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  • Use double integrals to locate the center of mass of a two-dimensional object.
  • Use double integrals to find the moment of inertia of a two-dimensional object.
  • Use triple integrals to locate the center of mass of a three-dimensional object.

We have already discussed a few applications of multiple integrals, such as finding areas, volumes, and the average value of a function over a bounded region. In this section we develop computational techniques for finding the center of mass and moments of inertia of several types of physical objects, using double integrals for a lamina (flat plate) and triple integrals for a three-dimensional object with variable density. The density is usually considered to be a constant number when the lamina or the object is homogeneous; that is, the object has uniform density.

Center of mass in two dimensions

The center of mass is also known as the center of gravity if the object is in a uniform gravitational field. If the object has uniform density, the center of mass is the geometric center of the object, which is called the centroid. [link] shows a point P as the center of mass of a lamina. The lamina is perfectly balanced about its center of mass.

A surface is delicately balanced on a fine point.
A lamina is perfectly balanced on a spindle if the lamina’s center of mass sits on the spindle.

To find the coordinates of the center of mass P ( x , y ) of a lamina, we need to find the moment M x of the lamina about the x -axis and the moment M y about the y -axis . We also need to find the mass m of the lamina. Then

x = M y m and y = M x m .

Refer to Moments and Centers of Mass for the definitions and the methods of single integration to find the center of mass of a one-dimensional object (for example, a thin rod). We are going to use a similar idea here except that the object is a two-dimensional lamina and we use a double integral.

If we allow a constant density function, then x = M y m and y = M x m give the centroid of the lamina.

Suppose that the lamina occupies a region R in the x y -plane , and let ρ ( x , y ) be its density (in units of mass per unit area) at any point ( x , y ) . Hence, ρ ( x , y ) = lim Δ A 0 Δ m Δ A , where Δ m and Δ A are the mass and area of a small rectangle containing the point ( x , y ) and the limit is taken as the dimensions of the rectangle go to 0 (see the following figure).

A lamina R is shown on the x y plane with a point (x, y) surrounded by a small rectangle marked Mass = Delta m and Area = Delta A.
The density of a lamina at a point is the limit of its mass per area in a small rectangle about the point as the area goes to zero.

Just as before, we divide the region R into tiny rectangles R i j with area Δ A and choose ( x i j * , y i j * ) as sample points. Then the mass m i j of each R i j is equal to ρ ( x i j * , y i j * ) Δ A ( [link] ). Let k and l be the number of subintervals in x and y , respectively. Also, note that the shape might not always be rectangular but the limit works anyway, as seen in previous sections.

A lamina is shown on the x y plane with a point (x* sub ij, y* sub ij) surrounded by a small rectangle marked R sub ij.
Subdividing the lamina into tiny rectangles R i j , each containing a sample point ( x i j * , y i j * ) .

Hence, the mass of the lamina is

m = lim k , l i = 1 k j = 1 l m i j = lim k , l i = 1 k j = 1 l ρ ( x i j * , y i j * ) Δ A = R ρ ( x , y ) d A .

Let’s see an example now of finding the total mass of a triangular lamina.

Finding the total mass of a lamina

Consider a triangular lamina R with vertices ( 0 , 0 ) , ( 0 , 3 ) , ( 3 , 0 ) and with density ρ ( x , y ) = x y kg/m 2 . Find the total mass.

A sketch of the region R is always helpful, as shown in the following figure.

A triangular lamina is shown on the x y plane bounded by the x and y axes and the line x + y = 3. The point (1, 1) is marked and is surrounded by a small squared marked d m = p(x, y) dA.
A lamina in the x y -plane with density ρ ( x , y ) = x y .

Using the expression developed for mass, we see that

m = R d m = R ρ ( x , y ) d A = x = 0 x = 3 y = 0 y = 3 x x y d y d x = x = 0 x = 3 [ x y 2 2 | y = 0 y = 3 x ] d x = x = 0 x = 3 1 2 x ( 3 x ) 2 d x = [ 9 x 2 4 x 3 + x 4 8 ] | x = 0 x = 3 = 27 8 .

The computation is straightforward, giving the answer m = 27 8 kg .

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Source:  OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2
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