CC Density, Mass, and Center of Mass

Section 6.4 Density, Mass, and Center of Mass

Motivating Questions

How are mass, density, and volume related?
How is the mass of an object with varying density computed?
What is is the center of mass of an object, and how are definite integrals used to compute it?

Studying the units of the integrand and variable of integration helps us understand the meaning of a definite integral. For instance, if

v (t)

is the velocity of an object moving along an axis, measured in feet per second, and

t

measures time in seconds, then both the definite integral and its Riemann sum approximation,

\int_{a}^{b} v (t) d t \approx \sum_{i = 1}^{n} v (t_{i}) Δ t,

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have units given by the product of the units of

v (t)

and

t :

(feet/sec) \cdot (sec) = feet .

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Thus,

\int_{a}^{b} v (t) d t

measures the total change in position of the moving object in feet.

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Unit analysis will be particularly helpful to us in what follows.

Example 6.48.

In each of the following scenarios, we consider the distribution of a quantity along an axis.

Suppose that the function $c (x) = 200 + 100 e^{- 0.1 x}$ models the density of traffic on a straight road, measured in cars per mile, where $x$ is the number of miles east of a major interchange, and consider the definite integral

$\int_{0}^{2} (200 + 100 e^{- 0.1 x}) d x .$
1. What are the units of the product $c (x) \cdot Δ x ?$
2. What are the units of the definite integral and its Riemann sum approximation given by
  
  $\int_{0}^{2} c (x) d x \approx \sum_{i = 1}^{n} c (x_{i}) Δ x ?$
3. Evaluate the definite integral $\int_{0}^{2} c (x) d x = \int_{0}^{2} (200 + 100 e^{- 0.1 x}) d x$ and write one sentence to explain the meaning of the value you find.
On a 6 foot long shelf filled with books, the function $B$ models the distribution of the weight of the books, in pounds per inch, where $x$ is the number of inches from the left end of the bookshelf. Let $B (x)$ be given by the rule $B (x) = 0.5 + \frac{1}{(x + 1)^{2}} .$
1. What are the units of the product $B (x) \cdot Δ x ?$
2. What are the units of the definite integral and its Riemann sum approximation given by
  
  $\int_{12}^{36} B (x) d x \approx \sum_{i = 1}^{n} B (x_{i}) Δ x ?$
3. Evaluate the definite integral $\int_{0}^{72} B (x) d x = \int_{0}^{72} (0.5 + \frac{1}{(x + 1)^{2}}) d x$ and write one sentence to explain the meaning of the value you find.

Solution.

1. Since $c (x)$ has units of cars per mile, and $Δ x$ has units of miles, then $c (x) \cdot Δ x$ has units of cars.
2. From the answer to the previous part, $c (x) \cdot Δ x$ has units of cars, so the summation and integral will be the sum of things with units of cars, and therefore also have units of cars.
3. Using the exponent rule for integrating:
  
  $\int_{0}^{2} c (x) d x = \int_{0}^{2} (200 + 100 e^{- 0.1 x}) d x = {200 x - 1000 e^{- 0.1 x} |}_{0}^{2} = (200 (2) - 1000 e^{- 0.1 (2)}) - (200 (0) - 1000 e^{- 0.1 (0)}) = 581$
  
  This represents that, on the two mile stretch of road, there are 581 cars.
1. Since $B (x)$ has units of pounds per inch, and $Δ x$ has units of inches, then $B (x) \cdot Δ x$ has units of pounds.
2. From the answer to the previous part, $B (x) \cdot Δ x$ has units of pounds, so the summation and integral will be the sum of things with units of pounds, and therefore also have units of pounds.
3. Using a power rule, we get that
  
  $\int_{0}^{72} B (x) d x = \int_{0}^{72} (0.5 + \frac{1}{(x + 1)^{2}}) d x = {0.5 x - \frac{1}{x + 1} |}_{0}^{72} = (0.5 (72) - \frac{1}{72 + 1}) - (0.5 (0) - \frac{1}{0 + 1}) = 36.986$
  
  This represents that, on the 72 inches of book shelf, there are 36.986 pounds of books.

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Subsection 6.4.1 Density

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The mass of an object, typically measured in metric units such as grams or kilograms, is a measure of the amount of material in the object. The density of an object measures the distribution of mass per unit volume. For instance, if a brick has mass 3 kg and volume 0.002 m

^{3},

then the density of the brick is

\frac{3 kg}{0.002 m^{3}} = 1500 \frac{kg}{m^{3}} .

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As another example, the mass density of water is 1000 kg/m

^{3} .

Each of these relationships demonstrate the following general principle.

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Density Formula.

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For an object of constant density

d,

with mass

m

and volume

V,

d = \frac{m}{V}, or m = d \cdot V .

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But what happens when the density is not constant?

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The formula

m = d \cdot V

is reminiscent of two other equations that we have used in our work: for a body moving in a fixed direction, distance = rate

\cdot

time, and, for a rectangle, its area is given by

A = l \cdot w .

These formulas hold when the principal quantities involved, such as the rate the body moves and the height of the rectangle, are constant. When these quantities are not constant, we have turned to the definite integral for assistance. By working with small slices on which the quantity of interest (such as velocity) is approximately constant, we can use a definite integral to add up the values on the pieces.

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For example, if we have a nonnegative velocity function that is not constant, over a short time interval

Δ t,

we know that the distance traveled is approximately

v (t) Δ t

since

v (t)

is almost constant on a small interval. Similarly, if we are thinking about the area under a nonnegative function

f

whose value is changing, on a short interval

Δ x,

the area under the curve is approximately the area of the rectangle whose height is

f (x)

and whose width is

Δ x :

f (x) Δ x .

Both of these principles are represented visually in Figure 6.49.

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Figure 6.49. At left, estimating a small amount of distance traveled, $v (t) Δ t,$ and at right, a small amount of area under the curve, $f (x) Δ x .$

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In a similar way, if the density of some object is not constant, we can use a definite integral to compute the overall mass of the object. We will focus on problems where the density varies in only one dimension, say along a single axis.

Example 6.50.

