Calculus II – Fall 2014
Lecture 4 – Double Integrals
Eitan Angel
University of Colorado
Wednesday, December 3, 2014
E. Angel (CU)
Calculus II
3 Dec
1 / 32
Introduction
We extend the idea of a definite integral to double and triple integrals
of functions of two and three variables. These ideas are used to
compute volumes, masses, centroids, and more general regions than
we can compute with single variable calculus.
E. Angel (CU)
Calculus II
3 Dec
2 / 32
Introduction
We extend the idea of a definite integral to double and triple integrals
of functions of two and three variables. These ideas are used to
compute volumes, masses, centroids, and more general regions than
we can compute with single variable calculus.
Recall that the definite integral of a single variable function solves the
area problem.
E. Angel (CU)
Calculus II
3 Dec
2 / 32
Introduction
We extend the idea of a definite integral to double and triple integrals
of functions of two and three variables. These ideas are used to
compute volumes, masses, centroids, and more general regions than
we can compute with single variable calculus.
Recall that the definite integral of a single variable function solves the
area problem.
The double integral of a two variable function solves the
corresponding volume problem: For a (nonnegative) continuous
function f defined on some region R in the xy-plane, what is the
volume contained between the surface z = f (x, y) and the region R?
E. Angel (CU)
Calculus II
3 Dec
2 / 32
Introduction
We extend the idea of a definite integral to double and triple integrals
of functions of two and three variables. These ideas are used to
compute volumes, masses, centroids, and more general regions than
we can compute with single variable calculus.
Recall that the definite integral of a single variable function solves the
area problem.
The double integral of a two variable function solves the
corresponding volume problem: For a (nonnegative) continuous
function f defined on some region R in the xy-plane, what is the
volume contained between the surface z = f (x, y) and the region R?
We will integrate over both rectangular and more general regions,
where either the left and right, or top and bottom bounds of the
region are given by general continuous curves.
E. Angel (CU)
Calculus II
3 Dec
2 / 32
Riemann Sum Definition (Single Variable)
We should first recall the basic facts concerning definite integrals of single
variable functions.
Let f be defined on a closed interval [a, b].
E. Angel (CU)
Calculus II
3 Dec
3 / 32
Riemann Sum Definition (Single Variable)
We should first recall the basic facts concerning definite integrals of single
variable functions.
Let f be defined on a closed interval [a, b].
We take a partition P of [a, b] into subintervals [xi−1 , xi ] where
a = x0 < x1 < · · · < xn−1 < xn = b
E. Angel (CU)
Calculus II
3 Dec
3 / 32
Riemann Sum Definition (Single Variable)
We should first recall the basic facts concerning definite integrals of single
variable functions.
Let f be defined on a closed interval [a, b].
We take a partition P of [a, b] into subintervals [xi−1 , xi ] where
a = x0 < x1 < · · · < xn−1 < xn = b
We choose points x∗i in [xi−1 , xi ] and let ∆xi = xi − xi−1 and
kP k = max{∆xi }.
E. Angel (CU)
Calculus II
3 Dec
3 / 32
Riemann Sum Definition (Single Variable)
We should first recall the basic facts concerning definite integrals of single
variable functions.
Let f be defined on a closed interval [a, b].
We take a partition P of [a, b] into subintervals [xi−1 , xi ] where
a = x0 < x1 < · · · < xn−1 < xn = b
We choose points x∗i in [xi−1 , xi ] and let ∆xi = xi − xi−1 and
kP k = max{∆xi }.
Then we form the Riemann sum
n
X
f (x∗i )∆xi
i=1
and take the limit of such sums as kP k → 0 to obtain the definite
integral of f from a to b.
E. Angel (CU)
Calculus II
3 Dec
3 / 32
Riemann Sum Definition (Single Variable)
The definite integral of f from a to b is defined as
Z
b
f (x)dx = lim
a
E. Angel (CU)
kP k→0
Calculus II
n
X
f (x∗i )∆xi
i=1
3 Dec
4 / 32
Riemann Sum Definition (Single Variable)
The definite integral of f from a to b is defined as
Z
b
f (x)dx = lim
a
kP k→0
n
X
f (x∗i )∆xi
i=1
Also recall that the definite integral is the solution to the area
problem:
E. Angel (CU)
Calculus II
3 Dec
4 / 32
Riemann Sum Definition (Single Variable)
The definite integral of f from a to b is defined as
Z
b
f (x)dx = lim
kP k→0
a
n
X
f (x∗i )∆xi
i=1
Also recall that the definite integral is the solution to the area
problem:
Definition
If y = f (x) is nonnegative and integrable over a closed interval [a, b], then
the area under the curve y = f (x) over [a, b] is the integral of f
from a to b,
Z b
f (x)dx
A=
a
E. Angel (CU)
Calculus II
3 Dec
4 / 32
Riemann Sum Definition
We proceed similarly to define the double integral of a function of two
variables that is defined on the closed rectangle
R = [a, b] × [c, d] = {(x, y) ∈ R2 | a ≤ x ≤ b, c ≤ y ≤ d}
E. Angel (CU)
Calculus II
3 Dec
5 / 32
Riemann Sum Definition
We proceed similarly to define the double integral of a function of two
variables that is defined on the closed rectangle
R = [a, b] × [c, d] = {(x, y) ∈ R2 | a ≤ x ≤ b, c ≤ y ≤ d}
The first step is to take a partition P of R into subrectangles. We
can partition the intervals [a, b] and [c, d] as follows:
a = x0 < x1 < · · · < xi−1 < xi < · · · < xm = b
c = y0 < y1 < · · · < yi−1 < yi < · · · < yn = d
E. Angel (CU)
Calculus II
3 Dec
5 / 32
Riemann Sum Definition
We proceed similarly to define the double integral of a function of two
variables that is defined on the closed rectangle
R = [a, b] × [c, d] = {(x, y) ∈ R2 | a ≤ x ≤ b, c ≤ y ≤ d}
The first step is to take a partition P of R into subrectangles. We
can partition the intervals [a, b] and [c, d] as follows:
a = x0 < x1 < · · · < xi−1 < xi < · · · < xm = b
c = y0 < y1 < · · · < yi−1 < yi < · · · < yn = d
This partitions R into mn subrectangles
Rij = [xi−1 , xi ] × [yj−1 , yj ] = {(x, y) | xi−1 ≤ x ≤ xi , yj−1 ≤ y ≤ yj }
and if ∆xi = xi − xi−1 and ∆yj = yj − yj−1 , then the area of Rij is
∆Aij = ∆xi ∆yj .
