x, y) : c ≤ y ≤ d, h1 (y) ≤ x ≤ h2 (y)}
where h1 (y) and h2 (y) are continuous on [c, d].
15
Some examples of Type II region:
16
Here is one example which is NOT Type II:
Theorem 7 (Double Integral over Type I Domain). If f is continuous on a
Type I domain D such that
D = {(x, y) : a ≤ x ≤ b, g1 (x) ≤ y ≤ g2 (x)}
then
Z bZ g2 (x)
ZZ
f (x, y) dy dx.
f (x, y) dA =
a g1 (x)
D
Z bZ g2 (x)
ZZ
f (x, y) dA =
D
f (x, y) dy dx
a g1 (x)
In the inner integral we regard x as being constant not only in f (x, y) but
also in the limits of the integration, g1 (x) and g2 (x).
Theorem 8 (Double Integral over Type II Domain). If f is continuous on
a Type II domain D such that
D = {(x, y) : c ≤ y ≤ d, h1 (y) ≤ x ≤ h2 (y)}
then
Z dZ h2 (y)
ZZ
f (x, y) dA =
D
f (x, y) dx dy.
c
17
h1 (y)
Z dZ h2 (y)
ZZ
f (x, y) dA =
D
f (x, y) dx dy
c
h1 (y)
In the inner integral we regard y as being constant not only in f (x, y) but
also in the limits of the integration, h1 (y) and h2 (y).
5. Example on Double Integral
ZZ
(x + 2y) dA where D is the region bounded by
Example 4. Evaluate
D
• the parabola y = 2x2 , and
• the parabola y = 1 + x2 .
Solution.
Step 1. Identify the region.
We note that D is a Type I region:
D = {(x, y) : −1 ≤ x ≤ 1, 2x2 ≤ y ≤ 1 + x2 }.
18
Step 2. Set up the iterated integral.
Z 1 Z 1+x2
ZZ
(x + 2y) dy dx.
(x + 2y) dA =
−1
D
2x2
Step 3. Evaluate the inner integral.
Z 1+x2
(x + 2y) dy
2x2
=
y=1+x2
xy + y 2 y=2x2
= x(1 + x2 ) + (1 + x2 )2 − x(2x2 ) − (2x2 )2
= −3x4 − x3 + 2x2 + x + 1
Step 4. Complete the computation.
ZZ
(x + 2y) dA
D
Z 1
=
(−3x4 − x3 + 2x2 + x + 1) dx
−1
1
x5 x4
x3 x2
= −3 −
+2 +
+x
5
4
3
2
−1
32
=
.
15
TIPS:
For Type I region, it is helpful to draw a vertical arrow which starts at the
lower boundary y = g1 (x) and ends at the upper boundary y = g2 (x). This
corresponds to the inner integral.
19
For Type II region, it is helpful to draw a horizontal arrow which starts
at the left boundary x = h1 (y) and ends at the right boundary x = h2 (y).
This corresponds to the inner integral.
6. More Example on Double Integral
ZZ
Example 5. Evaluate
xy dA where D is the region bounded by
D
• the line y = x − 1 and
• the parabola y 2 = 2x + 6.
Solution.
The region D can be of Type I or II:
But we prefer D as Type II (Why?):
D = {(x, y) : −2 ≤ y ≤ 4,
1 2
y − 3 ≤ x ≤ y + 1}.
2
Z 4 Z y+1
ZZ
xy dA =
D
xy dx dy
−2
1 2
y −3
2
Z 4
y5
− + 4y 3 + 2y 2 − 8y
4
−2
6
4
1
y
y3
4
2
=
− + y + 2 − 4y
2
24
3
−2
= 36.
1
=
2
20
dy
Example 6. Find the volume of the tetrahedron T bounded by the planes
• x + 2y + z = 2,
• x = 2y,
• x = 0 and
• z = 0.
Solution. The required volume V lies under the graph z = 2 − x − 2y and
above
D = {(x, y) : 0 ≤ x ≤ 1,
21
x
x
≤ y ≤ 1 − }.
2
2
ZZ
(2 − x − 2y) dA
V olume =
D
Z 1 Z 1−x/2
(2 − x − 2y) dy dx
=
0
x/2
Z 1
=
Z0 1
y=1−x/2
2y − xy − y 2 y=x/2 dx
(x2 − 2x + 1) dx
0
3
1
x
2
=
−x +x
3
0
1
=
.
3
=
7. Additivity & Area of a Plane Region
If D is the union of domains D1 , . . . , Dn that do not overlap except possibly
on boundary curves, then
Theorem 9 (Additivity With Respect to Domain).
ZZ
f (x, y) dA
D
ZZ
ZZ
f (x, y) dA + · · · +
=
D1
f (x, y) dA.
Dn
When do we apply the above property of additivity?
If D is neither of Type I nor II, we hope to decompose it into finitely many
domains of Type I or II.
22
We can use double integral to compute area of a region D on the plane:
Theorem 10 (Area of plane region). Let f (x, y) = 1 over a given region D.
Then the area of D is
ZZ
1 dA.
A(D) =
D
Proof.
ZZ
1 dA is the volume of the solid which is a cylinder whose base is A(D)
D
and height 1.
Another way of computing the volume of a cylinder is
area of base × height
which is
A(D) · 1
in this case.
Together, we have
ZZ
A(D) =
1 dA,
D
23
as required.
Example 7. Find the area of the following region D on the xy-plane:
Solution. We partition the region D into D1 and D2 as follows:
We can describe both D1 and D2 as Type II regions:
√
y − 1 ≤ x ≤ y − y3}
D1 = {(x, y) : 0 ≤ y ≤ 1,
D2 = {(x, y) : −1 ≤ y ≤ 0,
− 1 ≤ x ≤ y − y3}
ZZ
Area(D) =
1 dA
D
ZZ
ZZ
=
1 dA +
D1
24
1 dA
D2
Z 1 Z y−y3
ZZ
1 dA =
√
0
D1
Z 1
=
1 dx dy
y−1
√
(y − y 3 −
y + 1) dy
0
1
2
y 4 y 3/2
y
−
−
+y
=
2
4
3/2
0
7
=
.
