SECTION 12.4

APPLICATIONS OF DOUBLE INTEGRALS

P212.4

APPLICATIONS OF DOUBLE INTEGRALS

We have already seen one application of double integrals: computing volumes.

Another geometric application is finding areas of surfaces and this will be done in the next chapter.

In this section we explore physical applications such as Computing mass Electric charge Center of mass Moment of inertia.

P312.4

DENSITY AND MASS

In Chapter 7, we used single integrals to compute moments and the center of mass of a thin plate or lamina with constant density.

Now, equipped with the double integral, we can consider a lamina with variable density.

P412.4

DENSITY

Suppose the lamina occupies a region D of the xy-plane.

Also, let its density (in units of mass per unit area) at a point (x, y) in D be given by (x, y), where is a continuous function on D.

P512.4

MASS

This means that:

where: Δm and ΔA are the mass

and area of a small rectangle that contains (x, y).

The limit is taken as the dimensions of the rectangle approach 0.

See Figure 1.

( , ) limm

x yA

P612.4

MASS

To find the total mass m of the lamina, we: Divide a rectangle R containing D into subrectangles

Rij of equal size. See Figure 2. Consider (x, y) to be 0

outside D.

P712.4

MASS

If we choose a point (xij*, yij*) in Rij , then the mass of the part of the lamina that occupies Rij is approximately

(xij*, yij*) ∆Aij

where ∆Aij is the area of Rij.

P812.4

MASS

If we add all such masses, we get an approximation to the total mass:

* *

1 1

( , )k l

ij ij iji j

m x y A

P912.4

MASS

If we now increase the number of subrectangles, we obtain the total mass m of the lamina as the limiting value of the approximations:

* *

max , 01 1

lim ( , )

( , )

i j

k l

ij ij ijx yi j

D

m x y A

x y dA

P1012.4

DENSITY AND MASS

Physicists also consider other types of density that can be treated in the same manner. For example, an electric charge is distributed over a

region D and the charge density (in units of charge per unit area) is given by (x, y) at a point (x, y) in D.

Then, the total charge Q is given by:( , )

D

Q x y dA

P1112.4

Example 1

Charge is distributed over the triangular region D in Figure 3 so that the charge density at (x, y) is (x, y) = xy, measured in coulombs per square meter (C/m2).

Find the total charge.

P1212.4

Example 1 SOLUTION

From Equation 2 and Figure 3, we have:

1 1

0 1

121

01

( , )

D

x

y

y x

Q x y dA

xy dy dx

yx dx

x

P1312.4

Example 1 SOLUTION

The total charge is C5

24

1 2 2

0

1 2 312 0

13 4

0

[1 (1 ) ]2

(2 )

1 2 5

2 3 4 24

xx dx

x x dx

x x

P1412.4

MOMENTS AND CENTERS OF MASS

In Section 7.6, we found the center of mass of a lamina with constant density.

Here, we consider a lamina with variable density.

Suppose the lamina occupies a region D and has density function (x, y). Recall from Chapter 7 that we defined the moment

of a particle about an axis as the product of its mass and its directed distance from the axis.

P1512.4

MOMENTS AND CENTERS OF MASS

We divide D into small rectangles as in Figure 2.

Then, the mass of Rij is approximately:

(xij*, yij*) ∆Aij

So, we can approximate the moment of Rij with respect to the x-axis by:

[(xij*, yij*) ∆Aij] yij*

P1612.4

MOMENT ABOUT X-AXIS

If we now add these quantities and take the limit as the number of subrectangles becomes large, we obtain the moment of the entire lamina about the x-axis:

* * *

max , 01 1

lim ( , )

( , )

i j

m n

x ij ij ij ijx y

i j

D

M y x y A

y x y dA

P1712.4

MOMENT ABOUT Y-AXIS

Similarly, the moment about the y-axis is:

* * *

max , 01 1

lim ( , )

( , )

i j

m n

y ij ij ij ijx y

i j

D

M x x y A

x x y dA

P1812.4

CENTER OF MASS

As before, we define the center of mass so that and

The physical significance is that: The lamina behaves as if its entire mass is

concentrated at its center of mass.