E. Angel (CU)
Calculus II
3 Dec
5 / 32
Riemann Sum Definition
∗ ) in R , and by analogy with the
Now choose some point (x∗ij , yij
ij
single variable Riemann sum, we form the double Riemann sum
m X
n
X
∗
f (x∗ij , yij
)∆Aij
i=1 j=1
E. Angel (CU)
Calculus II
3 Dec
6 / 32
Riemann Sum Definition
∗ ) in R , and by analogy with the
Now choose some point (x∗ij , yij
ij
single variable Riemann sum, we form the double Riemann sum
m X
n
X
∗
f (x∗ij , yij
)∆Aij
i=1 j=1
The double sum means that for each subrectangle we evaluate f at
the chosen point and multiply by the area of the subrectangle, then
add the results for each subrectangle.
E. Angel (CU)
Calculus II
3 Dec
6 / 32
Riemann Sum Definition
∗ ) in R , and by analogy with the
Now choose some point (x∗ij , yij
ij
single variable Riemann sum, we form the double Riemann sum
m X
n
X
∗
f (x∗ij , yij
)∆Aij
i=1 j=1
The double sum means that for each subrectangle we evaluate f at
the chosen point and multiply by the area of the subrectangle, then
add the results for each subrectangle.
The final piece we need to define the double integral is the norm of
the partition, which is the length of the longest diagonal of all the
subrectangles Rij , and is denoted kP k. If we let kP k → 0 then the
partition becomes finer.
E. Angel (CU)
Calculus II
3 Dec
6 / 32
Riemann Sum Definition
Now we can make the following definition:
Definition
The double integral of f over the rectangle R is
ZZ
f (x, y) dA = lim
R
kP k→0
m X
n
X
∗
f (x∗ij , yij
)∆Aij
i=1 j=1
when the limit exists.
E. Angel (CU)
Calculus II
3 Dec
7 / 32
Riemann Sum Definition
Now we can make the following definition:
Definition
The double integral of f over the rectangle R is
ZZ
f (x, y) dA = lim
kP k→0
R
m X
n
X
∗
f (x∗ij , yij
)∆Aij
i=1 j=1
when the limit exists.
Since ∆Aij = ∆xi ∆yj , we often use the notation
ZZ
ZZ
f (x, y) dA =
f (x, y) dx dy.
R
R
A function f is called integrable if the limit in the definition exists. Any
continuous function is integrable.
E. Angel (CU)
Calculus II
3 Dec
7 / 32
Riemann Sum Approximations
Example
√
Let f (x, y) = 2x y and R be the rectangle [0, 4] × [0, 4]. Estimate
RR
R f dA using four subrectangles of R.
E. Angel (CU)
Calculus II
3 Dec
8 / 32
Riemann Sum Approximations
Example
√
Let f (x, y) = 2x y and R be the rectangle [0, 4] × [0, 4]. Estimate
RR
R f dA using four subrectangles of R.
We will use the partition into equal size subrectangles x0 = 0, x1 = 2, and
x2 = 4 while y0 = 0, y1 = 2, and y2 = 4.
E. Angel (CU)
Calculus II
3 Dec
8 / 32
Riemann Sum Approximations
Example
√
Let f (x, y) = 2x y and R be the rectangle [0, 4] × [0, 4]. Estimate
RR
R f dA using four subrectangles of R.
We will use the partition into equal size subrectangles x0 = 0, x1 = 2, and
x2 = 4 while y0 = 0, y1 = 2, and y2 = 4.
Hence the subrectangles are
R00 = [0, 2] × [0, 2]
R10 = [2, 4] × [0, 2]
R01 = [0, 2] × [2, 4]
R11 = [2, 4] × [2, 4]
E. Angel (CU)
Calculus II
3 Dec
8 / 32
Riemann Sum Approximations
Example
√
Let f (x, y) = 2x y and R be the rectangle [0, 4] × [0, 4]. Estimate
RR
R f dA using four subrectangles of R.
The Riemann sum approximation is then
2 X
2
X
∗
f (x∗ij , yij
)∆Aij
i=1 j=1
E. Angel (CU)
Calculus II
3 Dec
8 / 32
Riemann Sum Approximations
Example
√
Let f (x, y) = 2x y and R be the rectangle [0, 4] × [0, 4]. Estimate
RR
R f dA using four subrectangles of R.
The Riemann sum approximation is then
2 X
2
X
∗
f (x∗ij , yij
)∆Aij
i=1 j=1
Let’s first obtain a lower bound using four rectangles.
Since f is an increasing function with respect to both x and y,
x∗ij = xi−1
∗
and yij
= yj−1
will produce the minimum value of f on each Rij .
E. Angel (CU)
Calculus II
3 Dec
8 / 32
Riemann Sum Approximations
Example
√
Let f (x, y) = 2x y and R be the rectangle [0, 4] × [0, 4]. Estimate
RR
R f dA using four subrectangles of R.
Again, as the lowest value for f on each subrectangle is obtained at the
lower left corner and the area of each subrectangle is 4
lower estimate =
2
2 X
X
(min of f on Rij ) · (Area of Rij )
i=1 j=1
E. Angel (CU)
Calculus II
3 Dec
8 / 32
Riemann Sum Approximations
Example
√
Let f (x, y) = 2x y and R be the rectangle [0, 4] × [0, 4]. Estimate
RR
R f dA using four subrectangles of R.