12
Z 0 Z y−y3
ZZ
1 dA =
1 dx dy
−1
Z 0
D2
=
−1
(y − y 3 + 1) dy
−1
2
0
y
y4
=
−
+y
2
4
−1
3
=
.
4
ZZ
Area(D) =
1 dA
D
ZZ
ZZ
=
1 dA +
D1
=
7
3
+
12 4
=
4
.
3
1 dA
D2
25
MA1104 Week 6
Double Integral over Polar Regions &
Triple Integrals
1. Polar Regions
Any point on the xy-plane can be represented by an ordered pair (r, θ) where
• r is the distance from the origin to the point
• θ is the angle from the positive x-axis to the straight line joining the
origin and the point.
The following figure shows the relationship between polar coordinates and
the rectangle coordinates:
Theorem 1 (Relationship between (r, θ) and (x, y)).
x = r cos θ,
r=
p
x2 + y 2 ,
y = r sin θ.
y
θ = tan−1 ,
x
1
provided x 6= 0.
Some examples of regions that are given in terms of its polar coordinates:
2. Double Integral over Polar Rectangles
In this section, we shall look at the formula for computing double integral
over a polar rectangle. But first, let us see what we mean by a polar rectangle.
2
Definition 2 (Polar Rectangle). A polar rectangle is a region
R = {(r, θ) : a ≤ r ≤ b, α ≤ θ ≤ β}.
The goal is to compute
ZZ
f (x, y) dA
R
where R is a polar rectangle
R = {(r, θ) : a ≤ r ≤ b, α ≤ θ ≤ β}.
We partition the polar rectangle R as follows:
By definition, we have
ZZ
XX
f (x, y) dA ≈
f (ri∗ cos θj∗ , ri∗ sin θj∗ )Area(Rij )
R
i
j
3
How do we calculate Area(Rij )?
The area of the ‘little polar rectangle’ is approximately
4r(r4θ).
Hence
ZZ
f (x, y) dA ≈
XX
f (ri∗ cos θj∗ , ri∗ sin θj∗ ) 4r(r4θ).
R
≈
XX
f (ri∗ cos θj∗ , ri∗ sin θj∗ )r 4r4θ.
Z βZ b
=
f (r cos θ, r sin θ)r dr dθ.
α
a
Theorem 3 (Change to Polar Coordinates in Double Integral). If f is continuous on a polar rectangle R given by
R = {(r, θ) : 0 ≤ a ≤ r ≤ b, α ≤ θ ≤ β}
where 0 ≤ β − α ≤ 2π, then
ZZ
Z βZ b
f (x, y) dA =
f (r cos θ, r sin θ)r dr dθ.
R
α
a
4
To summarize: When converting to polar coordinates,
• write x = r cos θ, y = r sin θ
• use the appropriate limits of integration for r and θ
• replace dA by r dr dθ (do not forget the additional r in r dr dθ)
3. Examples: Double Integral over Polar Rectangles
ZZ
(3x + 4y 2 ) dA where R is the region in the upper
Example 1. Evaluate
R
half-plane bounded by the circles
• x2 + y 2 = 1 and
• x2 + y 2 = 4.
Solution. The region R is shown below:
Hence
ZZ
(3x + 4y 2 ) dA
R
5
Z πZ 2
=
Z0 π Z1 2
=
0
Z π
=
0
(3r cos θ + 4r2 sin2 θ)r dr dθ
(3r2 cos θ + 4r3 sin2 θ) dr dθ
1
3
r=2
r cos θ + r4 sin2 θ r=1 dθ
Z π
(7 cos θ + 15 sin2 θ) dθ
Z0 π 15
=
7 cos θ + (1 − cos 2θ) dθ
2
0
π
15θ 15
−
sin 2θ
= 7 sin θ +
2
4
0
15π
=
.
2
=
Example 2. Find the volume of the solid bounded by the plane z = 0 and
the paraboloid z = 1 − x2 − y 2 .
Solution. The solid lies under the paraboloid and above the circular disk D
given by x2 + y 2 ≤ 1.
6
In polar coordinates, D is given by
D = {(r, θ) : 0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π}.
ZZ
(1 − x2 − y 2 ) dA
Volume =
D
Z 2π
Z 1
(1 − r2 )r dr dθ
0Z 2π 0 Z 1
3
(r − r ) dr
dθ
=
=
0
0
2
1
r
π
r4
= 2π
= .
−
2
4 0
2
7
4. Double Integral over General Polar Regions
In practice, polar rectangles can be restrictive. In this section, we want to define double integrals over more general regions in terms of polar coordinates.
These regions come in two different forms:
D = {(r, θ) : a ≤ r ≤ b, g1 (r) ≤ θ ≤ g2 (r)}.
D = {(r, θ) : α ≤ θ ≤ β, h1 (θ) ≤ r ≤ h2 (θ)}.
8
Theorem 4 (Polar Regions I). If f is continuous on a polar region D given
by
D = {(r, θ) : 0 ≤ a ≤ r ≤ b, g1 (r) ≤ θ ≤ g2 (r)}
then
Z b Z g2 (r)
ZZ
f (x, y) dA =
D
f (r cos θ, r sin θ)r dθ dr
a
g1 (r)
Theorem 5 (Polar Regions II). If f is continuous on a polar region D given
by
D = {(r, θ) : α ≤ θ ≤ β, h1 (θ) ≤ r ≤ h2 (θ)}.
Z β Z h2 (θ)
ZZ
f (r cos θ, r sin θ)r dr dθ
f (x, y) dA =
D
α
h1 (θ)
5. Examples: Double Integral over General Polar Regions
Example 3. Find the volume of the solid that lies under the paraboloid z =
x2 + y 2 , above the xy-plane, and inside the cylinder x2 + y 2 = 2x.
Solution.
9
The solid lies above the disk D whose boundary equation is
r = 2 cos θ.
D = {(r, θ) :
−
π
π
≤ θ ≤ , 0 ≤ r ≤ 2 cos θ}.
