Double Integrals - Examples -
c
CNMiKnO PG - 1
Double Integrals - Techniques and Examples
Iterated integrals on a rectangle
If function f is continuous on an integral [a, b] × [c, d], then:
ZZ
[a,b]×[c,d]
f(x, y) dxdy =
b
Z
a
d
Z
c
f(x, y) dy
dx =
d
Z
c
b
Z
a
f(x, y) dx
dy.
Notation
Instead of
b
R
a
d
R
c
f(x, y) dy
dx we may also write
b
R
a
dx
d
R
c
f(x, y) dy.
Instead of
d
R
c
b
R
a
f(x, y) dx
dy we may also write
d
R
c
dy
b
R
a
f(x, y) dx.
Example 1. Calculate
RR
R
x
y
2
dxdy, where R = [1, 2] × [4, 6].
Solution:
RR
R
x
y
2
dxdy =
RR
[1,2]×[4,6]
x
y
2
dxdy =
2
R
1
6
R
4
x
y
2
dy
dx =
2
R
1
([−
x
y
]
y=6
y=4
) dx =
2
R
1
(
x
4
−
x
6
) dx =
2
R
1
x
12
dx = [
x
2
24
]
x=2
x=1
=
4−1
24
=
3
24
=
1
8
.
A double integral of a function with separable variables
If function f is of form f(x, y) = g(x) ·h(y) and g is continuous in [a, b] and h is continuous in
[c, d], then:
ZZ
[a,b]×[c,d]
f(x, y) dxdy =
b
Z
a
g(x)dx
·
d
Z
c
h(y) dy
.
Example 2. Calculate
RR
R
x
y
2
dxdy, where R = [1, 2] × [4, 6], separating variables.
Solution:
RR
R
x
y
2
dxdy =
RR
R
x ·
1
y
2
dxdy =
2
R
1
x dx
·
6
R
4
dy
y
2
= ( [
x
2
2
]
x=2
x=1
) · ( [−
1
y
]
y=6
y=4
) =
(
4−1
2
) · (−
1
6
+
1
4
) =
3
2
·
−2+3
12
=
3
24
=
1
8
.
A double integral over a simple region
If f is a continuous function on the vertically simple region
D = {(x, y) : a ≤ x ≤ b, g(x) ≤ y ≤ h(x)},
Double Integrals - Examples -
c
CNMiKnO PG - 2
then
ZZ
D
f(x, y) dP =
b
Z
a
h(x)
Z
g(x)
f(x, y) dy
dx.
If f is a continuous function on the horizontally simple region
D = {(x, y) : c ≤ y ≤ d, p(y) ≤ x ≤ q(y)},
then
ZZ
D
f(x, y) dP =
d
Z
c
q(y)
Z
p(y)
f(x, y) dx
dy.
Example 3. Evaluate
RR
D
(x+y) dxdy over a region
bounded by curves xy = 6 and x + y = 7. Sketch
a diagram of the region.
Solution: From the s ystem of equations of xy = 6
and x + y = 7 (or: y =
6
x
, y = 7 − x) we obtain
two intersection points: A = (1, 6) and B = (6, 1).
Region D is vertically simple, so:
ZZ
D
(x + y) dxdy =
6
Z
1
y=7−x
Z
y=
6
x
(x + y) dy
dx =
6
Z
1
[xy +
y
2
2
]
y=7−x
y=
6
x
dx
=
6
Z
1
x(7 − x) +
(7 − x)
2
2
− x ·
6
x
−
36
2x
2
dx
=
6
Z
1
(−
x
2
2
−
18
x
2
+
37
2
) dx = [−
x
3
6
+
18
x
+
37x
2
]
6
1
=
125
3
.
Example 4. Evaluate
RR
D
(x − y) dxdy over a region bounded by curves
x = y
2
and x =
y
2
2
+ 1. Sketch a diagram of the region.
Solution: From the system of equations of x = y
2
and x =
y
2
2
+ 1 we obtain
two intersection points: (−
√
2, 2) and (
√
2, 2). Region D is horizontally
simple, so:
ZZ
D
(x − y) dxdy =
√
2
Z
−
√
2
y
2
2
+1
Z
y
2
(x − y) dx
dy =
√
2
Z
−
√
2
[
x
2
2
− xy]
x=
y
2
2
+1
x=y
2
dy =
Double Integrals - Examples -
c
CNMiKnO PG - 3
=
√
2
Z
−
√
2
(
y
2
2
+ 1)
2
2
− (
y
2
2
+ 1)y −
y
4
2
+ y
3
dy =
√
2
Z
−
√
2
−
3y
4
8
+
y
3
2
+
y
2
2
− y +
1
2
dy =
= [ −
3y
5
40
+
y
4
8
+
y
3
6
−
y
2
2
+
y
2
]
√
2
−
√
2
=
16
√
2
15
.