Again, as the lowest value for f on each subrectangle is obtained at the
lower left corner and the area of each subrectangle is 4
lower estimate =
2
2 X
X
(min of f on Rij ) · (Area of Rij )
i=1 j=1
= [f (0, 0) + f (0, 2) + f (2, 0) + f (2, 2)] · 4
E. Angel (CU)
Calculus II
3 Dec
8 / 32
Riemann Sum Approximations
Example
√
Let f (x, y) = 2x y and R be the rectangle [0, 4] × [0, 4]. Estimate
RR
R f dA using four subrectangles of R.
Again, as the lowest value for f on each subrectangle is obtained at the
lower left corner and the area of each subrectangle is 4
lower estimate =
2
2 X
X
(min of f on Rij ) · (Area of Rij )
i=1 j=1
= [f (0, 0) + f (0, 2) + f (2, 0) + f (2, 2)] · 4
√
√
= [0 + 0 + 0 + 4 2] · 4 = 16 2
E. Angel (CU)
Calculus II
3 Dec
8 / 32
Riemann Sum Approximations
Example
√
Let f (x, y) = 2x y and R be the rectangle [0, 4] × [0, 4]. Estimate
RR
R f dA using four subrectangles of R.
Similarly, an upper bound is obtained by
upper estimate =
2 X
2
X
(max of f on Rij ) · (Area of Rij )
i=1 j=1
E. Angel (CU)
Calculus II
3 Dec
8 / 32
Riemann Sum Approximations
Example
√
Let f (x, y) = 2x y and R be the rectangle [0, 4] × [0, 4]. Estimate
RR
R f dA using four subrectangles of R.
Similarly, an upper bound is obtained by
upper estimate =
2 X
2
X
(max of f on Rij ) · (Area of Rij )
i=1 j=1
= [f (2, 2) + f (2, 4) + f (4, 2) + f (4, 4)] · 4
E. Angel (CU)
Calculus II
3 Dec
8 / 32
Riemann Sum Approximations
Example
√
Let f (x, y) = 2x y and R be the rectangle [0, 4] × [0, 4]. Estimate
RR
R f dA using four subrectangles of R.
Similarly, an upper bound is obtained by
upper estimate =
2 X
2
X
(max of f on Rij ) · (Area of Rij )
i=1 j=1
= [f (2, 2) + f (2, 4) + f (4, 2) + f (4, 4)] · 4
√
√
√
= [4 2 + 8 + 8 2 + 16] · 4 = 32 2 + 48
E. Angel (CU)
Calculus II
3 Dec
8 / 32
Riemann Sum Approximations
Example
√
Let f (x, y) = 2x y and R be the rectangle [0, 4] × [0, 4]. Estimate
RR
R f dA using four subrectangles of R.
Similarly, an upper bound is obtained by
upper estimate =
2 X
2
X
(max of f on Rij ) · (Area of Rij )
i=1 j=1
= [f (2, 2) + f (2, 4) + f (4, 2) + f (4, 4)] · 4
√
√
√
= [4 2 + 8 + 8 2 + 16] · 4 = 32 2 + 48
The average of the upper and lower estimate is
√
(upper estimate + lower estimate)/2 = 32 2 + 48 ≈ 93
E. Angel (CU)
Calculus II
3 Dec
8 / 32
Interpretation As Volume
Just as single integrals of positive functions can be interpreted as
areas, double integrals of positive function can be interpreted as
volumes.
E. Angel (CU)
Calculus II
3 Dec
9 / 32
Interpretation As Volume
Just as single integrals of positive functions can be interpreted as
areas, double integrals of positive function can be interpreted as
volumes.
Suppose f (x, y) ≥ 0 and f is defined on R = [a, b] × [c, d]. The
graph of f is the surface with equation z = f (x, y). Let S be the
solid that lies above R and under the graph of f , that is,
S = {(x, y, z) ∈ R3 | 0 ≤ z ≤ f (x, y), (x, y) ∈ R}
E. Angel (CU)
Calculus II
3 Dec
9 / 32
Interpretation As Volume
Just as single integrals of positive functions can be interpreted as
areas, double integrals of positive function can be interpreted as
volumes.
Suppose f (x, y) ≥ 0 and f is defined on R = [a, b] × [c, d]. The
graph of f is the surface with equation z = f (x, y). Let S be the
solid that lies above R and under the graph of f , that is,
S = {(x, y, z) ∈ R3 | 0 ≤ z ≤ f (x, y), (x, y) ∈ R}
We can approximate the volume of the part of S that lies above Rij
∗ ). The
by a thin rectangular box with base Rij and height f (x∗ij , yij
volume of the thin box is
∗
Vij = f (x∗ij , yij
)∆Aij
E. Angel (CU)
Calculus II
3 Dec
9 / 32
Interpretation As Volume
Therefore we can approximate the volume of S by adding up the
volume of all such boxes
m X
n
X
∗
V ≈
f (x∗ij , yij
)∆Aij
i=1 j=1
E. Angel (CU)
Calculus II
3 Dec
10 / 32
Interpretation As Volume
Therefore we can approximate the volume of S by adding up the
volume of all such boxes
m X
n
X
∗
V ≈
f (x∗ij , yij
)∆Aij
i=1 j=1
Notice that the right hand side is the Riemann sum. If we use a finer
partition (more rectangles), we should expect that the Riemann sum
more closely approximates the volume. We obtain finer and finer
partitions by taking the limit kP k → 0.
E. Angel (CU)
Calculus II
3 Dec
10 / 32
Interpretation As Volume
Therefore we can approximate the volume of S by adding up the
volume of all such boxes
m X
n
X
∗
V ≈
f (x∗ij , yij
)∆Aij
i=1 j=1
Notice that the right hand side is the Riemann sum. If we use a finer
partition (more rectangles), we should expect that the Riemann sum
more closely approximates the volume. We obtain finer and finer
partitions by taking the limit kP k → 0.