2
2
ZZ
(x2 + y 2 ) dA
Volume =
D
Z π/2 Z 2 cos θ
=
−π/2
r2 r drdθ
0
Z π/2 4 2 cos θ
r
=
dθ
4 0
−π/2
Z π/2
=
−π/2
10
4 cos4 θ dθ
2
Z π/2 cos 2θ + 1
4
dθ
Volume =
2
−π/2
Z π/2
cos2 2θ + 2 cos 2θ + 1 dθ
=
−π/2
Z π/2 =
−π/2
1
Volume =
2
cos 4θ + 1
+ 2 cos 2θ + 1
2
dθ
Z π/2
(cos 4θ + 4 cos 2θ + 3) dθ
−π/2
π/2
1 sin 4θ
sin 2θ
=
+4
+ 3θ
2
4
2
−π/2
=
3π
.
2
6. Triple Integral over Rectangular Box
In this section, we shall introduce triple integral of three-variable functions
f (x, y, z). Just like double integrals, we first define it over regions where each
of the underlying variables are bounded by constants.
Consider a rectangular box:
B = {(x, y, z) : a ≤ x ≤ b, c ≤ y ≤ d, r ≤ z ≤ s}.
We want to define:
11
ZZZ
f (x, y, z) dV.
B
We achieve this by the following steps:
Step 1. Divide B into sub-boxes by:
• dividing the interval [a, b] into l subintervals [xi−1 , xi ] of equal length
4x
• dividing the interval [c, d] into m subintervals [yj−1 , yj ] of equal length
4y
• dividing the interval [r, s] into n subintervals [zk−1 , zk ] of equal length
4z
For every index i, j, k, we denote the corresponding sub-box by Bink :
12
Step 2. Form the Riemann (triple) sum:
l X
m X
n
X
∗
∗
f (x∗ijk , yijk
, zijk
)4V
i=1 j=1 k=1
∗
∗
) is in Bijk .
, zijk
where the sample point (x∗ijk , yijk
Step 3. Take limit:
The triple integral of f over the box B is defined to be
ZZZ
f (x, y, z) dV
B
=
l X
m X
n
X
lim
l,m,n→∞
∗
∗
f (x∗ijk , yijk
, zijk
)4V
i=1 j=1 k=1
if the limit exists and is independent of the choice of the partition and the
sample points.
Theorem 6 (Fubini’s Theorem for Triple Integral). If f is continuous on the
rectangular box B = [a, b] × [c, d] × [r, s] then
Z sZ dZ b
ZZZ
f (x, y, z) dx dy dz.
f (x, y, z) dV =
r
B
c
a
Furthermore, the iterated integral may be evaluated in any order
ZZZ
xyz 2 dV where B is the rect-
Example 4. Evaluate the triple integral
B
angular box
B = {(x, y, z) : 0 ≤ x ≤ 1, − 1 ≤ y ≤ 2, 0 ≤ z ≤ 3}.
Solution. We choose to integrate with respect to x, then y and then z:
ZZZ
2
xyz dV
B
Z 3Z 2 Z 1
=
0
13
−1
0
xyz 2 dx dy dz
=
=
=
=
=
=
Z 3 Z 2 2 2 x=1
x yz
dy dz
2
0
−1
x=0
Z 3Z 2 2
yz
dy dz
0
−1 2
Z 3 2 2 y=2
y z
dz
4 y=−1
0
Z 3 2
3z
dz
4
0
3 3
z
4 0
27
.
4
7. Triple Integral over Type 1, 2, 3 Regions
Rectangular boxes are too simple for many practical purposes. We shall now
consider more general solids in space, known here as Type 1, Type 2 or Type
3 regions. These regions are quite similar to each other. In fact, they are
equivalent to each other, up to relabellings of the underlying variables.
Definition 7 (Type 1 Region). A solid region E is of Type 1 if it lies between
the graphs of two continuous functions of x and y, that is
E = {(x, y, z) : (x, y) ∈ D, u1 (x, y) ≤ z ≤ u2 (x, y)}
where D is the projection of E onto the xy-plane.
We can think of the projection D as the “shadow” of the solid E when
viewing the object from the top. This way, the bottom surface of E is the
surface z = u1 (x, y) whereas the top surface of E is z = u2 (x, y).
Below is an example of a Type 1 region:
14
To compute a triple integral over a Type 1 region, we first integrate the
function with respect to z from u1 (x, y) to u2 (x, y), followed by the double
integral over the projection D.
Theorem 8 (Triple Integral over Type 1 region E).
Z Z "Z
ZZZ
u2 (x,y)
f (x, y, z) dz dA.
f (x, y, z) dV =
E
#
D
u1 (x,y)
Definition 9 (Type 2 Region). A solid region E is of Type 2 if it lies between
the graphs of two continuous functions of y and z, that is
E = {(x, y, z) : (y, z) ∈ D, u1 (y, z) ≤ x ≤ u2 (y, z)}
where D is the projection of E onto the yz-plane.
We can think of the projection D as the “shadow” of the solid E when
viewing the object from the front. This way, the back surface of E is the
surface x = u1 (y, z) whereas the front surface of E is x = u2 (y, z).
Below is an example of a Type 2 region:
15
To compute a triple integral over a Type 2 region, we first integrate the
function with respect to x from u1 (y, z) to u2 (y, z), followed by the double
integral over the projection D.
Theorem 10 (Triple Integral over Type 2 region E).
Z Z "Z
ZZZ
u2 (y,z)
f (x, y, z) dx dA.
f (x, y, z) dV =
E
#
D
u1 (y,z)
Definition 11 (Type 3 Region). A solid region E is of Type 3 if it lies
between the graphs of two continuous functions of x and z, that is
E = {(x, y, z) : (x, z) ∈ D, u1 (x, z) ≤ y ≤ u2 (x, z)}
where D is the projection of E onto the xz-plane.
We can think of the projection D as the “shadow” of the solid E when
viewing the object from the right. This way, the left surface of E is the
surface y = u1 (x, z) whereas the right surface of E is x = u2 (y, z).
Below is an example of a Type 3 region:
16
To compute a triple integral over a Type 3 region, we first integrate the
function with respect to y from u1 (x, z) to u2 (x, z), followed by the double
integral over the projection D.
Theorem 12 (Triple Integral over Type 3 region E).
ZZZ
Z Z "Z
u2 (x,z)
f (x, y, z) dV =
E
#
f (x, y, z) dy dA.