Let’s consider a thin bar of length

b

whose left end is at the origin, where

x = 0,

and assume that the bar has constant cross-sectional area of 1 cm

^{2} .

We let

ρ (x)

represent the mass density function of the bar, measured in grams per cubic centimeter. That is, given a location

x,

ρ (x)

tells us approximately how much mass will be found in a one-centimeter wide slice of the bar at

x .

Figure 6.51. A thin bar of constant cross-sectional area 1 cm $^{2}$ with density function $ρ (x)$ g/cm $^{3} .$

The volume of a thin slice of the bar of width

Δ x,

as pictured in Figure 6.51, is the cross-sectional area times

Δ x .

Since the cross-sections each have constant area 1 cm

^{2},

it follows that the volume of the slice is

1 Δ x

^{3} .

And because mass is the product of density and volume, we see that the mass of this slice is approximately

{mass}_{slice} \approx ρ (x) \frac{g}{{cm}^{3}} \cdot 1 Δ x {cm}^{3} = ρ (x) \cdot Δ x g .

Therefore, the total mass of the bar is measured by the corresponding Riemann sum (and the integral that it approximates)

\sum_{i = 1}^{n} ρ (x_{i}) Δ x \approx \int_{0}^{b} ρ (x) d x = {mass}_{total} .

(As usual, the Riemann sum is an approximation, while the integral will be the exact mass.)

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In the example above, computing the mass of the rod was made easier by the fact that the rod had constant cross-sectional area. This meant that the volume of any thin slice was approximately

1 {cm}^{2} \cdot Δ x cm = Δ x {cm}^{3}

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indepdently of the value of

x,

which allowed us to conclude that the mass of a thin slice at

x

was approximately

ρ (x) \cdot Δ x g .

In general, it may be the case that the volume of our thin slices also varies with

x .

In these situations, if

A (x)

gives the area of a cross section at

x,

then the volume of a thin slice at

x

is approximately

A (x) Δ x

and thus the mass of the slice will be approximately

ρ (x) \cdot A (x) Δ x .

From this expression, we pass to a definite integral in the usual way to reach the following.

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Calculating Mass using a Definite Integral.

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For an object whose mass is distributed along a single axis according to the function

ρ (x)

and whose cross-sectional area at

x

is given by the function

A (x),

the total mass

M

of the object between

x = a

and

x = b

M = \int_{a}^{b} ρ (x) A (x) d x

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Before embarking on an example, we require a short aside on units. The formula above implicitly assumes that the object in question is three-dimensional, since its cross sections have area

A (x)

and therefore its thin slices have volume

A (x) Δ x \approx A (x) d x .

However, we can and often will consider mass and density in one- and two-dimensional settings. In these cases, cross sections don’t have area, but rather they are points (in the one-dimensional case) or lengths (in the two-dimensional one). To avoid clashing with the usual convention that density is mass per unit volume, or

\frac{m}{V},

we leave the formula above in terms of three-dimensional objects, albeit with the caveat that for some problems terms like "area" and "volume" must be interpreted in context.

Example 6.52.

Consider the triangle below, which has density given by

ρ (x) = 2 x {g/cm}^{2}

for

0 \leq x \leq 4 .

That is, the density of the triangle along a vertical slice at

x

2 x {g/cm}^{2} .

Figure 6.53. A triangle with density function $ρ (x) = 2 x {g/cm}^{2} .$

We’ll compute the total mass

M

of the triangle using a definite integral. To accomplish this, we first find the height of a vertical cross section at

x

for

0 \leq x \leq 4 .

The equations of the lines representing the top and bottom edges of the triangle are

y = - \frac{x}{2} + 2

and

y = \frac{x}{2} - 2;

therefore, the height of a vertical cross section at

x

- \frac{x}{2} + 2 - (\frac{x}{2} - 2) = - x + 4.

Thus the mass of a thin slice at

x

2 x {g/cm}^{2} \cdot (- x + 4) d x {cm}^{2} = 2 x (- x + 4) d x g

It follows that

M = \int_{0}^{4} 2 x (- x + 4) d x

which evaluates to

\frac{64}{3} g .

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Calculating Mass using a Definite Integral.

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For an object of constant cross-sectional area whose mass is distributed along a single axis according to the function

ρ (x)

(whose units are units of mass per unit of length), the total mass,

M,

of the object between

x = a

and

x = b

is given by

M = \int_{a}^{b} ρ (x) d x .

Example 6.54.

Consider the following situations in which mass is distributed in a non-constant manner.

Suppose that a thin rod with constant cross-sectional area of 1 cm $^{2}$ has its mass distributed according to the density function $ρ (x) = 2 e^{- 0.2 x},$ where $x$ is the distance in cm from the left end of the rod, and the units of $ρ (x)$ are g/cm. If the rod is 10 cm long, determine the exact mass of the rod.
Consider the cone that has a base of radius 4 m and a height of 5 m.
Picture the cone with the center of its base at the origin and think of the cone as a solid of revolution.
1. Write and evaluate a definite integral whose value is the volume of the cone.
2. Next, suppose that the cone has uniform density of 800 kg/m $^{3} .$ What is the mass of the solid cone?
3. Now suppose that the cone’s density is not uniform, but rather that the cone is most dense at its base. In particular, assume that the density of the cone is uniform across cross sections parallel to its base, but that in each such cross section that is a distance $x$ units from the origin, the density of the cross section is given by the function $ρ (x) = 400 + \frac{200}{1 + x^{2}},$ measured in kg/m $^{3} .$ Determine and evaluate a definite integral whose value is the mass of this cone of non-uniform density. Do so by first thinking about the mass of a given slice of the cone $x$ units away from the base; remember that in such a slice, the density will be essentially constant.
Let a thin rod of constant cross-sectional area 1 cm $^{2}$ and length 12 cm have its mass be distributed according to the density function $ρ (x) = \frac{1}{25} (x - 15)^{2},$ measured in g/cm. Find the exact location $x$ at which to cut the bar so that the two pieces will each have identical mass.

Hint.