Theorem
If f (x, y) ≥ 0 and f is continuous on the rectangle R, then the volume V
of the solid that lies above R and under the surface z = f (x, y) is
ZZ
V =
f (x, y) dA
R
E. Angel (CU)
Calculus II
3 Dec
10 / 32
Evaluation of Double Integrals
The partial derivatives of a function f (x, y) are calculated by holding one
of the variables fixed and differentiating with respect to the other variable.
The reverse of this process is partial integration. The integrals
Z
b
Z
f (x, y) dx
a
d
f (x, y) dy
c
denote the partial definite integrals.
E. Angel (CU)
Calculus II
3 Dec
11 / 32
Evaluation of Double Integrals
The partial derivatives of a function f (x, y) are calculated by holding one
of the variables fixed and differentiating with respect to the other variable.
The reverse of this process is partial integration. The integrals
Z
b
Z
f (x, y) dx
d
f (x, y) dy
a
c
denote the partial definite integrals.
Example
1
3yx3 3x y dx = 3y
x dx =
=y
3 x=0
0
0
2
Z 2
Z 2
3x2 y 2 3x2 y dy = 3x2
y dy =
= 6x2
2
0
0
y=0
Z
E. Angel (CU)
1
2
Z
1
2
Calculus II
3 Dec
11 / 32
Evaluation of Double Integrals
Notice that the partial definite integral with respect to x is a function of y
and can thus be integrated with respect to y, and vice-versa. This
two-step integration is called iterated integration.
Z
dZ b
Z
d Z b
f (x, y) dx dy =
c
a
Z bZ
f (x, y) dx dy
c
d
a
Z b Z
f (x, y) dy dx =
a
c
d
f (x, y) dy dx
a
c
Now we should evaluate some iterated integrals.
E. Angel (CU)
Calculus II
3 Dec
12 / 32
Evaluation of Double Integrals
Example
Evaluate
Z 1Z 2
2
2
(16 − x − 2y )dy dx
0
Z
and
0
E. Angel (CU)
2Z 1
0
Calculus II
(16 − x2 − 2y 2 )dx dy
0
3 Dec
13 / 32
Evaluation of Double Integrals
Example
Evaluate
Z 1Z 2
2
(16 − x − 2y )dy dx
0
and
0
Z
0
1Z 2
2
2
Z
1 Z 2
(16 − x − 2y )dy dx =
0
2Z 1
Z
2
0
E. Angel (CU)
(16 − x2 − 2y 2 )dx dy
0
2
2
(16 − x − 2y )dy dx
0
Calculus II
0
3 Dec
13 / 32
Evaluation of Double Integrals
Example
Evaluate
Z 1Z 2
2
(16 − x − 2y )dy dx
0
and
0
Z
0
1Z 2
2
Z
2
0
2
2
(16 − x − 2y )dy dx
0
Z
0
1
=
0
E. Angel (CU)
(16 − x2 − 2y 2 )dx dy
0
1 Z 2
(16 − x − 2y )dy dx =
0
2Z 1
Z
2
Calculus II
2y 3
16y − x y −
3
2
2
dx
y=0
3 Dec
13 / 32
Evaluation of Double Integrals
Example
Evaluate
Z 1Z 2
2
(16 − x − 2y )dy dx
0
and
0
Z
0
1Z 2
2
2
Z
0
(16 − x2 − 2y 2 )dx dy
0
1 Z 2
(16 − x − 2y )dy dx =
0
2Z 1
Z
2
2
(16 − x − 2y )dy dx
0
0
1
2y 3
=
16y − x y −
3
0
Z 1
80
=
− 2x2 dx
3
0
Z
E. Angel (CU)
2
Calculus II
2
2
dx
y=0
3 Dec
13 / 32
Evaluation of Double Integrals
Example
Evaluate
Z 1Z 2
2
(16 − x − 2y )dy dx
0
and
0
Z
0
1Z 2
2
2
Z
0
(16 − x2 − 2y 2 )dx dy
0
1 Z 2
(16 − x − 2y )dy dx =
0
2Z 1
Z
2
2
2
(16 − x − 2y )dy dx
0
0
2
2y 3
=
16y − x y −
dx
3 y=0
0
Z 1
80
2
=
− 2x dx
3
0
1
80 2
80x 2x3
78
=
−
=
− =
3
3 0
3
3
3
Z
E. Angel (CU)
1
Calculus II
2
3 Dec
13 / 32
Evaluation of Double Integrals
Example
Evaluate
Z 1Z 2
2
2
(16 − x − 2y )dy dx
0
Z
and
0
E. Angel (CU)
2Z 1
0
Calculus II
(16 − x2 − 2y 2 )dx dy
0
3 Dec
14 / 32
Evaluation of Double Integrals
Example
Evaluate
Z 1Z 2
2
(16 − x − 2y )dy dx
0
and
0
Z
0
2Z 1
2
2
Z
2 Z 1
(16 − x − 2y )dx dy =
0
2Z 1
Z
2
0
E. Angel (CU)
(16 − x2 − 2y 2 )dx dy
0
2
2
(16 − x − 2y )dx dy
0
Calculus II
0
3 Dec
14 / 32
Evaluation of Double Integrals
Example
Evaluate
Z 1Z 2
2
(16 − x − 2y )dy dx
0
and
0
Z
0
2Z 1
2
Z
2
2 Z 1
(16 − x − 2y )dx dy =
0
2Z 1
Z
2
0
0
2
2
(16 − x − 2y )dx dy
0
Z
0
2
=
0
E. Angel (CU)
(16 − x2 − 2y 2 )dx dy
Calculus II
x3
16x −
− 2y 2 x
3
1
dy
x=0
3 Dec
14 / 32
Evaluation of Double Integrals
Example
Evaluate
Z 1Z 2
2
(16 − x − 2y )dy dx
0
and
0
Z
0
2Z 1
2
2
Z
2 Z 1
(16 − x − 2y )dx dy =
0
2Z 1
Z
2
0
(16 − x2 − 2y 2 )dx dy
0
2
(16 − x − 2y )dx dy
0
0
2
x3
=
16x −
− 2y 2 x
3
0
Z 2
47
2
=
− 2y dy
3
0
Z
E. Angel (CU)
2
Calculus II
1
dy
x=0
3 Dec
14 / 32
Evaluation of Double Integrals
Example
Evaluate
Z 1Z 2
2
(16 − x − 2y )dy dx
0
and
0
Z
0
2Z 1
2
2
Z
2 Z 1
(16 − x − 2y )dx dy =
0
2Z 1
Z
2
0
(16 − x2 − 2y 2 )dx dy
0
2
2
(16 − x − 2y )dx dy
0
0
1
x3
2
dy
=
16x −
− 2y x
3
0
x=0
Z 2
47
2
=
− 2y dy
3
0
2
94 16
78
47y 2y 3
=
−
=
−
=
3
3 0
3
3
3
Z
E. Angel (CU)
2
Calculus II
3 Dec
14 / 32
Evaluation of Double Integrals
At this point it is fair to ask for a geometric interpretation of the iterated
integral. The volume beneath the surface z = 16 − x2 − 2y 2 and above
the rectangle R = [0, 1] × [0, 2] can be obtained by the method of slicing:
Z 1
V =
A(x) dx
0
where A(x) is the area of a vertical cross section of the surface taken
perpendicular to the x-axis.