D
u1 (x,z)
8. Examples: Triple Integral over Type 1, 2, 3 Regions
ZZZ
Example 5. Evaluate
by the four planes
z dV where E is the solid tetrahedron bounded
E
• x = 0,
• y = 0,
• z = 0 and
• x + y + z = 1.
17
Solution.
The solid E can be described as a Type 1 region:
E = {(x, y, z) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 − x, 0 ≤ z ≤ 1 − x − y}.
Z Z Z 1−x−y
ZZZ
z dV
E
=
z dz dA
Z 1−x−y
=
z dz dy dx
0
0
0
Z 1 Z 1−x 2 z=1−x−y
z
=
dy dx
2 z=0
0
0
Z Z 1−x
1 1
=
(1 − x − y)2 dy dx
2 0
0
Z 1D Z0 1−x
18
ZZZ
z dV
E
1
=
2
=
1
6
Z 1
0
Z 1
1−x
(1 − x − y)3
−
dx
3
y=0
(1 − x)3 dx
0
1
1
(1 − x)4
=
−
6
4
0
=
1
.
24
RRR √
Example 6. Evaluate E x2 + z 2 dV where E is the solid bounded by the
paraboloid y = x2 + z 2 and the plane y = 4.
Solution. The solid E is shown below:
We shall describe E as a Type 3 Region whose projection D onto the xzplane:
19
ZZZ √
x2 + z 2 dV
E
√
Z Z Z 4
=
x2 + z 2 dy
dA
x2 +z 2
D
√
(4 − x2 − z 2 ) x2 + z 2 dA
ZZ
=
D
The resulting double integral is easier to evaluate using polar coordinates.
ZZZ √
x2 + z 2 dV
√
(4 − x2 − z 2 ) x2 + z 2 dA
ZZ
=
E
D
Z 2π Z 2
=
0
(4 − r2 )rr dr dθ
0
Z 2π
=
Z 2
4
(4r − r ) dr
dθ
0
2
0
2
4r3 r5
= 2π
−
3
5 0
=
128π
.
15
20
MA1104 Week 7
Triple Integral in Cylindrical & Spherical
Coordinates
1. Some Interpretations of Triple Coordinates
Just like double integrals, there are many interpretations of triple integral.
So we do ourselves no favour by just thinking of triple integrals in one interpretation. The key is that any interpretations depend on the units given to
x, y, z and f (x, y, z).
Triple Integral as mass of a solid E:
• x, y, z are measured in meters;
• f (x, y, z) is the mass density measure in kg per meter3 .
Triple Integral as volume E:
• f (x, y, z) is unitless and is set to constant 1; and
• x, y, z are measured in meters, say.
Theorem 1 (Volume of a Solid). Let f (x, y, z) = 1 for all (x, y, z) ∈ E where
E is a solid. Then the triple integral of f over E gives the volume V of E,
that is
ZZZ
Volume of E =
1 dV.
E
Example 7: Find the volume of the solid E bounded by the graphs of
z = 4 − y 2 , x + z = 4, x = 0 and z = 0.
Solution. First we sketch the graphs of x + z = 4, x = 0 and z = 0:
1
The sketch of the solid E is shown below:
As a Type 2 solid, the projection on the yz-plane:
2
ZZZ
Volume of E =
dV
E
Z Z Z 4−z
dx dA
=
D 0
Z 2 Z 4−y2 Z 4−z
=
dx dz dy
−2 0
Z 2 Z 4−y2
0
(4 − z) dz dy
=
−2
= ... =
0
128
.
5
2. Cylindrical Coordinates
Cylindrical coordinate is just another coordinate system that one can use
to represent points in the xyz-space. Basically, we can think of cylindrical
coordinate as the combination of polar coordinate and the z-coordinate:
Cylindrical Coordinates = Polar Coordinates + z-coordinate
Formally, a point P (x, y, z) can be represented by the ordered triple (r, θ, z),
where:
• r and θ are polar coordinates of the projection of P onto the xy-plane.
• z is the directed distance from the xy-plane to P .
3
To convert from cylindrical coordinate to rectangular coordinates, we use the
equations:
x = r cos θ
y = r sin θ
z=z
To convert from rectangular to cylindrical coordinates, we use the equations:
r 2 = x2 + y 2
tan θ = xy , provided x 6= 0
z=z
Example 1. Describe the surface whose equation in cylindrical coordinates
is |z| = r.
Solution. By squaring both sides of the equation |z| = r, we have
z 2 = r2
= x2 + y 2 ,
which is the equation of a cone.
4
Example 2. Describe the surface whose equation in cylindrical coordinates
is r = c, where c is a positive constant.
p
Solution. Since r = x2 + y 2 , the equation r = c is just
p
x2 + y 2 = c.
Squaring both sides:
x 2 + y 2 = c2
is the equation of a cylinder:
3. Triple Integrals in Cylindrical Coordinates
5
In this section, we shall see examples of triple integrals in cylindrical coordinates. First, let us think about the reason why we wish to change to
cylindrical coordinates.
Question:
When do we prefer cylindrical coordinates when computing
RRR
f (x, y, z) dV ?
E
Answer: When the projection D of the solid E onto the xy-plane can be
conveniently described in terms of polar coordinates. One example such solid
is given in the figure below:
Notice that this is a Type 1 solid given by
E = {(x, y, z) : (x, y) ∈ D, u1 (x, y) ≤ z ≤ u2 (x, y)},
where D can be described in terms of polar co-ordinates:
D = {(r, θ) : α ≤ θ ≤ β, h1 (θ) ≤ r ≤ h2 (θ)}.
Combining, we can write:
E = {(r, θ, z) : α ≤ θ ≤ β, h1 (θ) ≤ r ≤ h2 (θ),
u1 (r cos θ, r sin θ) ≤ z ≤ u2 (r cos θ, r sin θ)}.
The following theorem shows us how to evaluate a triple integral by converting it into an iterated integrals in the cylindrical coordinates.
6
Theorem 2 (Triple Integral in Cylindrical Coordinates). Suppose E is a
solid described as above. Then
ZZZ
f (x, y, z) dV =
E
Z β Z h2 (θ) Z u2 (r cos θ,r sin θ)
f (r cos θ, r sin θ, z)r dz dr dθ.