Remember that $M = \int_{a}^{b} ρ (x) d x .$
Think about slices: what’s the volume of a slice? what’s the mass of a slice?
Consider $\int_{0}^{b} ρ (x) d x .$

Answer.

$M = 0.1 - 0.1 e^{- 4} \approx 0.0981684$ grams.
1. $V = \int_{0}^{5} π (4 - \frac{4}{5} x)^{2} d x = \frac{80 π}{3} \approx 83.7758 m^{3} .$
2. $M = \int_{0}^{5} (400 + \frac{200}{1 + x^{2}}) \cdot π (4 - \frac{4}{5} x)^{2} d x$ $= \frac{32}{5} π (25015 - 5 \ln (17576) + 72 \tan^{- 1} (5) \approx 33597.4664 kg$
$b \approx 3.0652 .$

Solution.

Since the mass, $M,$ is given by $M = \int_{a}^{b} ρ (x) d x,$ it follows that

$M = \int_{0}^{10} 2 e^{- 0.2 x} = {\frac{2}{- 0.2} e^{- 0.2 x} |}_{0}^{20} = - 0.1 e^{- 4} + 0.1 e^{0},$

and hence $M = 0.1 - 0.1 e^{- 4} \approx 0.0981684$ grams.
Consider the cone that has a base of radius 4 m and a height of 5 m.
1. With the cone having its base centered at the origin and being a solid of revolution about the $x$ -axis, the cone is generated by the line $y = 4 - \frac{4}{5} x .$ Using the disk method, it follows that the cone’s volume is
  
  $V = \int_{0}^{5} π (4 - \frac{4}{5} x)^{2} d x = \frac{80 π}{3} \approx 83.7758 m^{3} .$
2. If the cone has uniform density 800 kg/m $^{3},$ we know that the cone’s mass is the product of its density and volume, and thus
  
  $M = ρ \cdot V = 800 \cdot \frac{80 π}{3} = \frac{6400 π}{3} \approx 67020.6433 kg .$
3. If the cone has non-uniform density, but is uniform across cross sections parallel to its base such that in each cross section a distance $x$ units from the origin has its density given by $ρ (x) = 400 + \frac{200}{1 + x^{2}}$ kg/m $^{3},$ we naturally consider slices of the cone that are perpendicular to the $x$ -axis. For each such slice, the radius is $r (x) = 4 - \frac{4}{5} x,$ which makes the volume of the slice
  
  $V_{slice} \approx π (4 - \frac{4}{5} x)^{2} Δ x .$
  
  Since the density is constant on the slice with value $ρ (x) = 400 + \frac{200}{1 + x^{2}},$ it follows that the mass of the slice is
  
  $M_{slice} \approx (400 + \frac{200}{1 + x^{2}}) \cdot π (4 - \frac{4}{5} x)^{2} Δ x .$
  
  Using a definite integral to let $Δ x \to 0$ and add up the values of all of the slices, we thus find the mass of the cone with this density distribution to be
  
  $\begin{aligned} M & = \int_{0}^{5} (400 + \frac{200}{1 + x^{2}}) \cdot π (4 - \frac{4}{5} x)^{2} d x \\ = \frac{32}{5} π (25015 - 5 \ln (17576) + 72 \tan^{- 1} (5) \approx 33597.4664 kg \\ \approx 33597.4664 kg . \end{aligned}$
The total mass of the bar is given by $M = \int_{0}^{12} \frac{1}{25} (x - 15)^{2} d x = \frac{1116}{25} .$ To find where to cut the bar in two so that the two pieces have equal mass, we have to determine the value of $b$ such that

$\int_{0}^{b} \frac{1}{25} (x - 15)^{2} d x = \frac{1}{2} \cdot \frac{1116}{25} .$

Since $\int_{0}^{b} \frac{1}{25} (x - 15)^{2} d x = \frac{1}{75} ((b - 15)^{3} - (- 15)^{3}),$ we need to have $b$ satisfy the equation

$\frac{1}{75} ((b - 15)^{3} - (- 15)^{3}) = \frac{1}{2} \cdot \frac{1116}{25} .$

Thus, $(b - 15)^{3} - (- 15)^{3} = 75 \cdot \frac{558}{25} = 3 \cdot 558 = 1674,$ so $(b - 15)^{3} = 1675 - 3375 = - 1700 .$ It follows that $b - 15 = \sqrt[3]{- 1700} \approx - 11.9348,$ so $b \approx 3.0652 .$

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Subsection 6.4.2 Radial Density

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In some 2D or 3D objects, density depends on distance from a center point. In this case, we will apply the principle from the previous section (that mass is the integral of density), but must make a slight adjustment to account for the fact that layers of the same thickness will have different volumes depending on their distance from the center.

Example 6.55.

Consider a 1 meter long metal pipe covered with a layer of insulation.

How would we find the total mass of the pipe? We would need to add up the masses of the metal and the insulation, and these masses can be found by taking density times volume:

{mass}_{layer} = ρ_{layer} V_{layer}

But note that the density of the layer depends on its distance from the center: close to the center of the pipe, the hollow core of the pipe has 0 density, at a radius of

r_{1}

the metal pipe has a high density, and at a radius of

r_{2}

the insulation has a low density. Let us use

ρ (r)

to denote the density as a function of distance

r

from the center.

Next, to get

V_{layer}

we can use approximate volume as circumference times thickness times length, so

V \approx (2 π r_{i}) (Δ r) (1 meter) = (2 π r_{i}) (Δ r) .

Thus the mass of a single layer is approximately

{mass}_{i} \approx ρ (r_{i}) (2 π r_{i}) (Δ r) .

Summing both layers, we get that

mass \approx \sum_{i = 1}^{2} ρ (r_{i}) (2 π r_{i}) (Δ r)

is the total mass of the pipe.

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We can generalize the formula we found in the previous example to any object for which density depends on radius. That more general formula is

mass \approx \sum_{i = 1}^{n} ρ (r_{i}) (2 π r_{i}) (Δ r)

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where

ρ (r)

is the density of the object a distance of

r_{i}

from the center.

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Similar to the earlier applications in this chapter: we will take the limit as

Δ r \to 0,

and doing so turns the sum into an integral.