E. Angel (CU)
Calculus II
3 Dec
15 / 32
Evaluation of Double Integrals
At this point it is fair to ask for a geometric interpretation of the iterated
integral. The volume beneath the surface z = 16 − x2 − 2y 2 and above
the rectangle R = [0, 1] × [0, 2] can be obtained by the method of slicing:
Z 1
V =
A(x) dx
0
where A(x) is the area of a vertical cross section of the surface taken
perpendicular to the x-axis. To find the cross sectional area at a fixed
value of x, where 0 ≤ x ≤ 1, we integrate height with x held constant:
Z 2
80
A(x) =
(16 − x2 − 2y 2 ) dy =
− 2x2
3
0
E. Angel (CU)
Calculus II
3 Dec
15 / 32
Evaluation of Double Integrals
At this point it is fair to ask for a geometric interpretation of the iterated
integral. The volume beneath the surface z = 16 − x2 − 2y 2 and above
the rectangle R = [0, 1] × [0, 2] can be obtained by the method of slicing:
Z 1
V =
A(x) dx
0
where A(x) is the area of a vertical cross section of the surface taken
perpendicular to the x-axis. To find the cross sectional area at a fixed
value of x, where 0 ≤ x ≤ 1, we integrate height with x held constant:
Z 2
80
A(x) =
(16 − x2 − 2y 2 ) dy =
− 2x2
3
0
Notice that in the previous example, we found
Z 1Z 2
Z 1
80
2
2
2
− 2x dx
(16 − x − 2y )dy dx =
3
0
0
0
E. Angel (CU)
Calculus II
3 Dec
15 / 32
Evaluation of Double Integrals
In the previous example, it was not coincidence that we obtained the same
value upon integrating in different orders. This can be seen geometrically
by the method of slicing: integrating by cross sections perpendicular to the
x-axis must produce the same result as integrating by cross sections
perpendicular to the y-axis.
E. Angel (CU)
Calculus II
3 Dec
16 / 32
Evaluation of Double Integrals
In the previous example, it was not coincidence that we obtained the same
value upon integrating in different orders. This can be seen geometrically
by the method of slicing: integrating by cross sections perpendicular to the
x-axis must produce the same result as integrating by cross sections
perpendicular to the y-axis.
Theorem
Let R = [a, b] × [c, d]. If f (x, y) is continuous on R, then
ZZ
Z
dZ b
f (x, y) dA =
R
E. Angel (CU)
Z bZ
f (x, y) dx dy =
c
a
f (x, y) dy dx
a
Calculus II
d
c
3 Dec
16 / 32
Evaluation of Double Integrals
In the previous example, it was not coincidence that we obtained the same
value upon integrating in different orders. This can be seen geometrically
by the method of slicing: integrating by cross sections perpendicular to the
x-axis must produce the same result as integrating by cross sections
perpendicular to the y-axis.
Theorem
Let R = [a, b] × [c, d]. If f (x, y) is continuous on R, then
ZZ
Z
dZ b
f (x, y) dA =
R
Z bZ
f (x, y) dx dy =
c
a
d
f (x, y) dy dx
a
c
So we can evaluate double integrals over rectangles by converting to either
iterated integral.
E. Angel (CU)
Calculus II
3 Dec
16 / 32
Evaluation of Double Integrals
Example
Evaluate
ZZ
x3 y dA
R
where R = [−1, 2] × [0, 3].
E. Angel (CU)
Calculus II
3 Dec
17 / 32
Evaluation of Double Integrals
Example
Evaluate
ZZ
x3 y dA
R
where R = [−1, 2] × [0, 3].
We can evaluate using iterated integrals in either order. Let’s integrate
with respect to y first:
E. Angel (CU)
Calculus II
3 Dec
17 / 32
Evaluation of Double Integrals
Example
Evaluate
ZZ
x3 y dA
R
where R = [−1, 2] × [0, 3].
We can evaluate using iterated integrals in either order. Let’s integrate
with respect to y first:
ZZ
Z 2Z 3
3
x y dA =
x3 y dy dx
R
E. Angel (CU)
−1
0
Calculus II
3 Dec
17 / 32
Evaluation of Double Integrals
Example
Evaluate
ZZ
x3 y dA
R
where R = [−1, 2] × [0, 3].
We can evaluate using iterated integrals in either order. Let’s integrate
with respect to y first:
ZZ
Z 2Z 3
3
x y dA =
x3 y dy dx
R
−1 0
2 3 2 3
x y
Z
=
−1
E. Angel (CU)
2
dx
0
Calculus II
3 Dec
17 / 32
Evaluation of Double Integrals
Example
Evaluate
ZZ
x3 y dA
R
where R = [−1, 2] × [0, 3].