α
h1 (θ)
u1 (r cos θ,r sin θ)
Remarks:
• Note that we replace dV by r dz dr dθ.
• Just like converting to polar coordinates, do not forget the additional
r.
RRR p
Example 3. Evaluate E x2 + y 2 dV where E is the solid that lies
• within the cylinder x2 + y 2 = 1;
• below the plane z = 4; and
• above the paraboloid z = 1 − x2 − y 2 .
Solution. The solid E is given in the figure below:
In terms of cylindrical coordinates, we have
E = {(r, θ, z) : 0 ≤ θ ≤ 2π, 0 ≤ r ≤ 1, 1 − r2 ≤ z ≤ 4}.
Hence,
7
Z 2π Z 1 Z 4
ZZZ p
x2 + y 2 dV
r r dz dr dθ
=
0
E
0
Z 2π Z 1
=
Z0 2π Z0 1
=
0
(3r2 + r4 ) dr dθ
2
4
(3r + r ) dr
dθ
=
r2 (4 − (1 − r2 )) dr dθ
0
Z 1
Z 2π
1−r2
0
0
1
r5
12π
= 2π r +
.
=
5 0
5
3
Example 4. Evaluate
Z 2 Z √4−x2 Z 2
−2
√
− 4−x2
(x2 + y 2 ) dz dy dx.
√
x2 +y 2
Solution. The solid E of integration is:
√
√
E = {(x, y, z) : −2 ≤ x ≤ 2, − 4 − x2 ≤ y ≤ 4 − x2 ,
p
x2 + y 2 ≤ z ≤ 2}.
8
In cylindrical coordinates,:
E = {(r, θ, z) : 0 ≤ θ ≤ 2π, 0 ≤ r ≤ 2, r ≤ z ≤ 2}.
Therefore,
Z 2 Z √4−x2 Z 2
−2
√
√
− 4−x2
(x2 + y 2 ) dz dy dx
x2 +y 2
ZZZ
(x2 + y 2 ) dV
=
Z 2π Z 2 ZE2
=
0
0
r
Z 2
Z 2π
3
r (2 − r) dr
dθ
=
r2 r dz dr dθ
0
0
4
2
r
r5
= 2π
−
2
5 0
=
16π
5
4. Spherical Coordinates
Spherical coordinate is yet another coordinate system for representing points
in the xyz-space.
Notice that any point P in the xyz-space can be determined by:
• ρ = |OP | is the distance from the origin to P (ρ, θ, φ). where
• θ is the same angle as in cylindrical coordinates.
• φ is the angle between the positive z-axis and the line segment OP .
9
The triple (ρ, θ, φ) is called the spherical coordinate of the point (x, y, z).
The range for these coordinates are given as follows:
• ρ≥0
• 0 ≤ θ ≤ 2π
• 0≤φ≤π
So how do we relate (x, y, z) to (ρ, θ, φ)?
The relationship between rectangular and spherical coordinates can be seen
from this figure.
Note that the distance formula yields
ρ 2 = x2 + y 2 + z 2
10
From the triangle OP Q and OP P 0 ,
z = ρ cos φ,
r = ρ sin φ.
Since x = r cos θ, y = r sin θ, to convert spherical to rectangular coordinates,
we use the equations:
x = ρ sin φ cos θ
y = ρ sin φ sin θ
z = ρ cos φ
√
Example 5. The point (0, 2 3, −2) is given in rectangular coordinates. Find
the spherical coordinates of this point.
Solution. We need to determine ρ, φ and θ.
ρ =
p
q
=
= 4.
x2 + y 2 + z 2
√
02 + (2 3)2 + (−2)2
Recall that z = ρ cos φ. Thus
−2 = 4 cos φ
φ = cos−1 (−0.5) =
2π
.
3
Next, from x = ρ sin φ cos θ, we have
0 = 4 sin
2π
cos θ
3
cos θ = 0
π 3π
θ =
,
.
2
2
Since y = ρ sin φ sin θ, we have
11
, then y = 4 sin 2π
sin 3π
< 0, which is impossible.
• If θ = 3π
2
3
2
√
• If θ = π2 , then y = 4 sin 2π
sin π2 = 2 3.
3
Hence θ = π2 .
The point of using spherical coordinates:
Spherical coordinates simplify equations for
regions involving spheres or cones.
↓
Spherical coordinates simplify triple integrals involving spheres or cones.
Example 6. For example, the sphere with center the origin and radius c has
the simple equation ρ = c.
Example 7. The graph of the equation θ = c is a vertical half-plane.
Example 8. The equation φ = c represents a half-cone with the z-axis as its
axis.
12
5. Triple Integral in Spherical Coordinates
ZZZ
In this section, we shall compute triple integrals
f (x, y, z) dV in spherE
ical coordinates, where
• E is a spherical wedge
• E is more general region in terms of spherical coordinates
Spherical wedge: A spherical wedge is a solid E in the xyz-space where
all the spherical co-ordinates are bounded by constants:
E = {(ρ, θ, φ) : a ≤ ρ ≤ b, α ≤ θ ≤ β, c ≤ φ ≤ d}
Here, all a, b, α, β, c and d are constants.
The following theorem shows us how to convert a triple integral over a spherical wedge into an iterated intreated integrals in spherical coordinates:
13
Theorem 3 (Triple Integral in Spherical Coordinates).
ZZZ
f (x, y, z) dV =
E
Z dZ βZ b
c
α
f (ρ sin φ cos θ, ρ sin φ sin θ, ρ cos φ)ρ2 sin φ dρ dθ dφ
a
where E is the spherical wedge
E = {(ρ, θ, φ) : a ≤ ρ ≤ b, α ≤ θ ≤ β, c ≤ φ ≤ d}
We convert a triple integral from rectangular coordinates to spherical coordinates by:
• writing
x = ρ sin φ cos θ,
y = ρ sin φ sin θ,
z = ρ cos φ.
• using appropriate limits of integration
• replacing dV by ρ2 sin φ dρ dθ dφ. For now, we will just this formula
as given. Later on, we shall see in the course that the additional term
ρ2 sin φ is the ‘Jacobian’ of the underlying transformation.
