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16.6 Integrals for Mass Calculations

mass of shape in 2D: \(\iint_{R} \rho(x,y) dA\)

Note: \(\rho\) represents the density.

3D: \(\iiint_{D} \rho(x,y,z) dV\)

We will find the center of mass for 2D and 3D objects in this section.

We first need the mass moment.

\(\text{mass moment} = \text{mass} \cdot \text{distance from the axis of rotation}\)

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The moment of the mass about the y-axis is:

\[ M_y = \text{mass} \cdot \text{distance from y-axis} \]

where the distance from the y-axis is the \(x\)-coord of the point.

Coordinate graph showing a point with mass m at coordinates (x, y) relative to the origin.

So,

\[ M_y = m \cdot x \]

Similarly,

\[ M_x = m \cdot y \]

What about a 2D plate of certain shape?

For a general 2D shape in the coordinate plane, we want to find:

  • \(M_y = ?\)
  • \(M_x = ?\)
A 2D coordinate system with an irregular closed loop representing a plate of a certain shape.
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Moments and Center of Mass for a Region

Divide the region into small rectangles.

A region R in the xy-plane divided into a grid of small rectangles. One rectangle is highlighted.

Find \( M_y \) and \( M_x \) for one rectangle, then integrate to accumulate all.

A small rectangle with width dx and height dy.
mass density \( \rho(x, y) \)

Then mass of this small piece is:

\[ m = \rho(x, y) dA \]\[ = \rho(x, y) dy dx \]\[ = \rho(x, y) dx dy \]

The small piece is at distance \( x \) from y-axis.

\[ M_y = \underbrace{x}_{\text{dist}} \cdot \underbrace{\rho(x, y) dA}_{\text{mass}} \quad \text{and} \quad M_x = y \cdot \rho(x, y) dA \]

Then accumulate by double integral, so, for the entire plate:

\[ M_y = \iint_R x \cdot \rho(x, y) dA \]
\[ M_x = \iint_R y \cdot \rho(x, y) dA \]
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Center of Mass

If the entire plate is collapsed into a point w/ same mass, where should the point be placed to keep the same \( M_y \) and \( M_x \)?

A region R in the xy-plane.
\( \rightarrow \)
A single point at (x-bar, y-bar) in the xy-plane.
location: \( (\bar{x}, \bar{y}) \)
"x bar"
x-coord of center of mass

has

\( M_y = \iint_R x \rho dA \)

\( M_x = \iint_R y \rho dA \)

has

\( M_y = m \bar{x} \)

\( M_x = m \bar{y} \)

Equate them:

\( m \bar{x} = \iint_R x \rho dA \) \( \rightarrow \)
\[ \bar{x} = \frac{1}{m} \iint_R x \rho dA \]
\( m \bar{y} = \iint_R y \rho dA \) \( \rightarrow \)
\[ \bar{y} = \frac{1}{m} \iint_R y \rho dA \]
\[ m = \iint_R \rho dA \]
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Example: Center of Mass for a Rectangular Region

Consider a region $R$ defined by:

\[ R = \{ (x, y) : 0 \le x \le 1, \quad 0 \le y \le 5 \} \]

with density function:

\[ \rho(x, y) = 2e^{-\frac{1}{2}y} \]

Visual Representation

The region is a vertical rectangle in the first quadrant. The center of mass is calculated to be at approximately $(\frac{1}{2}, 1.6)$.

Coordinate graph showing a rectangle from x=0 to 1 and y=0 to 5, with center of mass marked at (0.5, 1.6).

1. Calculate Total Mass ($m$)

Start with mass:

\[ m = \iint_R \rho \, dA = \int_0^1 \int_0^5 2e^{-\frac{1}{2}y} \, dy \, dx = \dots = 4(1 - e^{-5/2}) \]

2. Calculate Center of Mass Coordinates

The x-coordinate of the center of mass $\bar{x}$ is:

\[ \bar{x} = \frac{1}{m} \iint_R x \rho \, dA = \frac{1}{4(1 - e^{-5/2})} \int_0^1 \int_0^5 x \cdot 2e^{-\frac{1}{2}y} \, dy \, dx = \dots = \frac{1}{2} \]

The y-coordinate of the center of mass $\bar{y}$ is (using integration by parts):

\[ \bar{y} = \frac{1}{m} \iint_R y \rho \, dA = \frac{1}{4(1 - e^{-5/2})} \int_0^1 \int_0^5 y \cdot 2e^{-\frac{1}{2}y} \, dy \, dx \]
\[ = \dots = \frac{-14e^{-5/2} + 4}{2(1 - e^{-5/2})} \approx 1.6 \]
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Example: Center of Mass in Polar Coordinates

Consider a region $R$ between circles of radii 1 and 2 centered at the origin in the first quadrant (QI) with density:

\[ \rho(x, y) = \sqrt{x^2 + y^2} \]

Since these are circles, let's set up the problem in polar coordinates:

\[ R = \{ (r, \theta) : 0 \le \theta \le \frac{\pi}{2}, \quad 1 \le r \le 2 \} \]

Density in polar coordinates: $\rho = r$.