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Calculating Mass using a Definite Integral for Radially Symmetric Objects.

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For a 2D disk (respectively, 3D cylinder), if the density of the object depends only on the distance from the center point (respectively, center axis) according to the function

ρ (r),

then the mass per unit length (respectively, mass of the object) is given by the formula

mass = \int_{0}^{R} ρ (r) (2 π r) d r = \int_{0}^{R} 2 π r ρ (r) d r

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where

R

is the outer radius of the object.

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Note that this formula is almost identical to the one we use for calculating mass of an object with constant cross-sectional area, except that this formula incorporates a factor of

2 π r

(the circumference of a circle) inside the integral.

Example 6.56.

After an oil spill, scientists measure that the concentration of oil depends on the distance from the source according to the function

ρ (r) = 150 e^{- r^{2}}

where

r

is in miles, and

ρ (r)

is in barrels per square mile. If the spill has reached a maximum radius of 5 miles, how much oil was spilled?

Hint.

Set up the integral like for the mass of a disk, then integrate. To evaluate the integral use the

u

-substitution

u = - r^{2} .

Answer.

471.23

barrels.

Solution.

Since the density depends on the radius, we will use the mass of a disk formula

\int_{0}^{R} ρ (r) (2 π r) d r .

For us, the maximum radius is

R = 5

miles, so the total amount of oil is

\int_{0}^{5} (150 e^{- r^{2}}) (2 π r) d r = {- 150 π e^{- r^{2}} |}_{0}^{5} (- 150 π e^{- (5)^{2}}) - (- 150 π e^{- (0)^{2}}) = 471.23 barrels

Example 6.57.

The 2D region bounded by

y = 6 - x,

x = 0,

x = 3,

and

y = 0

(pictured below), is rotated around the

x

-axis to make a solid of revolution.

Figure 6.58. The 2D region before it is rotated around the $x$ -axis.

When this object is carved out of material, its density depends on the distance from the

x

-axis according to

ρ_{r a d i a l} (r) = 9 - r .

Fix a value of $x$ in the interval $[0, 3] .$ Set up and evaluate the integral to find $ρ_{a x i s} (x),$ the mass per unit length of the circular cross-section at that $x$ -value. (Your answer will be in terms of $x .$ )
Use your answer to the previous question and the techniques in the “Calculating Mass using a Definite Integral” section to find the total mass of the object.

Hint.

Each cross-section is a disk with outer radius $y = 6 - x .$
Use your formula for $ρ_{a x i s} (x)$ and the formula from "Calculating Mass using a Definite Integral":
$M = \int_{a}^{b} ρ (x) d x$

Answer.

$ρ_{a x i s} (x) = π (\frac{2}{3} x^{3} - 3 x^{2} - 36 x + 180)$
$M = 364.5 π \approx 1145.11$

Solution.

Since we have circular cross-sections and their density only depends on the distance to the center of the circle (which is the $x$ -axis), we will integrate as we do for a disk. Note that our outer radius depends on the $x$ -value, and in fact, the outer radius is $y = 6 - x .$ Therefore,

$ρ_{a x i s} (x) = \int_{0}^{6 - x} (2 π r) ρ_{r a d i a l} (r) d r = \int_{0}^{6 - x} (2 π r) (9 - r) d r = {(9 π r^{2} - \frac{2 π}{3} r^{3}) |}_{r = 0}^{r = 6 - x} = (9 π (6 - x)^{2} - \frac{2 π}{3} (6 - x)^{3}) - (0) = π (\frac{2}{3} x^{3} - 3 x^{2} - 36 x + 180)$
What we found in the previous part is the mass per unit length, so just as in "Calculating Mass using a Definite Integral", we simply need to integrate $ρ_{a x i s} (x)$ from the endpoints $a = 0$ to $b = 3 :$

$M = \int_{0}^{3} ρ_{a x i s} (x) d x = \int_{0}^{3} π (\frac{2}{3} x^{3} - 3 x - 36 x + 180) d x = {π (\frac{1}{6} x^{4} - x^{3} - 18 x^{2} + 180 x) |}_{0}^{3} = π (\frac{1}{6} (3)^{4} - (3)^{3} - 18 (3)^{2} + 180 (3)) - (0) = 364.5 π \approx 1145.11$

Example 6.59.

Suppose that a circular city’s population density is given by

ρ (r) = 1000 - 100 r,

where

r

is the distance from the city’s center in miles and

ρ (r)

is in units of people per square mile. If the city has a radius of 6 miles, find the population of the city.

Hint.

Set up the integral like for the mass of a disk, then integrate.

Answer.

67858

people.

Solution.

Since the density depends on the radius, we will use the mass of a disk formula

\int_{0}^{R} ρ (r) (2 π r) d r .

For us, the maximum radius is

R = 6

miles, so the total population is

\int_{0}^{6} ((1000 - 100 r) (2 π r) d r = {1000 π r^{2} - \frac{200 π}{3} r^{3} |}_{0}^{6} (1000 π (6)^{2} - \frac{200 π}{3} (6)^{3}) - (0) \approx 67858 people

Example 6.60.

When the density of a spherical object only depends on the distance from the center, the formula for the object’s mass is

mass = \int_{0}^{R} 4 π r^{2} ρ (r) d r

which is identical to the formula for a circular object, except that the circumference of a circle (

C = 2 π r

) is replaced by the surface area of a sphere (

A = 4 π r^{2}

The earth is denser near its core. Suppose that the density of the earth was given by

ρ (r) = (- 0.0016) r + 13

where

r

has units of kilometers and

ρ

is in trillions of kilograms per cubic kilometer (in more normal units,

1 trillion \frac{k g}{k m^{3}} = 1 \frac{g}{c m^{3}}

The earth has a radius of approximately 6300 kilometers. Find the total mass of the earth.
The volume of a sphere is $V = \frac{4}{3} π R^{3} .$ Find the volume of the earth, and calculate the average density of the earth. (Average density is the total mass divided by total volume.)
Goofus sees that at the center of the earth, the density is $ρ (0) = 13,$ and at the surface the density is $ρ (6300) = 2.92,$ and therefore concludes that the average density of the earth is $\frac{13 + 2.92}{2} = 7.96 .$ Is Goofus correct? If he is not, explain why his calculation is incorrect.
The earth’s inner core has a radius of 1200 kilometers. Find the average density of the earth’s inner core.
What percent of the earth’s mass is in the inner core? What percent of the earth’s volume is in the inner core?