This formula can be extended to include more general spherical region such
as
E = {(ρ, θ, φ) : c ≤ φ ≤ d, α ≤ θ ≤ β, g1 (θ, φ) ≤ ρ ≤ g2 (θ, φ)}.
The formula for triple integral over such region is
RRR
E
f (x, y, z) dV =
R d R β R g2 (θ,φ)
c
Example 9. Evaluate
α
g1 (θ,φ)
2
RRR
B
2
f (ρ sin φ cos θ, ρ sin φ sin θ, ρ cos φ)ρ2 sin φ dρ dθ dφ
2 3/2
e(x +y +z )
dV where B is the unit ball:
B = {(x, y, z) : x2 + y 2 + z 2 ≤ 1}.
14
Solution. Note that B is a spherical wedge:
B = {(ρ, θ, φ) : 0 ≤ ρ ≤ 1, 0 ≤ θ ≤ 2π, 0 ≤ φ ≤ π}.
Also, notice that
x2 + y 2 + z 2 = ρ 2 .
Therefore,
ZZZ
2
2
2 3/2
e(x +y +z )
dV
B
Z π Z 2π Z 1
=
0
Z π
=
0
0
2 3/2
e(ρ )
ρ2 sin φ dρ dθ dφ
0
Z 2π Z 1
2 ρ3
ρ e dρ
dθ
sin φ dφ
0
0
= [− cos φ]π0 (2π)
1
1 ρ3
e
3
0
4
= π(e − 1).
3
Example 10. Use spherical coordinates to find the volume of the solid that
lies:
p
• above the cone z = x2 + y 2 , and
• below the sphere x2 + y 2 + z 2 = z
15
Solution. In spherical coordinates, we have
z=
p
x2 + y 2
ρ cos φ = ρ sin φ
sin φ = cos φ
π
φ= .
4
x2 + y 2 + z 2 = z
ρ2 = ρ cos φ
ρ = cos φ.
In the last step, we have assumed ρ 6= 0. Why can we do that?
Therefore, E in spherical coordinates is :
E = {(ρ, θ, φ) : 0 ≤ ρ ≤ cos φ, 0 ≤ θ ≤ 2π, 0 ≤ φ ≤ π/4}.
The volume of E is
ZZZ
dV
V =
E
Z 2π Z π/4 Z cos φ
=
0
0
ρ2 sin φ dρ dφ dθ
0
Z 2π Z π/4 Z cos φ
V =
0
0
Z 2π
V
0
Z π/4
dθ
=
0
2π
=
3
0
Z π/4
3 ρ=cos φ !
ρ
sin φ
dφ
3 ρ=0
sin φ cos3 φ dφ
0
π/4
2π
cos4 φ
=
−
3
4
0
=
ρ2 sin φ dρ dφ dθ
π
.
8
16
The following figure shows how the volume is being ‘swept out’ as we compute
the iterated integrals:
17
MA1104 Week 8
Change of Variables & Jacobian
1. Plane Transformation
A plane transformation T : (u, v) 7→ (x, y) from the uv-plane to the xy-plane
is a set of equations
x = x(u, v),
y = y(u, v).
An example of a plane transformation the function that maps polar coordinates to rectangular coordinates:
x = r cos θ,
y = r sin θ
Here, the plane transformation maps a point (r, θ) on the rθ-plane to a point
(x, y) on the xy-plane.
Now, in general, T transforms S into a region R in the xy-plane which is
called the image of S, consisting of the images of all points in S.
Let us now focus on the following question:
Question: Given a closed and bounded region S, how do we determine its
image R under transformation T ?
1
Answer. The key is to examine the boundary.
T maps the boundary of the domain into the boundary of the image.
S closed and bounded =⇒ R closed and bounded.
Therefore, once we know the boundary of the image, we will be able to
determine the image R.
Example 1. A transformation T is defined by:
x = u2 − v 2 ,
y = 2uv.
Find the image of the square
S = {(u, v)|0 ≤ u ≤ 1, 0 ≤ v ≤ 1}.
Solution. Consider the four boundaries of S.
2
Side 1:
S1 : v = 0,
0 ≤ u ≤ 1.
⇓T
x = u2 , y = 0.
y = 0, 0 ≤ x ≤ 1.
So the image of S1 is the line y = 0, 0 ≤ x ≤ 1 on the xy-plane.
Side 2:
S2 : u = 1,
0 ≤ v ≤ 1.
⇓T
x = 1 − v 2 , y = 2v, 0 ≤ v ≤ 1.
2
x = 1 − y4 , 0 ≤ y ≤ 2.
3
The image of S2 :
Side 3:
S3 : v = 1,
0 ≤ u ≤ 1.
⇓T
x = u2 − 1, y = 2u, 0 ≤ u ≤ 1.
2
x = y4 − 1, 0 ≤ y ≤ 2.
The image of S3 :
Side 4:
S4 : u = 0,
0 ≤ v ≤ 1.
4
⇓T
x = −v 2 , y = 0.
y = 0, −1 ≤ x ≤ 0.
The image of S4 :
y2
,
4
0≤y≤2
y2
− 1,
4
0 ≤ y ≤ 2.
x=1−
x=
5
2. Change of Variable Formula: 2D Case
In this section, we shall be concerned with the following:
• How does a plane transformation T : (u, v) → (x, y) affect an area?
• Derive a formula that converts a double integral in (x, y) to a double
integral in (u, v).
Let us first look at the area question:
Given T : (u, v) 7→ (x, y), if S is a rectangle on the uv-plane, what is the
area of the image of S under T ?
To fix the idea,
• We take a small rectangle S on the uv-plane, whose area is 4u4v.
• Suppose R is the image of S under T on the xy-plane.
Let 4A be the area of R. How can we compute the area 4A of R?
6
Let A0 , B 0 , C 0 , D0 are the images of A, B, C, D respectively.
Let r(u, v) denote the position vector of the image (x, y) of the point (u, v)
under T :
r(u, v) = x(u, v)i + y(u, v)j.