( , )x y

ymx Mxmy M

P1912.4

CENTER OF MASS

Thus, the lamina balances horizontally when supported at its center of mass.

See Figure 4.

P2012.4

The coordinates of the center of mass of a lamina occupying the region D and having density function (x, y) are:

where the mass m is given by:

Formulas 5

( , )x y

1( , )y

D

Mx x x y dA

m m

( , )D

m x y dA

1( , )x

D

My y x y dA

m m

P2112.4

Example 2

Find the mass and center of mass of a triangular lamina with vertices

(0, 0), (1, 0), (0, 2)

and if the density function is

(x, y) = 1 + 3x + y

P2212.4

Example 2 SOLUTION

The triangle is shown in Figure 5. Note that the equation of the upper boundary is:

y = 2 – 2x

P2312.4

Example 2 SOLUTION

The mass of the lamina is:

1 2 2

0 0

2 221

00

131 2

00

( , ) (1 3 )

32

84 (1 ) 4

3 3

x

D

y x

y

m x y dA x y dy dx

yy xy dx

xx dx x

P2412.4

Example 2 SOLUTION

Then, Formulas 5 give:

1 2 2 238 0 0

2 221 2

00

12 41 33

2 00

1( , ) ( 3 )

33

8 2

3 3( )

2 2 4 8

x

D

y x

y

x x x y dA x x xy dy dxm

yxy x y x dx

x xx x dx

P2512.4

Example 2 SOLUTION

1 2 2 238 0 0

2 22 2 31

00

1 2 314 0

12 43

0

1( , ) ( 3 )

33

8 2 2 3

(7 9 3 5 )

1 117 9 5

4 2 4 16

x

D

y x

y

y y x y dA y xy y dy dxm

y y yx dx

x x x dx

x xx x

P2612.4

Example 2 SOLUTION

The center of mass is at the point .3 11

, 8 16

P2712.4

Example 3

The density at any point on a semicircular lamina is proportional to the distance from the center of the circle.

Find the center of mass of the lamina.

P2812.4

Example 3 SOLUTION

Let’s place the lamina as the upper half of the circle x2 + y2 = a2. See Figure 6. Then, the distance from a

point (x, y) to the center of the circle (the origin) is:

2 2x y

P2912.4

Example 3 SOLUTION

Therefore, the density function is:

where K is some constant.Both the density function and the shape of the

lamina suggest that we convert to polar coordinates. Then, and the region D is given by:

0 ≤ r ≤ a, 0 ≤ ≤

2 2( , )x y K x y

2 2x y r

P3012.4

Example 3 SOLUTION

Thus, the mass of the lamina is:

2 2

0 0

2

0 0

3 3

0

( , )

( )

3 3

D D

a

a

a

m x y dA K x y dA

Kr r dr d

K d r dr

r K aK

P3112.4

Example 3 SOLUTION

Both the lamina and the density function are symmetric with respect to the y-axis. So, the center of mass must lie on the y-axis, that is,

= 0x

Fig. 16.5.6, p. 1018

P3212.4

Example 3 SOLUTION

The y-coordinate is given by:

3 0 0

33 0 0

4

03

0

4

3

1 3( , ) sin ( )

3sin

3[ cos ]

4

3 2 3

4 2

a

D

a

a

y y x y dA r Kr r dr dm K a

d r dra

r

a

a a

a

P3312.4

Example 3 SOLUTION

Thus, the center of mass is located at the point (0, 3a/(2)).

P3412.4

MOMENT OF INERTIA

The moment of inertia (also called the second moment) of a particle of mass m about an axis is defined to be mr2, where r is the distance from the particle to the axis. We extend this concept to a lamina with density

function (x, y) and occupying a region D by proceeding as we did for ordinary moments.