Hint.

Set up and evaluate the integral given by the equation

$mass = \int_{0}^{R} 4 π r^{2} ρ (r) d r$
Use the volume formula to get the volume, then divide mass by volume to get the average density.
Goofus is incorrect.
Perform the same calculations as parts (a.) and (b.) but for the smaller radius.
Divide your answers to the previous parts.

Answer.

The mass of the earth is roughly $5.7 \times 10^{12}$ trillions of kilograms.
The volume of the earth is $V = 1.05 \times 10^{12}$ cubic kilometers. Its average density is $5.44$ trillions of kilograms per cubic kilometer.
Goofus is incorrect.
The density is $11.56$ trillion kilograms per cubic kilometer.
Approximately $1.47 %$ of the earth’s mass is in the inner core.

Approximately $0.689 %$ of the earth’s volume is in the inner core.

Solution.

We must take the integral:

$mass = \int_{0}^{R} 4 π r^{2} ρ (r) d r = \int_{0}^{6300} 4 π r^{2} (- 0.0016 r + 13) d r = {4 π (- 0.0004 r^{4} + \frac{13}{3} r^{3}) |}_{0}^{6300} \approx 5.7 \times 10^{12}$

so the mass of the earth is roughly $5.7 \times 10^{12}$ trillions of kilograms.
The volume formula for a sphere says that the volume of the earth is $V = \frac{4}{3} π (6300)^{3} = 1.05 \times 10^{12}$ cubic kilometers. Therefore, its average density is $\frac{5.7 \times 10^{12}}{1.05 \times 10^{12}} = 5.44$ trillions of kilograms per cubic kilometer.
Goofus is incorrect. His calculation is wrong because he is not taking into account the factor of $r^{2}$ inside the integral, which counts the lower-density parts on the outside of the earth more than the higher-density parts near the center. For instance, his average treated innermost 1 kilometer of the earth as having the same volume as the outermost 1 kilometer of earth, even though the latter is much larger in terms of volume.
We will essentially repeat our calculations from parts (a.) and (b.), except with 1200 as the maximum radius instead of 6300. Therefore, the volume of the inner core is $V = \frac{4}{3} π (1200)^{3} = 7.24 \times 10^{9}$ cubic killometers, and the mass is

$mass = \int_{0}^{R} 4 π r^{2} ρ (r) d r = \int_{0}^{1200} 4 π r^{2} (- 0.0016 r + 13) d r = {4 π (- 0.0004 r^{4} + \frac{13}{3} r^{3}) |}_{0}^{1200} \approx 8.37 \times 10^{10}$

so the density is $(8.37 \times 10^{10}) / (7.24 \times 10^{9}) = 11.56$ trillion kilograms per cubic kilometer.
In the previous parts, we saw that the mass of the inner core was $8.37 \times 10^{10}$ and the mass of the whole earth was $5.7 \times 10^{12},$ so dividing gives that approximately $1.47 %$ of the earth’s mass is in the inner core.

In the previous parts, we saw that the volume of the inner core was $7.24 \times 10^{9}$ and the volume of the whole earth was $1.05 \times 10^{12},$ so dividing gives that approximately $0.689 %$ of the earth’s volume is in the inner core.

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Subsection 6.4.3 Weighted Averages

🔗

The concept of an average is a natural one, and one that we have used repeatedly as part of our understanding of the meaning of the definite integral. If we have

n

values

a_{1},

a_{2},

\dots,

a_{n},

we know that their average is given by

\frac{a_{1} + a_{2} + \dots + a_{n}}{n},

🔗

and for a quantity being measured by a function

f

on an interval

[a, b],

the average value of the quantity on

[a, b]

\frac{1}{b - a} \int_{a}^{b} f (x) d x .

🔗

As we continue to think about problems involving the distribution of mass, it is natural to consider the idea of a weighted average, where certain quantities involved are counted more in the average.

🔗

A common use of weighted averages is in the computation of a student’s GPA, where grades are weighted according to credit hours. Let’s consider the scenario in Table 6.61.

class	grade	grade points	credits
chemistry	B+	3.3	5
calculus	A-	3.7	4
history	B-	2.7	3
psychology	B-	2.7	3

Table 6.61. A college student’s semester grades.

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If all of the classes were of the same weight (i.e., the same number of credits), the student’s GPA would simply be calculated by taking the average

\frac{3.3 + 3.7 + 2.7 + 2.7}{4} = 3.1 .

🔗

But since the chemistry and calculus courses have higher weights (of 5 and 4 credits respectively), we actually compute the GPA according to the weighted average

\frac{3.3 \cdot 5 + 3.7 \cdot 4 + 2.7 \cdot 3 + 2.7 \cdot 3}{5 + 4 + 3 + 3} = 3.1 \overset{―}{6} .

🔗

The weighted average reflects the fact that chemistry and calculus, as courses with higher credits, have a greater impact on the students’ grade point average. Note particularly that in the weighted average, each grade gets multiplied by its weight, and we divide by the sum of the weights.

🔗

In the following example, we explore further how weighted averages can be used to find the balancing point of a physical system.

Example 6.62.

For quantities of equal weight, such as two children on a teeter-totter, the balancing point is found by taking the average of their locations. When the weights of the quantities differ, we use a weighted average of their respective locations to find the balancing point.