Notice that the position vectors of the points A0 , B 0 , C 0 and D0 are:
−−→0
OA
−−→0
OB
−−→0
OC
−−→0
OD
= r(u0 , v0 )
= r(u0 + 4u, v0 )
= r(u0 + 4u, v0 + 4v)
= r(u0 , v0 + 4v)
The key is to approximate the area 4A of R using the paralleogram spanned
−−→
−−→
by the vectors a = A0 B 0 and b = A0 D0
The region R is roughly the parallelogram spanned by the vectors a and b.
7
4A ≈ Area of the parallelogram
spanned by a and b
So
4A ≈ ||a × b||
Now,
a =
=
=
−−
→
A0 B 0
−−→0 −−→0
OB − OA
r(u0 + 4u, v0 ) − r(u0 , v0 )
b =
=
=
−−
→
A0 D0
−−→0 −−→0
OD − OA
r(u0 , v0 + 4v) − r(u0 , v0 ).
Since 4u is small, we have
ru (u0 , v0 ) ≈
r(u0 + 4u, v0 ) − r(u0 , v0 )
4u
a = r(u0 + 4u, v0 ) − r(u0 , v0 ) ≈ 4uru (u0 , v0 ).
Since 4v is small, we have
rv (u0 , v0 ) ≈
r(u0 , v0 + 4v) − r(u0 , v0 )
4v
b = r(u0 , v0 + 4v) − r(u0 , v0 ) ≈ 4vrv (u0 , v0 ).
It follows that
4A ≈ ||a × b|| ≈ ||4uru × 4vrv ||
≈ ||ru × rv || 4u4v.
8
Next, lets calculate ||ru × rv || .....
Note that r(u, v) = x(u, v)i + y(u, v)j, and so
ru =
∂y
∂x
i+
j,
∂u
∂u
rv =
∂x
∂y
i+
j.
∂v
∂v
Computing the cross product, we obtain
ru × rv =
=
i
j
∂x
∂u
∂x
∂v
∂y
∂u
∂y
∂v
∂x
∂u
∂x
∂v
∂y
∂u
∂y
∂v
k
0
0
∂x
∂u
∂y
∂u
k=
∂x
∂v
∂y
∂v
k
Thus,
||ru × rv || = absolute value of
∂x ∂y ∂x ∂y
−
∂u ∂v ∂v ∂u
This motivates the following definition:
Definition 1. The Jacobian of the transformation T given by x = x(u, v)
and y = y(u, v) is
∂(x, y)
=
∂(u, v)
∂x
∂u
∂y
∂u
∂x
∂v
∂y
∂v
=
∂x ∂y ∂x ∂y
−
∂u ∂v ∂v ∂u
Using the above definition,
||ru × rv || = absolute value of
∂x ∂y ∂x ∂y
−
∂u ∂v ∂v ∂u
= absolute value of the Jacobian
=
∂(x, y)
∂(u, v)
9
With the above notation, we can give an approximation to the area 4A of
R:
4A ≈ ||ru × rv || 4u4v
∂(x, y)
4u4v
∂(u, v)
=
To summarize:
Let dA be the image of the rectangle du dv under
the plane transformation T given by x = x(u, v),
y = y(u, v). Then
dA =
∂(x, y)
du dv.
∂(u, v)
Using this formula for dA, we can now compute a double integral in dA =
dx dy by changing it to a double integral in du dv:
Theorem 2 (Change of Variable in Double Integral).
ZZ
f (x, y) dA
R
ZZ
f (x(u, v), y(u, v))
=
S
∂(x, y)
du dv.
∂(u, v)
Example 2. The formula for double integral over polar rectangle is a special
case of the formula using the Jacobian.
Solution. Suppose the polar rectangle R is given by
R = {(r, θ) : a ≤ r ≤ b, α ≤ θ ≤ β}.
Here the transformation T from the rθ-plane to the xy-plane is given by
x = r cos θ,
y = r sin θ.
The geometry of the transformation is shown here.
10
The Jacobian of T is
∂x
∂r
∂y
∂r
∂(x, y)
=
∂(r, θ)
∂x
∂θ
∂y
∂θ
cos θ −r sin θ
sin θ r cos θ
=
= r cos2 θ + r sin2 θ = r.
ZZ
f (x, y) dA
R
ZZ
f (x(u, v), y(u, v))
=
S
∂(x, y)
du dv
∂(u, v)
Z βZ b
=
f (r cos θ, r sin θ)|r| dr dθ
α
a
Z βZ b
=
f (r cos θ, r sin θ)r dr dθ
α
a
3. Examples: 2D Jacobian
11
Example 3. Use Change of Variables x = uv −1 , y = uv to compute
ZZ
(x2 + y 2 ) dA
R
where R is the region in the first quadrant (x ≥ 0, y ≥ 0) satisfying 1 ≤
xy ≤ 4, 1 ≤ y/x ≤ 4.
Solution. The region R:
The region S:
S = {(u, v) : 1 ≤ u ≤ 2, 1 ≤ v ≤ 2}
The Jacobian of T is
∂(x, y)
=
∂(u, v)
v −1 −uv −2
v
u
By Change of Variable Formula,
12
=
2u
.
v
ZZ
(x2 + y 2 ) dA
R
Z Z 2
u
2u
2
=
+ (uv)
du dv
v
v
S
Z 2Z 2
u3 (v −3 + v) du dv
= 2
1
Z 2
1
Z 2
−3
= 2
u du
(v + v) dv
1
1
4
2
2 !
2 − 14
v
1 −2
225
= 2
− v
=
.
4
2
2
16
1
3
Example 4. Evaluate the integral
ZZ
e(x+y)/(x−y) dA
R
where R is the trapezoidal region with vertices (1, 0), (2, 0), (0, −2) and
(0, −1).
Solution. It is difficult to integrate e(x+y)/(x−y) , so we make a change of
variables suggested by the form of this function:
v = x − y.
u = x + y,
Solving the equation for x and y, we obtain a transformation T from the
uv-plane to the xy-plane:
1
x = (u + v),
2
Region R on the xy-plane.
1
y = (u − v).
2
13
The sides of R are given by the lines:
y = 0,
x − y = 2,
x = 0,
x − y = 1.