We can evaluate using iterated integrals in either order. Let’s integrate
with respect to y first:
ZZ
Z 2Z 3
3
x y dA =
x3 y dy dx
R
−1 0
2 3 2 3
x y
Z
=
−1
Z 2
=
−1
E. Angel (CU)
2
dx
0
9 4 2
9
9 3
x dx =
x
= 18 −
2
8
8
−1
Calculus II
3 Dec
17 / 32
Properties
The double integral satisfies linearity:
ZZ
ZZ
cf (x, y) dA = c
f (x, y) dA
R
R
ZZ
ZZ
[f (x, y) + g(x, y)] dA =
R
ZZ
f (x, y) dA +
R
g(x, y) dA
R
Also, if a region R in the plane can be subdivided into regions R1 and R2 ,
then the solid between z = f (x, y) and R is subdivided into two solids,
and the following holds:
ZZ
ZZ
f (x, y) dA =
R
E. Angel (CU)
ZZ
f (x, y) dA +
R1
f (x, y) dA
R2
Calculus II
3 Dec
18 / 32
Nonconstant Limits of Integration
Now let’s learn how to evaluate iterated integrals with nonconstant limits
of integration:
Z bZ
Z b "Z
g2 (x)
g2 (x)
f (x, y) dy dx =
a
E. Angel (CU)
g1 (x)
#
f (x, y) dy dx
a
Calculus II
g1 (x)
3 Dec
19 / 32
Nonconstant Limits of Integration
Now let’s learn how to evaluate iterated integrals with nonconstant limits
of integration:
Z bZ
Z b "Z
g2 (x)
g2 (x)
f (x, y) dy dx =
a
Z
g1 (x)
f (x, y) dy dx
a
d Z h2 (y)
Z
g1 (x)
d
"Z
h2 (y)
f (x, y) dx dy =
c
E. Angel (CU)
h1 (y)
#
#
f (x, y) dx dy
c
Calculus II
h1 (y)
3 Dec
19 / 32
Nonconstant Limits of Integration
Now let’s learn how to evaluate iterated integrals with nonconstant limits
of integration:
Z bZ
Z b "Z
g2 (x)
g2 (x)
f (x, y) dy dx =
a
Z
g1 (x)
f (x, y) dy dx
a
d Z h2 (y)
Z
g1 (x)
d
"Z
h2 (y)
f (x, y) dx dy =
c
h1 (y)
#
#
f (x, y) dx dy
c
h1 (y)
In the first case we first integrate f with respect to y, then evaluate along
the variable limits g1 (x) and g2 (x). In the second case we integrate f with
respect to x, then evaluate along the variable limits h1 (y) and h2 (y).
E. Angel (CU)
Calculus II
3 Dec
19 / 32
Nonconstant Limits of Integration
An example of the first type:
Example
Z
0
E. Angel (CU)
1Z x
x2
xy 2 dy dx =
Z
0
1
xy 3
3
Calculus II
x
dx
y=x2
3 Dec
20 / 32
Nonconstant Limits of Integration
An example of the first type:
Example
Z
0
1Z x
xy 2 dy dx =
x2
Z
0
Z
=
0
E. Angel (CU)
1
1
xy 3
3
x
dx
y=x2
x(x3 ) x(x2 )3
−
3
3
Calculus II
dx
3 Dec
20 / 32
Nonconstant Limits of Integration
An example of the first type:
Example
Z
0
1Z x
xy 2 dy dx =
x2
1
Z
0
1
Z
=
0
=
E. Angel (CU)
1
3
Z
xy 3
3
x
dx
y=x2
x(x3 ) x(x2 )3
−
3
3
dx
1
(x4 − x7 ) dx
0
Calculus II
3 Dec
20 / 32
Nonconstant Limits of Integration
An example of the first type:
Example
Z
0
1Z x
xy 2 dy dx =
x2
1
Z
0
1
Z
=
0
=
1
3
Z
xy 3
3
x
dx
y=x2
x(x3 ) x(x2 )3
−
3
3
dx
1
(x4 − x7 ) dx
0
1 x5 x8
=
−
3 5
8
E. Angel (CU)
Calculus II
1
0
3 Dec
20 / 32
Nonconstant Limits of Integration
An example of the first type:
Example
Z
0
1Z x
xy 2 dy dx =
x2
1
Z
0
1
Z
=
0
=
1
3
Z
xy 3
3
x
dx
y=x2
x(x3 ) x(x2 )3
−
3
3
dx
1
(x4 − x7 ) dx
0
1 x5 x8
=
−
3 5
8
1 1 1
=
−
3 5 8
E. Angel (CU)
Calculus II
1
0
3 Dec
20 / 32
Nonconstant Limits of Integration
An example of the second type:
Example
Z
3/2 Z 3−y
Z
y dx dy =
1
E. Angel (CU)
y
1
3/2
3−y
dy
[xy] x=y
Calculus II
3 Dec
21 / 32
Nonconstant Limits of Integration
An example of the second type:
Example
Z
3/2 Z 3−y
Z
3/2
y dx dy =
1
y
1
Z
3−y
dy
[xy] x=y
3/2
[(3 − y)y − (y)y] dy
=
1
E. Angel (CU)
Calculus II
3 Dec
21 / 32
Nonconstant Limits of Integration
An example of the second type:
Example
Z
3/2 Z 3−y
Z
3/2
y dx dy =
1
y
1
Z
3−y
dy
[xy] x=y
3/2
[(3 − y)y − (y)y] dy
=
1
Z
=
3/2
(−2y 2 + 3y) dy
1
E. Angel (CU)
Calculus II
3 Dec
21 / 32
Nonconstant Limits of Integration
An example of the second type:
Example
Z
3/2 Z 3−y
Z
3/2
y dx dy =
1
y
1
Z
3−y
dy
[xy] x=y
3/2
[(3 − y)y − (y)y] dy
=
1
Z
=
3/2
(−2y 2 + 3y) dy
1
2y 3 3y 2
+
= −
3
2
E. Angel (CU)
Calculus II
3/2
1
3 Dec
21 / 32
Nonconstant Limits of Integration
An example of the second type:
Example
Z
3/2 Z 3−y
Z
3/2
y dx dy =
1
y
1
Z
3−y
dy
[xy] x=y
3/2
[(3 − y)y − (y)y] dy
=
1
Z
=
3/2
(−2y 2 + 3y) dy
1
3/2
2y 3 3y 2
+
= −
3
2
1 9
27
2 3
= −
+
− − +
4
8
3 2
E. Angel (CU)
Calculus II
3 Dec
21 / 32
Double Integrals Over Nonrectangular Regions
Definition
: A type I region is bounded on the left and
right by the vertical lines x = a and x = b
and above and below by continuous curves
y = g1 (x) and y = g2 (x), where
g1 (x) ≤ g2 (x) for a ≤ x ≤ b.