Suppose that a shelf is 6 feet long, with its left end situated at $x = 0 .$ If one book of weight 1 lb is placed at $x_{1} = 0,$ and another book of weight 1 lb is placed at $x_{2} = 6,$ what is the location of $\overset{―}{x},$ the point at which the shelf would (theoretically) balance on a fulcrum?
Now, say that we place four books on the shelf, each weighing 1 lb: at $x_{1} = 0,$ at $x_{2} = 2,$ at $x_{3} = 4,$ and at $x_{4} = 6 .$ Find $\overset{―}{x},$ the balancing point of the shelf.
How does $\overset{―}{x}$ change if we change the location of the third book? Say the locations of the 1-lb books are $x_{1} = 0,$ $x_{2} = 2,$ $x_{3} = 3,$ and $x_{4} = 6 .$
Next, suppose that we place four books on the shelf, but of varying weights: at $x_{1} = 0$ a 2-lb book, at $x_{2} = 2$ a 3-lb book, at $x_{3} = 4$ a 1-lb book, and at $x_{4} = 6$ a 1-lb book. Use a weighted average of the locations to find $\overset{―}{x},$ the balancing point of the shelf. How does the balancing point in this scenario compare to that found in (b)?
What happens if we change the location of one of the books? Say that we keep everything the same in (d), except that $x_{3} = 5 .$ How does $\overset{―}{x}$ change?
What happens if we change the weight of one of the books? Say that we keep everything the same in (d), except that the book at $x_{3} = 4$ now weighs 2 lbs. How does $\overset{―}{x}$ change?
Experiment with a couple of different scenarios of your choosing where you move one of the books to the left, or you decrease the weight of one of the books.
Write a couple of sentences to explain how adjusting the location of one of the books or the weight of one of the books affects the location of the balancing point of the shelf. Think carefully here about how your changes should be considered relative to the location of the balancing point $\overset{―}{x}$ of the current scenario.

Hint.

Find the average location.
Note that you are averaging 4 locations.
Compare (b).
If the book at location $x_{1}$ weighs 2 pounds, it’s like there are two books at $x_{1},$ so $2 x_{1}$ must play a role in the average.
Use a weighted average.
Compare (e).
Experiment.
Compare the effects of moving locations left and adding weight to the left of the balancing point.

Answer.

$\overset{―}{x} = \frac{x_{1} + x_{2}}{2} = 3 .$
$\overset{―}{x} = \frac{x_{1} + x_{2} + x_{3} + x_{4}}{4} = 3 .$
$\overset{―}{x} = \frac{x_{1} + x_{2} + x_{3} + x_{4}}{4} = 2.75 .$
$\overset{―}{x} = \frac{2 x_{1} + 3 x_{2} + 1 x_{3} + 1 x_{4}}{7} = \frac{16}{7} .$
$\overset{―}{x} = \frac{2 x_{1} + 3 x_{2} + 1 x_{3} + 1 x_{4}}{7} = \frac{17}{7} .$
$\overset{―}{x} = \frac{2 x_{1} + 3 x_{2} + 1 x_{3} + 2 x_{4}}{7} = \frac{22}{7} .$
Answers will vary.
If we have an existing arrangement and balancing point, moving one of the locations to the left will move the balancing point to the left; similarly, moving one of the locations to the right will move the balancing point to the right. If instead we add weight to an existing location, if that location is left of the balancing point, the balancing point will move left; the behavior is similar if on the right.

Solution.

The shelf will balance at $\overset{―}{x} = \frac{x_{1} + x_{2}}{2} = \frac{0 + 6}{2} = 3 .$
The shelf will balance at the average of the locations: $\overset{―}{x} = \frac{x_{1} + x_{2} + x_{3} + x_{4}}{4} = \frac{0 + 2 + 4 + 6}{4} = 3 .$
Changing the location of $x_{3}$ to the left will move the average to the left, so the new location for $\overset{―}{x}$ is $\overset{―}{x} = \frac{x_{1} + x_{2} + x_{3} + x_{4}}{4} = \frac{0 + 2 + 3 + 6}{4} = \frac{11}{4} = 2.75 .$
We can think of the weighted books almost as if there were that number of 1-lb books at each location. The balancing point is given by the weighted average

$\overset{―}{x} = \frac{2 x_{1} + 3 x_{2} + 1 x_{3} + 1 x_{4}}{7} = \frac{2 \cdot 0 + 3 \cdot 2 + 1 \cdot 4 + 1 \cdot 6}{7} = \frac{16}{7} .$

Note particularly that we divide by the total weight of the books, not the total number.
Here we find

$\overset{―}{x} = \frac{2 x_{1} + 3 x_{2} + 1 x_{3} + 1 x_{4}}{7} = \frac{2 \cdot 0 + 3 \cdot 2 + 1 \cdot 5 + 1 \cdot 6}{7} = \frac{17}{7},$

which moves the balancing point slightly to the right.
In this situation, we find the balancing point moves significantly to the right:

$\overset{―}{x} = \frac{2 x_{1} + 3 x_{2} + 1 x_{3} + 2 x_{4}}{7} = \frac{2 \cdot 0 + 3 \cdot 2 + 1 \cdot 4 + 2 \cdot 6}{7} = \frac{22}{7} .$
Try, for instance, having $x_{2} = 1$ and $x_{3} = 2$ and experiment with different weights.
If we have an existing arrangement and balancing point, moving one of the locations to the left will move the balancing point to the left; similarly, moving one of the locations to the right will move the balancing point to the right. If instead we add weight to an existing location, if that location is left of the balancing point, the balancing point will move left; the behavior is similar if on the right.

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Subsection 6.4.4 Center of Mass

🔗

In Example 6.62, we saw that the balancing point of a system of point-masses

⁴

In the example, we actually used weight rather than mass. Since weight is proportional to mass, the computations for the balancing point result in the same location regardless of whether we use weight or mass. The gravitational constant is present in both the numerator and denominator of the weighted average.

(such as books on a shelf) is found by taking a weighted average of their respective locations. In the example, we were computing the center of mass of a system of masses distributed along an axis, which is the balancing point of the axis on which the masses rest.

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Center of Mass (point-masses).

🔗

For a collection of

n

masses

m_{1},

\dots,

m_{n}

that are distributed along a single axis at the locations

x_{1},

\dots,

x_{n},

the center of mass is given by

\overset{―}{x} = \frac{x_{1} m_{1} + x_{2} m_{2} + \dots + x_{n} m_{n}}{m_{1} + m_{2} + \dots + m_{n}} .