Graph of a quarter-annulus in the first quadrant between r=1 and r=2, with center of mass marked.

1. Calculate Total Mass ($m$)

\[ m = \iint_R \rho \, dA = \int_0^{\pi/2} \int_1^2 r \cdot r \, dr \, d\theta = \int_0^{\pi/2} \int_1^2 r^2 \, dr \, d\theta = \dots = \frac{7\pi}{6} \]

2. Calculate Center of Mass Coordinates

Using $x = r \cos \theta$ and $dA = r \, dr \, d\theta$:

\[ \bar{x} = \frac{1}{m} \iint_R x \rho \, dA = \frac{1}{7\pi/6} \int_0^{\pi/2} \int_1^2 r \cos \theta \cdot r \cdot r \, dr \, d\theta = \dots = \frac{45}{14\pi} \approx 1.023 \]

Using $y = r \sin \theta$:

\[ \bar{y} = \frac{1}{m} \iint_R y \rho \, dA = \frac{1}{7\pi/6} \int_0^{\pi/2} \int_1^2 r \sin \theta \cdot r \cdot r \, dr \, d\theta = \dots = \frac{45}{14\pi} \approx 1.023 \]
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Transition from 2D to 3D

When moving from 2D to 3D, we transition from area elements to volume elements: \( dA \to dV \) and density functions \( \rho(x, y) \to \rho(x, y, z) \).

Consequently, double integrals become triple integrals.

Mass and Center of Mass Formulas in 3D

\[ m = \iiint_{D} \rho(x, y, z) \, dV \]

\[ \bar{x} = \frac{1}{m} \iiint_{D} x \rho(x, y, z) \, dV \]

\[ \bar{y} = \frac{1}{m} \iiint_{D} y \rho(x, y, z) \, dV \]

\[ \bar{z} = \frac{1}{m} \iiint_{D} z \rho(x, y, z) \, dV \]

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Example: Center of Mass of a Solid

Find the center of mass of a solid bounded by \( 3x + 2y + z = 6 \) and the coordinate planes, with density \( \rho(x, y, z) = 1 + x \).

The bounding surface is a plane in the first octant.

3D graph of a tetrahedron in the first octant bounded by axes and a plane with intercepts (2,0,0), (0,3,0), (0,0,6).

Plane Equation:

\( 3x + 2y + z = 6 \)

Setting up the Integrals

For the integrals, let's use the \( xy \)-plane as the "floor".

2D projection on the xy-plane showing a triangle with vertices at (0,0), (2,0), and (0,3).

Region in \( xy \)-plane:

\[ 0 \le x \le 2 \]

\[ 0 \le y \le 3 - \frac{3}{2}x \]

Limits for \( z \)

For \( z \), from the \( xy \)-plane to the bounding plane:

\( 0 \le z \le 6 - 3x - 2y \)

Calculating Mass

\[ m = \int_{0}^{2} \int_{0}^{3-\frac{3}{2}x} \int_{0}^{6-3x-2y} (1+x) \, dz \, dy \, dx = \dots = 9 \]

Note: \( \rho = (1+x) \) and \( dV = dz \, dy \, dx \).

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\[ \bar{x} = \frac{1}{m} \iiint_{D} x \rho dV \]
\[ = \frac{1}{9} \int_{0}^{2} \int_{0}^{3-\frac{3}{2}x} \int_{0}^{6-3x-2y} x(1+x) dz dy dx = \dots = 3/5 \]

\( \bar{y}, \bar{z} \) same idea \( \dots \dots \bar{y} = 7/10, \bar{z} = 7/5 \)

\[ M_y = \iint_{R} x \rho dA \]

first power, so this is a first moment

\[ I_y = \iint_{R} x^2 \rho dA \rightarrow \text{moment of inertia about } y\text{-axis} \]

second power

this is a second moment