Iterated integrals in a reversed order
Example 5. Sketch the region over which the
integration
3
R
1
x
R
−x+2
(2x + 1) dydx takes place and
write an equivalent integral with the order of
integration reversed. Evaluate both integrals.
Solution: First let us evaluate:
3
Z
1
x
Z
−x+2
(2x + 1) dydx =
3
Z
1
( [y(2x + 1)]
x
−x+2
) dx =
3
Z
1
( x (2x + 1) − (−x + 2)(2x + 1) ) dx
=
3
Z
1
(−2 − 2x + 4x
2
) dx = [−2x − x
2
+
4x
3
3
]
3
1
=
68
3
.
To reverse the order of integration, we need to di-
vide the region into two parts that are horizontally
simple. Now:
3
Z
1
x
Z
−x+2
(2x + 1) dydx =
3
Z
1
3
Z
y
(2x + 1) dxdy +
1
Z
−1
3
Z
−y+2
(2x + 1) dxdy
=
3
Z
1
[x + x
2
]
3
y
dy +
1
Z
−1
[x + x
2
]
3
−y+2
dy =
3
Z
1
(12 − y − y
2
) dy +
1
Z
−1
(6 + 5y − y
2
) dy
= [12y −
y
2
2
−
y
3
3
]
3
1
+ [6y +
5y
2
2
−
y
3
3
]
1
−1
=
34
3
+
34
3
=
68
3
.
Double Integrals - Examples -
c
CNMiKnO PG - 4
Polar coordinates
For any point P other than the origin, let r be the distance between P and the origin, and ϕ
an angle having its initial side on the positive x axis and its terminal side on the line segment
joining P and the origin. The pair (r, ϕ) is called a set of polar coordinates for the point P .
Every point (x, y) in the plane has both Cartesian and polar coordinates (r, ϕ):
x = r cos ϕ
y = r sin ϕ
.
We have the following result for polar coordinates:
Z
D
Z
f(x, y) dxdy =
Z
∆
Z
f (r cos ϕ, r sin ϕ) r drdϕ .
Example 6. Using polar coordinates, calculate
RR
D
xy
2
dxdy where
D : x
2
+ y
2
≤ 4, x ≥ 0.
Solution: The region of integration is a semicircle with radius equal
2. Therefore, the region in polar coordinates is given by
−π
2
≤ θ ≤
π
2
and 0 ≤ r ≤ 2.
After s ubstituting x and y with polar coordinates, we have:
ZZ
D
xy
2
dxdy =
π
2
Z
−
π
2
2
Z
0
(r cos θ) · (r sin θ)
2
r dr
dθ =
π
2
Z
−
π
2
2
Z
0
r
4
sin
2
θ cos θdr
dθ
=
π
2
Z
−
π
2
sin
2
θ cos θdθ
·
2
Z
0
r
4
dr
= [
sin
3
θ
3
]
π
2
−
π
2
· [
r
5
5
]
2
0
=
64
15
.
Example 7. Using polar coordinates, calculate
RR
D
(x
2
+ y
2
) dxdy, where D : x
2
+ y
2
−2y ≤ 0.
Solution (a): Let us represent the equation describing D in a different form:
x
2
+ y
2
− 2y ≤ 0
x
2
+ (y
2
− 2y + 1) − 1 ≤ 0
x
2
+ (y − 1)
2
≤ 1
Double Integrals - Examples -
c
CNMiKnO PG - 5
Such an equation describes a circle with the origin in (0, 1), so we cannot describe it with polar
coordinates as easily as in Example 6. Let us substitute x = r cos θ and y = r sin θ:
x
2
+ y
2
− 2y ≤ 0
r
2
cos
2
θ + r
2
sin
2
θ − 2r sin θ ≤ 0
r ≤ 2 sin θ
the integral is equal to:
ZZ
D
(x
2
+ y
2
) dxdy =
Z
π
0
2 sin θ
Z
0
r
2
(sin
2
θ + cos
2
θ)rdr
dθ =
π
Z
0
(
2 sin θ
Z
0
r
3
dr)dθ =
π
Z
0
[
r
4
4
]
2 sin θ
0
dθ
= 4
π
Z
0
sin
4
θdθ = 4[
3θ
8
−
sin 2θ
4
+
sin 4θ
32
]
π
0
=
3π
2
.