🔗

Now consider a thin bar over which density is distributed continuously. If the density is constant, it is obvious that the balancing point of the bar is its midpoint. But if density is not constant, we must compute a weighted average. Let’s say that the function

ρ (x)

tells us the density distribution along the bar, measured in g/cm. If we slice the bar into small sections, we can think of the bar as holding a collection of adjacent point-masses. The mass

m_{i}

of a slice of thickness

Δ x

at location

x_{i},

m_{i} \approx ρ (x_{i}) Δ x .

🔗

If we slice the bar into

n

pieces, we can approximate its center of mass by

\overset{―}{x} \approx \frac{x_{1} \cdot ρ (x_{1}) Δ x + x_{2} \cdot ρ (x_{2}) Δ x + \dots + x_{n} \cdot ρ (x_{n}) Δ x}{ρ (x_{1}) Δ x + ρ (x_{2}) Δ x + \dots + ρ (x_{n}) Δ x} .

🔗

Rewriting the sums in sigma notation, we have

\begin{matrix} (6.4) & \overset{―}{x} \approx \frac{\sum_{i = 1}^{n} x_{i} \cdot ρ (x_{i}) Δ x}{\sum_{i = 1}^{n} ρ (x_{i}) Δ x} . \end{matrix}

🔗

The greater the number of slices, the more accurate our estimate of the balancing point will be. The sums in Equation (6.4) can be viewed as Riemann sums, so in the limit as

n \to \infty,

we find that the center of mass is given by the quotient of two integrals.

🔗

Center of Mass (continuous mass distribution).

🔗

For a thin rod of density

ρ (x)

distributed along an axis from

x = a

x = b,

the center of mass of the rod is given by

\overset{―}{x} = \frac{\int_{a}^{b} x ρ (x) d x}{\int_{a}^{b} ρ (x) d x} .

🔗

Note that the denominator of

\overset{―}{x}

is the mass of the bar, and that this quotient of integrals is simply the continuous version of the weighted average of locations,

x,

along the bar.

Example 6.63.

Consider a thin bar of length 20 cm whose density is distributed according to the function

ρ (x) = 4 + 0.1 x,

where

x = 0

represents the left end of the bar. Assume that

ρ

is measured in g/cm and

x

is measured in cm.

Find the total mass, $M,$ of the bar.
Without doing any calculations, do you expect the center of mass of the bar to be equal to 10, less than 10, or greater than 10? Why?
Compute $\overset{―}{x},$ the exact center of mass of the bar.
What is the average density of the bar?
Now consider a different density function, given by $p (x) = 4 e^{0.020732 x},$ also for a bar of length 20 cm whose left end is at $x = 0 .$ Plot both $ρ (x)$ and $p (x)$ on the same axes. Without doing any calculations, which bar do you expect to have the greater center of mass? Why?
Compute the exact center of mass of the bar described in (e) whose density function is $p (x) = 4 e^{0.020732 x} .$ Check the result against the prediction you made in (e).

Hint.

Recall $M = \int_{a}^{b} ρ (x) d x .$
Think about whether there’s more mass to the left of 10 or to the right of 10.
Use the formula just developed in the preceding subsection of the text.
Recall that when density, $ρ$ is constant, $M = ρ \cdot V .$
Think about how the two density functions distribute mass. Note that they appear to distribute the same total mass, as the areas under the two curves appear to be equal.
Use $p (x)$ in place of $ρ (x)$ in standard calculations.

Answer.

$M = \int_{0}^{20} 4 + 0.1 x d x = 100$ g.
Greater than 10.
$\overset{―}{x} = \frac{\int_{0}^{20} x (4 + 0.1 x)) d x}{\int_{0}^{20} 4 + 0.1 x d x} = \frac{32}{3} .$
5 g/cm.
Slightly to the right of the center of mass for $ρ (x) .$
$\overset{―}{x} = \frac{\int_{0}^{20} x 4 e^{0.020732 x} d x}{\int_{0}^{20} 4 e^{0.020732 x} d x} \approx 10.6891,$

Solution.

$M = \int_{0}^{20} 4 + 0.1 x d x = 100$ g.
The center of mass will be greater than 10 since the mass is more concentrated as $x$ increases.
Using the formula we developed, $\overset{―}{x} = \frac{\int_{0}^{20} x ρ (x) d x}{\int_{0}^{20} ρ (x) d x},$ so

$\overset{―}{x} = \frac{\int_{0}^{20} x (4 + 0.1 x)) d x}{\int_{0}^{20} 4 + 0.1 x d x} = \frac{\frac{3200}{3}}{100} = \frac{32}{3},$

which is indeed just greater than 10.
Recall that average density is given by mass divided by length, so since the bar is 20 cm long and its mass is 100 g, the average density is 100/20 = 5 g/cm. This can also be computed using the formula for average value of a function:

$ρ_{AVG [0, 20]} = \frac{1}{20 - 0} \int_{0}^{20} ρ (x) d x = \frac{100}{20} .$
Since $p (x) = 4 e^{0.020732 x}$ lies below $ρ (x) = 4 + 0.1 x$ up until where they intersect at about $17.53,$ after which $p (x) > ρ (x),$ we see that $p (x)$ distributes mass more to the right. It also appears as if the two curves may bound the same total area, and thus the generate the same total mass. If that is the case, we expect $\overset{―}{x}$ for $p (x)$ to lie slightly to the right of the center of mass for $ρ (x) .$
We see that

$\overset{―}{x} = \frac{\int_{0}^{20} x 4 e^{0.020732 x} d x}{\int_{0}^{20} 4 e^{0.020732 x} d x} \approx 10.6891,$

which is indeed slightly right of $\overset{―}{x}$ for $ρ (x) .$

🔗

Subsection 6.4.5 Summary

🔗

For an object of constant density $D,$ with volume $V$ and mass $m,$ we have $m = D \cdot V .$
If an object with constant cross-sectional area (such as a thin bar) has its density distributed along an axis according to the function $ρ (x),$ then we can find the mass of the object between $x = a$ and $x = b$ by