E. Angel (CU)
Calculus II
3 Dec
22 / 32
Double Integrals Over Nonrectangular Regions
Definition
: A type I region is bounded on the left and
right by the vertical lines x = a and x = b
and above and below by continuous curves
y = g1 (x) and y = g2 (x), where
g1 (x) ≤ g2 (x) for a ≤ x ≤ b.
: A type II region is bounded below and
above by horizontal lines y = c and y = d
and is bounded on the left and right by
continuous curves x = h1 (y) and x = h2 (y),
where h1 (y) ≤ h2 (y) for c ≤ y ≤ d.
E. Angel (CU)
Calculus II
3 Dec
22 / 32
Double Integrals Over Nonrectangular Regions
We can evaluate integrals over type I and type II regions using iterated
integrals with nonconstant limits of integration.
Theorem
1
If R is a type I region on which f (x, y) is continuous, then
Z bZ
ZZ
g2 (x)
f (x, y) dA =
f (x, y) dy dx
R
2
a
g1 (x)
If R is a type II region on which f (x, y) is continuous, then
ZZ
Z
d Z h2 (y)
f (x, y) dA =
R
E. Angel (CU)
f (x, y) dx dy
c
Calculus II
h1 (y)
3 Dec
23 / 32
Finding Volume
Example
Find the volume of the prism whose base is the triangle in the xy-plane
bounded by the x-axis and the lines y = x and x = 1 and whose top lies in
the plane z = f (x, y) = 3 − x − y.
E. Angel (CU)
Calculus II
3 Dec
24 / 32
Finding Volume
Example
Find the volume of the prism whose base is the triangle in the xy-plane
bounded by the x-axis and the lines y = x and x = 1 and whose top lies in
the plane z = f (x, y) = 3 − x − y.
: The prism is pictured to the right. The
base is the triangle R. We will find the
volume two ways: First by considering R as
a type I region, then as a type II region.
E. Angel (CU)
Calculus II
3 Dec
24 / 32
Finding Volume
Example
Find the volume of the prism whose base is the triangle in the xy-plane
bounded by the x-axis and the lines y = x and x = 1 and whose top lies in
the plane z = f (x, y) = 3 − x − y.
E. Angel (CU)
Calculus II
3 Dec
25 / 32
Finding Volume
Example
Find the volume of the prism whose base is the triangle in the xy-plane
bounded by the x-axis and the lines y = x and x = 1 and whose top lies in
the plane z = f (x, y) = 3 − x − y.
For any x between 0 and 1, y may vary from
y = 0 to y = x. So
E. Angel (CU)
Calculus II
3 Dec
25 / 32
Finding Volume
Example
Find the volume of the prism whose base is the triangle in the xy-plane
bounded by the x-axis and the lines y = x and x = 1 and whose top lies in
the plane z = f (x, y) = 3 − x − y.
For any x between 0 and 1, y may vary from
y = 0 to y = x. So
Z 1Z x
V =
(3 − x − y) dy dx
0
E. Angel (CU)
0
Calculus II
3 Dec
25 / 32
Finding Volume
Example
Find the volume of the prism whose base is the triangle in the xy-plane
bounded by the x-axis and the lines y = x and x = 1 and whose top lies in
the plane z = f (x, y) = 3 − x − y.
For any x between 0 and 1, y may vary from
y = 0 to y = x. So
Z 1Z x
V =
(3 − x − y) dy dx
0
0
y=x
Z 1
y2
=
3y − xy −
dx
2 y=0
0
E. Angel (CU)
Calculus II
3 Dec
25 / 32
Finding Volume
Example
Find the volume of the prism whose base is the triangle in the xy-plane
bounded by the x-axis and the lines y = x and x = 1 and whose top lies in
the plane z = f (x, y) = 3 − x − y.
For any x between 0 and 1, y may vary from
y = 0 to y = x. So
Z 1Z x
V =
(3 − x − y) dy dx
0
0
y=x
Z 1
y2
=
3y − xy −
dx
2 y=0
0
Z 1
3x2
=
3x −
dx
2
0
E. Angel (CU)
Calculus II
3 Dec
25 / 32
Finding Volume
Example
Find the volume of the prism whose base is the triangle in the xy-plane
bounded by the x-axis and the lines y = x and x = 1 and whose top lies in
the plane z = f (x, y) = 3 − x − y.
For any x between 0 and 1, y may vary from
y = 0 to y = x. So
Z 1Z x
V =
(3 − x − y) dy dx
0
0
y=x
Z 1
y2
=
3y − xy −
dx
2 y=0
0
Z 1
3x2
=
3x −
dx
2
0
=1
E. Angel (CU)
Calculus II
3 Dec
25 / 32
Finding Volume
Example
Find the volume of the prism whose base is the triangle in the xy-plane
bounded by the x-axis and the lines y = x and x = 1 and whose top lies in
the plane z = f (x, y) = 3 − x − y.
E. Angel (CU)
Calculus II
3 Dec
26 / 32
Finding Volume
Example
Find the volume of the prism whose base is the triangle in the xy-plane
bounded by the x-axis and the lines y = x and x = 1 and whose top lies in
the plane z = f (x, y) = 3 − x − y.