Angle θ ranges from 0 to only π, because for θ ∈ (π, 2π] the radius would be negative – w hich
is impossible.
Solution (b): Since the c ircle is moved by a vector of ~v = (0, 1), then we can also move the
function x
2
+ y
2
by the same vector. The new function will be x
2
+ (y − 1)
2
. We can now use
the me thod from Example 6:
ZZ
D
(x
2
+ y
2
) dxdy =
2π
Z
0
1
Z
0
(r
2
cos
2
θ + (r sin θ − 1)
2
)rdr
dθ = ··· =
3π
2
.
Area of a bounded region in the plane
The area of a closed bounded plane region R is given by the formula
Area =
RR
R
1 dxdy.
Example 8. Calculate the area of a region bounded by curves
y =
1
x
, y =
√
x and a line x = 2. Sketch the region.
Solution: The area is equal to:
2
Z
1
(
y=
√
x
Z
y=
1
x
1 dy) dx =
2
Z
1
[y]
y=
√
x
y=
1
x
dx =
2
Z
1
(
√
x −
1
x
) dx = [
2
3
x
3
2
− ln |x|]
2
1
=
1
3
(−2 + 4
√
2 − ln 8).
Double Integrals - Examples -
c
CNMiKnO PG - 6
Volume
Let R be a a bounded region in the OXY plane and f be a function continuous on R. If f is
nonnegative and integrable on R, then the volume of the solid region between the graph of f
and R is given by
V olume =
RR
R
f(x, y) dxdy.
Let R be a a bounded region in the xy plane and g
1
, g
2
be continuous functions on R. If g
1
and g
2
are integrable on R such that g
1
(x, y) ≤ g
2
(x, y), then the volume of the solid region
between the graph of g
1
and g
2
is given by
V olume =
RR
R
(g
2
(x, y) − g
1
(x, y)) dxdy.
Example 9. Calculate the volume of a solid bounded by curves y = x
2
, y = 1, z = 0, z = 2y.
solution: The region of integration is bounded by y = x
2
and y = 1 and f(x, y) = 2y.
Therefore:
V olume =
x=1
Z
x=−1
(
y=1
Z
y=x
2
2y dy) dx =
x=1
Z
x=−1
[y
2
]
y=1
y=x
2
dx =
x=1
Z
x=−1
(1 − x
4
) dx = [x −
x
5
5
]
1
−1
= 1 −
1
5
− (−1 +
1
5
) = 2 −
2
5
=
8
5
.
Surface
Let S be the surface z = f (x, y) where the points (x, y) come from the given region R in the
OXY plane. Then
Area
S
=
RR
R
q
1 + (
∂f
∂x
)
2
+ (
∂f
∂y
)
2
dxdy,
where f and its first partial derivatives are continuous.
Example 10. Calculate the surface of a plane 2x + 2y + z = 8 bounded by the coordinate
system axes.
Solution: After transformations of the equation of a plane, we have
x
4
+
y
4
+
z
8
= 1, so the plane
intersects the coordinate system axes at points A = (4, 0, 0), B = (0, 4, 0) and C = (0, 0, 8).
Therefore, the region of integration is bounded by x = 0, y = 0, y = −x + 4. We also have
Double Integrals - Examples -
c
CNMiKnO PG - 7
f(x, y) = z = 8 − 2x − 2y, so
∂f
∂x
= −2 and
∂f
∂y
= −2. Therefore:
Surf ace =
x=4
Z
x=0
(
y=−x+4
Z
y=0
p
1 + (−2)
2
+ (−2)
2
dy) dx =
x=4
Z
x=0
(
y=−x+4
Z
y=0
√
9 dy) dx = 3
x=4
Z
x=0
(
y=−x+4
Z
y=0
1 dy) dx
= 3
x=4
Z
x=0
[y]
y=−x+4
y=0
dx = 3
x=4
Z
x=0
(−x + 4) dx = 3[−
x
2
2
+ 4x]
4
0
= 24.