P3512.4

MOMENT OF INERTIA

Thus, we: Divide D into small rectangles. Approximate the moment of inertia of each

subrectangle about the x-axis. Take the limit of the sum as the number of

subrectangles becomes large.

P3612.4

MOMENT OF INERTIA (X-AXIS)

The result is the moment of inertia of the lamina about the x-axis:

* 2 * *

max , 01 1

2

lim ( ) ( , )

( , )

i j

m n

x ij ij ij ijx y

i j

D

I y x y A

y x y dA

P3712.4

MOMENT OF INERTIA (Y-AXIS)

Similarly, the moment of inertia about the y-axis is:

* 2 * *

max , 01 1

2

lim ( ) ( , )

( , )

i j

m n

y ij ij ij ijx y

i j

D

I x x y A

x x y dA

P3812.4

MOMENT OF INERTIA (ORIGIN)

It is also of interest to consider the moment of inertia about the origin (also called the polar moment of inertia):

Note that I0 = Ix + Iy.

* 2 * 2 * *0 max , 0

1 1

2 2

lim [( ) ( ) ] ( , )

( ) ( , )

i j

m n

ij ij ij ij ijx yi j

D

I x y x y A

x y x y dA

P3912.4

Example 4

Find the moments of inertia Ix , Iy , and I0 of a homogeneous disk D with: Density (x, y) = Center the origin Radius a

P4012.4

Example 4 SOLUTION

The boundary of D is the circle

x2 + y2 = a2

In polar coordinates, D is described by:

0 ≤ ≤ 2, 0 ≤ r ≤ a

P4112.4

Example 4 SOLUTION

Let’s compute I0 first:22 2 2

0 0 0

2 3

0 0

4 4

0

( )

24 2

a

D

a

a

I x y dA r r dr d

d r dr

r a

P4212.4

Example 4 SOLUTION

Instead of computing Ix and Iy directly, we use the facts that Ix + Iy = I0 and Ix = Iy (from the symmetry of the problem).

Thus,4

0

2 4x y

I aI I

P4312.4

MOMENTS OF INERTIA

In Example 4, notice that the mass of the disk is:

m = density × area = (a2)

P4412.4

MOMENTS OF INERTIA

So, the moment of inertia of the disk about the origin (like a wheel about its axle) can be written as:

Thus, if we increase the mass or the radius of the disk, we thereby increase the moment of inertia.

42 2 21 1

0 2 2( )2

aI a a ma

P4512.4

MOMENTS OF INERTIA

In general, the moment of inertia plays much the same role in rotational motion that mass plays in linear motion. The moment of inertia of a wheel is what makes it

difficult to start or stop the rotation of the wheel. This is just as the mass of a car is what makes it

difficult to start or stop the motion of the car.

P4612.4

RADIUS OF GYRATION

The radius of gyration of a lamina about an axis is the number R such that

mR2 = Iwhere: m is the mass of the lamina. I is the moment of inertia about the given axis.

P4712.4

RADIUS OF GYRATION

Equation 9 says that: If the mass of the lamina were concentrated at a

distance R from the axis, then the moment of inertia of this “point mass” would be the same as the moment of inertia of the lamina.

P4812.4

RADIUS OF GYRATION

In particular, the radius of gyration with respect to the x-axis and the radius of gyration

with respect to the y-axis are given by:

yx

2 2x ymy I mx I

P4912.4

RADIUS OF GYRATION

Thus, is the point at which the mass of the lamina can be concentrated without changing the moments of inertia with respect to the coordinate axes. Note the analogy with the center of mass.

( , )x y

P5012.4

Example 5

Find the radius of gyration about the x-axis of the disk in Example 4. As noted, the mass of the disk is m = a2. So, from Equations 10, we have:

So, the radius of gyration about the x-axis is , which is half the radius of the disk.

4 212 4

2 4x aI a

ym a

12y a