We can reverse the order of integration, and the
volume integral becomes
E. Angel (CU)
Calculus II
3 Dec
26 / 32
Finding Volume
Example
Find the volume of the prism whose base is the triangle in the xy-plane
bounded by the x-axis and the lines y = x and x = 1 and whose top lies in
the plane z = f (x, y) = 3 − x − y.
We can reverse the order of integration, and the
volume integral becomes
Z 1Z 1
V =
(3 − x − y) dx dy
0
y
E. Angel (CU)
Calculus II
3 Dec
26 / 32
Finding Volume
Example
Find the volume of the prism whose base is the triangle in the xy-plane
bounded by the x-axis and the lines y = x and x = 1 and whose top lies in
the plane z = f (x, y) = 3 − x − y.
We can reverse the order of integration, and the
volume integral becomes
Z 1Z 1
V =
(3 − x − y) dx dy
0
Z
=
0
y
1
x2
3x −
− xy
2
E. Angel (CU)
x=1
dy
x=y
Calculus II
3 Dec
26 / 32
Finding Volume
Example
Find the volume of the prism whose base is the triangle in the xy-plane
bounded by the x-axis and the lines y = x and x = 1 and whose top lies in
the plane z = f (x, y) = 3 − x − y.
We can reverse the order of integration, and the
volume integral becomes
Z 1Z 1
V =
(3 − x − y) dx dy
0
y
x=1
x2
=
3x −
− xy
dy
2
0
x=y
Z 1
y2
1
2
=
3 − − y − 3y +
+y
dy
2
2
0
Z
1
E. Angel (CU)
Calculus II
3 Dec
26 / 32
Finding Volume
Example
Find the volume of the prism whose base is the triangle in the xy-plane
bounded by the x-axis and the lines y = x and x = 1 and whose top lies in
the plane z = f (x, y) = 3 − x − y.
We can reverse the order of integration, and the
volume integral becomes
Z 1Z 1
V =
(3 − x − y) dx dy
0
y
x=1
x2
=
3x −
− xy
dy
2
0
x=y
Z 1
y2
1
2
=
3 − − y − 3y +
+y
dy
2
2
0
=1
Z
1
E. Angel (CU)
Calculus II
3 Dec
26 / 32
Evaluating a Double Integral
Example
Calculate
ZZ
R
sin x
dA,
x
where R is the triangle in the xy-plane bounded by the x-axis, the line
y = x, and the line x = 1.
E. Angel (CU)
Calculus II
3 Dec
27 / 32
Evaluating a Double Integral
Example
Calculate
ZZ
R
sin x
dA,
x
where R is the triangle in the xy-plane bounded by the x-axis, the line
y = x, and the line x = 1.
The region of integration is shown to the right. If
we integrate first with respect to y and then with
respect to x,
Z 1Z x
Z 1
sin x
sin x y=x
dy dx =
dx
y
x
x y=0
0
0
0
Z 1
sin x dx = − cos(1) + 1
=
0
E. Angel (CU)
Calculus II
3 Dec
27 / 32
Evaluating a Double Integral
Example
Calculate
ZZ
R
sin x
dA,
x
where R is the triangle in the xy-plane bounded by the x-axis, the line
y = x, and the line x = 1.
But if we attempt to reverse the order of
integration and calculate
Z 1Z 1
sin x
dx dy,
x
0
y
R
then we run into a problem. ((sin x)/x) dx
cannot be expressed in terms of elementary
functions (i.e., there is no simple antiderivative).
E. Angel (CU)
Calculus II
3 Dec
28 / 32
Finding Limits of Integration
The procedure to determine limits of
integration (for a type I region) follows.
Regions that are more complicated can often
be split into pieces on which this type of
procedure works.
Sketch: Sketch the region of integration and
label the bounding curves.
E. Angel (CU)
Calculus II
3 Dec
29 / 32
Finding Limits of Integration
Find y-limits: Imagine a vertical line L
cutting through R in the direction of
increasing y. Mark the y-values where L
enters and leaves. These are the y-limits of
integration and are generally functions of x
(instead of constants).
E. Angel (CU)
Calculus II
3 Dec
30 / 32
Finding Limits of Integration
Find x-limits: Choose the x-limits that
include all the vertical lines through R. The
integral shown here is
ZZ
f (x, y) dA
R
Z
√
x=1 Z y= 1−x2
=
f (x, y) dy dx
x=0
E. Angel (CU)
y=1−x
Calculus II
3 Dec
31 / 32
Reversing the Order of Integration
Example
Sketch the region for the integral
Z 2 Z 2x
(4x + 2) dy dx
0
x2
and write an equivalent integral with the order of integration reversed.
E. Angel (CU)
Calculus II
3 Dec
32 / 32
Reversing the Order of Integration
Example
Sketch the region for the integral
Z 2 Z 2x
(4x + 2) dy dx
0
x2
and write an equivalent integral with the order of integration reversed.
The region is given by x2 ≤ y ≤ 2x and 0 ≤ x ≤ 2, i.e., the region bounded
by y = x2 , y = 2x, and between x = 0 and x = 2.
E. Angel (CU)
Calculus II
3 Dec
32 / 32
Reversing the Order of Integration
Example
Sketch the region for the integral
Z 2 Z 2x
(4x + 2) dy dx
0
x2
and write an equivalent integral with the order of integration reversed.
The region is given by x2 ≤ y ≤ 2x and 0 ≤ x ≤ 2, i.e., the region bounded
by y = x2 , y = 2x, and between x = 0 and x = 2.
To find the limits of integration in the reverse order, imagine a horizontal
line passing from left to right through the region. It enters at x = y/2 and
√
leaves at x = y. To include all such lines, let y run from y = 0 to y = 4.
The integral is
Z 4 Z √y
(4x + 2) dx dy
0
E. Angel (CU)
y/2
Calculus II
3 Dec
32 / 32