# Section 13.6: Applications of Double Integrals

```Section 13.6: Applications of Double Integrals
In this section, we explore physical applications of double integrals such as computing
mass, center of mass, and electrical charge.
Theorem: (Total Electrical Charge)
Suppose an electrical charge is distributed over a region D in the xy-plane and the charge
density at a point (x, y) ∈ D is given by σ(x, y). Then the total charge is given by
ZZ
Q=
σ(x, y)dA.
D
Example: Electric charge is distributed over the unit disk x2 + y 2 ≤ 1 so that the charge
density at (x, y) is σ(x, y) = 1 + x2 + y 2 (measured in coulombs per square meter). Find the
total charge on the disk.
Using polar coordinates, the total charge is
ZZ
Q =
σ(x, y)dA
D
Z 2π Z 1
=
(1 + r2 )rdrdθ
0
Z 10
= 2π
(r + r3 )dr
0
1
1 2 1 4 r + r = 2π
2
4
0
3π
=
C.
2
Theorem: (Center of Mass of a Lamina)
Suppose a lamina occupies some region D in the xy-plane and its density at a point (x, y) ∈ D
is given by ρ(x, y). Then the total mass of the lamina is given by
ZZ
m=
ρ(x, y)dA.
D
The coordinates (x̄, ȳ) of the center of mass of the lamina are
ZZ
ZZ
1
1
x̄ =
xρ(x, y)dA
ȳ =
yρ(x, y)dA.
m D
m D
Example:
√ Find the mass and center of mass of the lamina that occupies the region D bounded
by y = x, y = 0, and x = 1 with density function ρ(x, y) = 3x.
The mass of the lamina is
ZZ
ρ(x, y)dA
m =
Z
D
√
1Z
x
3xdydx
=
0
Z
=
=
0
1
3x3/2 dx
0
1
6 5/2 x 5
0
6
=
.
5
Using the formulas in the previous theorem,
ZZ
1
x̄ =
xρ(x, y)dA
m D
Z Z √x
5 1
=
3x2 dydx
6 0 0
Z
5 1 5/2
=
x dx
2 0
1
5 7/2 =
x 7
0
5
=
7
and
ZZ
1
yρ(x, y)dA
ȳ =
m D
Z Z √x
5 1
=
3xydydx
6 0 0
√x
Z
5 1 2 xy dx
=
4 0
0
Z 1
5
x2 dx
=
4 0
1
5 2 =
x
12 0
5
=
.
12
The mass of the lamina is m =
6
5
and the center of mass is (x̄, ȳ) =
5 5
,
7 12
.
```