Find S by converting the boundary R back to the uv-plane:
y = 0,
x − y = 2,
x = 0,
x − y = 1.
via
u = x + y, v = x − y
x = 21 (u + v), y = 21 (u − v)
u = x + y, v = x − y
x = 21 (u + v), y = 21 (u − v)
The images of the sides of R in the uv-plane under T −1 are:
u = v,
v = 2,
u = −v,
14
v = 1.
We can write S as
S = {(u, v) : 1 ≤ v ≤ 2, − v ≤ u ≤ v}.
The Jacobian of T is
∂x
∂u
∂y
∂u
∂(x, y)
=
∂(u, v)
=
∂x
∂v
∂y
∂v
1/2 1/2
1/2 −1/2
= −1/2 6= 0.
Then, Change of Variables Formula gives
ZZ
e
(x+y)/(x−y)
R
ZZ
ZZ
∂(x, y)
du dv
∂(u, v)
S
Z 2Z v
1
u/v
e
=
du dv
2
1
−v
dA =
eu/v
1 u/v u=v
dv
ve
u=−v
2
Z
1 2
(e − e−1 )v dv
=
2 1
e(x+y)/(x−y) dA =
R
=
3
(e − e−1 ).
4
4. Change of Variable Formula: 3D Case
In this section, we shall examine the change of variable formula in 3 dimensions. Basically, we want to answer the following questions:
15
• How does a space transformation T : (u, v, w) 7→ (x, y, z) affect a volume?
• How to convert a triple integral in x, y, z to that in u, v, w?
It turns out that the following holds (proof is omitted):
Let dV be the image of the rectangular box
du dv dw under the space transformation T given
by x = x(u, v, w), y = y(u, v, w), z = z(u, v, w).
Then
dV =
∂(x, y, z)
du dv dw.
∂(u, v, w)
Here, the Jacobian of space transformation T is the following 3 × 3 determinant:
∂x
∂u
∂y
∂u
∂z
∂u
∂(x, y, z)
=
∂(u, v, w)
∂x
∂v
∂y
∂v
∂z
∂v
∂x
∂w
∂y
∂w
∂z
∂w
With this, we can compute a triple integral in x, y, z by changing it to a
triple interval in u, v, w.
Theorem 3 (Change of Variable Formula in Triple Integral).
ZZZ
f (x, y, z) dV =
R
ZZZ
f (x(u, v, w), y(u, v, w), z(u, v, w))
S
∂(x, y, z)
du dv dw.
∂(u, v, w)
As shown in the following example, the Triple Integral in Spherical Coordinates that we learned previously is just an instance of the above Change of
Variable Formula.
16
Example 5. Recall that Triple Integral in Spherical Coordinates is given by
ZZZ
f (x, y, z) dV =
E
Z dZ βZ b
c
α
f (ρ sin φ cos θ, ρ sin φ sin θ, ρ cos φ)ρ2 sin φ dρ dθ dφ
a
E = {(ρ, θ, φ) : a ≤ ρ ≤ b, α ≤ θ ≤ β, c ≤ φ ≤ d}
Show that this formula is a special case of the Change of Variable Formula
for Triple Integral.
Solution. Note that
x = ρ sin φ cos θ,
y = ρ sin φ sin θ,
Partial derivatives of x:
∂x
= sin φ cos θ,
∂ρ
∂x
= −ρ sin φ sin θ,
∂θ
∂x
= ρ cos φ cos θ.
∂φ
Partial derivatives of y:
∂y
= sin φ sin θ,
∂ρ
∂y
= ρ sin φ cos θ,
∂θ
∂y
= ρ cos φ sin θ.
∂φ
Partial derivatives of z:
17
z = ρ cos φ.
∂z
= cos φ,
∂ρ
∂z
= 0,
∂θ
∂z
= −ρ sin φ.
∂φ
∂x
∂ρ
∂y
∂ρ
∂z
∂ρ
∂(x, y, z)
=
∂(ρ, θ, φ)
∂x
∂θ
∂y
∂θ
∂z
∂θ
∂x
∂φ
∂y
∂φ
∂z
∂φ
.
= ..
= −ρ2 sin φ.
Hence the Jacobian is
∂(x, y, z)
= ρ2 sin φ.
∂(ρ, θ, φ)
∂(x, y, z)
=
∂(ρ, θ, φ)
∂x
∂ρ
∂y
∂ρ
∂z
∂ρ
∂x
∂θ
∂y
∂θ
∂z
∂θ
= sin φ cos θ
∂x
∂φ
∂y
∂φ
∂z
∂φ
ρ sin φ cos θ ρ cos φ sin θ
0
−ρ sin φ
−(−ρ sin φ sin θ)
+ρ cos φ cos θ
18
sin φ sin θ ρ cos φ sin θ
cos φ
−ρ sin φ
sin φ sin θ ρ sin φ cos θ
cos φ
0
∂(x, y, z)
= sin φ cos θ −ρ2 sin2 φ cos θ
∂(ρ, θ, φ)
+ρ sin φ sin θ −ρ sin2 φ sin θ − ρ cos2 φ sin θ
+ρ cos φ cos θ (−ρ sin φ cos φ cos θ)
= −ρ2 sin3 φ cos2 θ − ρ2 sin φ sin2 θ
−ρ2 sin φ cos2 φ cos2 θ
= −ρ2 sin φ cos2 θ sin2 φ + cos2 φ
−ρ2 sin φ sin2 θ
= −ρ2 sin φ cos2 θ − ρ2 sin φ sin2 θ
= −ρ2 sin φ.
Example 6. Find the volume of the solid Q bounded by x + y + z = 1,
x + y + z = 2, x + 2y = 0, x + 2y = 1, y + z = 2 and y + z = 4.
Solution. Let
u = x + y + z, v = x + 2y, w = y + z.
x = u − w, y =
v−u+w
w−v+u
, z=
2
2
1
0
−1
−1/2 1/2 1/2
1/2 −1/2 1/2
∂(x, y, z)
=
∂(u, v, w)
= 1/2
Therefore,
ZZZ
Volume =
1 dV
Q
Z 2Z 1Z 4
=
1
0
2
1
dw dv du
2
1
=
·1·1·2
2
= 1.
19