$m = \int_{a}^{b} ρ (x) d x .$
If the density of an object (such as a disk) depends on the distance from one central point according to the function $ρ (r),$ then we can find the mass of the object by

$m = \int_{0}^{R} 2 π r ρ (x) d x .$
For a system of point-masses distributed along an axis, say $m_{1}, \dots, m_{n}$ at locations $x_{1}, \dots, x_{n},$ the center of mass, $\overset{―}{x},$ is given by the weighted average

$\overset{―}{x} = \frac{\sum_{i = 1}^{n} x_{i} m_{i}}{\sum_{i = 1}^{n} m_{i}} .$

If instead we have mass continuously distributed along an axis, such as by a density function $ρ (x)$ for a thin bar of constant cross-sectional area, the center of mass of the portion of the bar between $x = a$ and $x = b$ is given by

$\overset{―}{x} = \frac{\int_{a}^{b} x ρ (x) d x}{\int_{a}^{b} ρ (x) d x} .$

In each situation, $\overset{―}{x}$ represents the balancing point of the system of masses or of the portion of the bar.

🔗

Exercises 6.4.6 Exercises

🔗

1. Center of mass for a linear density function.

🔗

A rod has length 3 meters. At a distance

x

meters from its left end, the density of the rod is given by

δ (x) = 4 + 4 x

g/m.

🔗

(a) Complete the Riemann sum for the total mass of the rod (use

D x

in place of

Δ x

🔗

mass =

Σ

🔗

(b) Convert the Riemann sum to an integral and find the exact mass.

🔗

mass =

🔗

(include units

⁵

/webwork2_files/helpFiles/Units.html

)

🔗

2. Center of mass for a nonlinear density function.

🔗

A rod with uniform density (mass/unit length)

δ (x) = 8 + \sin (x)

lies on the

x

-axis between

x = 0

and

x = π .

Find the mass and center of mass of the rod.

🔗

mass =

🔗

center of mass =

🔗

3. Interpreting the density of cars on a road.

🔗

Suppose that the density of cars (in cars per mile) down a 20-mile stretch of the Pennsylvania Turnpike is approximated by

δ (x) = 350 (2 + \sin (4 \sqrt{x + 0.175})),

at a distance

x

miles from the Breezewood toll plaza. Sketch a graph of this function for

0 \leq x \leq 20 .

🔗

(a) Complete the Riemann sum that approximates the total number of cars on this 20-mile stretch (use

D x

instead of

Δ x

🔗

Number =

Σ

🔗

(b) Find the total number of cars on the 20-mile stretch.

🔗

Number =

🔗

4. Center of mass in a point-mass system.

🔗

A point mass of 1 grams located 5 centimeters to the left of the origin and a point mass of 2 grams located 8 centimeters to the right of the origin are connected by a thin, light rod. Find the center of mass of the system.

🔗

Center of Mass =

to the left of the origin
to the right of the origin
(at the origin)

🔗

(include units

⁶

/webwork2_files/helpFiles/Units.html

in your center of mass)

🔗

5. Center of mass in a continuous 1-dimensional object.

🔗

Let a thin rod of length

a

have density distribution function

ρ (x) = 10 e^{- 0.1 x},

where

x

is measured in cm and

ρ

in grams per centimeter.

If the mass of the rod is 30 g, what is the value of $a ?$
For the 30g rod, will the center of mass lie at its midpoint, to the left of the midpoint, or to the right of the midpoint? Why?
For the 30g rod, find the center of mass, and compare your prediction in (b).
At what value of $x$ should the 30g rod be cut in order to form two pieces of equal mass?

🔗

6. Combining masses and centers of masses for continuous 1-dimensional object.

🔗

Consider two thin bars of constant cross-sectional area, each of length 10 cm, with respective mass density functions

ρ (x) = \frac{1}{1 + x^{2}}

and

p (x) = e^{- 0.1 x} .

🔗

Find the mass of each bar.
Find the center of mass of each bar.
Now consider a new 10 cm bar whose mass density function is $f (x) = ρ (x) + p (x) .$
1. Explain how you can easily find the mass of this new bar with little to no additional work.
2. Similarly, compute $\int_{0}^{10} x f (x) d x$ as simply as possible, in light of earlier computations.
3. True or false: the center of mass of this new bar is the average of the centers of mass of the two earlier bars. Write at least one sentence to say why your conclusion makes sense.

🔗

7. Mass and center of mass for a solid of revolution.

🔗

Consider the curve given by

y = f (x) = 2 x e^{- 1.25 x} + (30 - x) e^{- 0.25 (30 - x)} .

Plot this curve in the window $x = 0 \dots 30,$ $y = 0 \dots 3$ (with constrained scaling so the units on the $x$ and $y$ axis are equal), and use it to generate a solid of revolution about the $x$ -axis. Explain why this curve could generate a reasonable model of a baseball bat.
Let $x$ and $y$ be measured in inches. Find the total volume of the baseball bat generated by revolving the given curve about the $x$ -axis. Include units on your answer.
Suppose that the baseball bat has constant weight density, and that the weight density is $0.6$ ounces per cubic inch. Find the total weight of the bat whose volume you found in (b).
Because the baseball bat does not have constant cross-sectional area, we see that the amount of weight concentrated at a location $x$ along the bat is determined by the volume of a slice at location $x .$ Explain why we can think about the function $ρ (x) = 0.6 π f (x)^{2}$ (where $f$ is the function given at the start of the problem) as being the weight density function for how the weight of the baseball bat is distributed from $x = 0$ to $x = 30 .$
Compute the center of mass of the baseball bat.

🔗

8. Mass of a disk.

🔗

The Math Olympics gives out prize medals, which are disks with varying density.

The first place medal has a radius of 2 inches, and the density of the disk is given by $ρ (r) = 9 - 2 r$ where $r$ is the distance from the center of the disk. Find the mass of the first place medal.
To make the second place medal, the Math Olympics takes a first-place medal and removes material from the outside until the radius is 1.8 inches. Find the mass of the second place medal.

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