TBIL-Cal Density, Mass, and Center of Mass (AI5)

Section 6.5 Density, Mass, and Center of Mass (AI5)

Learning Outcomes

Set up integrals to solve problems involving density, mass, and center of mass.
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Subsection 6.5.1 Activities

Activity 6.5.1.

Consider a rectangular prism with a 10 meters \(\times\) 10 meters square base and height 20 meters. Suppose the density of the material in the prism increases with height, following the function \(\delta(h)=10+h\) kg/m\(^3\text{,}\) where \(h\) is the height in meters.

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(a)

If one were to cut this prism, parallel to the base, into 4 pieces with height 5 meters, what would the volume of each piece be?

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(b)

Consider the piece sitting on top of the slice made at height \(h=5\text{.}\) Using a density of \(\delta(5)=15\) kg/m\(^3\text{,}\) and the volume you found in (a), estimate the mass of this piece.

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\(500\cdot 5=2500\) kg

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\(500\cdot 15=7500\) kg

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\(500\cdot 15\cdot 5=37500\) kg

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(c)

Is this estimate the actual mass of this piece?

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Activity 6.5.2.

Consider all 4 slices from Activity 6.5.1.

described in detail following the image — Figure 135. \(10\times 10\times 20\) prism sliced into 4 pieces.
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(a)

Fill out the following table.

\begin{equation*} \begin{array}{|c|c|c|c|} \hline h_i & \delta(h_i) & \text{Volume} & \text{Estimated Mass}\\ \hline h_4=15\ \text{m} & \delta(15)=25\ \text{kg/m}^3 & 500\ \text{m}^3 & \\ \hline h_3=10\ \text{m} & \delta(10)=20\ \text{kg/m}^3 & 500\ \text{m}^3 & \\ \hline h_2=5\ \text{m} & \delta(5)=15\ \text{kg/m}^3 & 500\ \text{m}^3 & 7500\ \text{kg}\\ \hline h_1=0\ \text{m} & \delta(0)=10\ \text{kg/m}^3 & 500\ \text{m}^3 & \\ \hline \end{array} \end{equation*}

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(b)

What is the estimated mass of the rectangular prism?

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Activity 6.5.3.

Suppose instead that we sliced the prism from Activity 6.5.1 into 5 pieces of height 4 meters.

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(a)

Fill out the following table.

\begin{equation*} \begin{array}{|c|c|c|c|} \hline h_i & \delta(h_i) & \text{Volume} & \text{Estimated Mass}\\ \hline h_5=16\ \text{m} & \delta(16)=26\ \text{kg/m}^3 & 400\ \text{m}^3 & \\ \hline h_4=12\ \text{m} & \delta(12)=22\ \text{kg/m}^3 & 400\ \text{m}^3 & \\ \hline h_3=8\ \text{m} & \delta(8)=18\ \text{kg/m}^3 & 400\ \text{m}^3 & \\ \hline h_2=4\ \text{m} & \delta(4)=14\ \text{kg/m}^3 & 400\ \text{m}^3 & \phantom{7500\ \text{kg}}\\ \hline h_1=0\ \text{m} & \delta(0)=10\ \text{kg/m}^3 & 400\ \text{m}^3 & \\ \hline \end{array} \end{equation*}

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(b)

What is the estimated mass of the rectangular prism?

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Activity 6.5.4.

Which of the estimates computed in Activity 6.5.2 and Activity 6.5.3 is a better estimate of the mass of the prism?

Activity 6.5.2, 4 pieces is a better estimate.

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Activity 6.5.3, 5 pieces is a better estimate.

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Activity 6.5.5.

Suppose now that we slice the prism from Activity 6.5.1 into slices of height \(\Delta h\) meters.

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(a)

Consider the piece sitting atop the slice made at height \(h_i\text{.}\) Using \(\delta(h_i)=10+h_i\) as the estimate for the density of this piece, what is the mass of this piece?

\(\displaystyle (10+h)100\cdot h_i\)

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\(\displaystyle (10+\Delta h)100\cdot h_i\)

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\(\displaystyle (10+h_i)100\cdot \Delta h\)

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\(\displaystyle (10+h_i)100\cdot h\)

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Activity 6.5.6.

Consider a cylindrical cone with a base radius of 15 inches and a height of 60 inches. Suppose the density of the cone is \(\delta(h)= 15+\sqrt{h}\) oz/in\(^3\text{.}\)

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(a)

Let \(r_2\) be the radius of the circular cross section of the cone, made at height 30 inches. Recall that \(\Delta ABC, \Delta AB'C'\) are similar triangles, what is \(r_2\text{?}\)

15 inches.

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7.5 inches.

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30 inches.

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60 inches.

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(b)

What is the volume of a cylinder with radius \(r_1=15\) inches and height \(30\) inches?

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(c)

What is the volume of a cylinder with radius \(r_2\) inches and height \(30\) inches?

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Activity 6.5.7.

Suppose that we estimate the mass of the cone from Activity 6.5.6 with 2 cylinders of height 30 inches.

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(a)

Fill out the following table.

\begin{equation*} \begin{array}{|c|c|c|c|} \hline h_i & \delta(h_i) & \text{Volume} & \text{Estimated Mass}\\ \hline h_2=30\ \text{in} & \delta(30)=15+\sqrt{30}\ \text{oz/in}^3 & \pi(7.5)^2\cdot30\ \text{in}^3 & \phantom{6500\ \text{kg}}\\ \hline h_1=0\ \text{in} & \delta(0)=15\ \text{oz/in}^3 & \pi(15)^2\cdot30\ \text{in}^3 & \\ \hline \end{array} \end{equation*}

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(b)

What is the estimated mass of the cone?

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Activity 6.5.8.

Suppose that we estimate the mass of the cone from Activity 6.5.6 with 3 cylinders of height 20 inches.

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(a)

Fill out the following table.

\begin{equation*} \begin{array}{|c|c|c|c|} \hline h_i & \delta(h_i) & \text{Volume} & \text{Estimated Mass}\\ \hline h_2=40\ \text{in} & \delta(40)=15+\sqrt{40}\ \text{oz/in}^3 & \pi(5)^2\cdot20\ \text{in}^3 & \phantom{6500\ \text{kg}}\\ \hline h_2=20\ \text{in} & \delta(20)=15+\sqrt{20}\ \text{oz/in}^3 & \pi(10)^2\cdot20\ \text{in}^3 & \phantom{6500\ \text{kg}}\\ \hline h_1=0\ \text{in} & \delta(0)=15\ \text{oz/in}^3 & \pi(15)^2\cdot20\ \text{in}^3 & \\ \hline \end{array} \end{equation*}

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(b)

What is the estimated mass of the cone?

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Activity 6.5.9.

Suppose that we estimate the mass of the cone from Activity 6.5.6 with cylinders of height \(\Delta h\text{.}\)

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(a)

Consider the piece sitting atop the slice made at height \(h_i\text{.}\) Using \(\delta(h_i)=15+\sqrt{h_i}\) as the estimate for the density of this cylinder, what is the mass of this cylinder?

\(\displaystyle (15+\sqrt{h})\pi r_i^2\cdot \Delta h\)

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\(\displaystyle (15+\sqrt{h_i})\pi r_i^2\cdot \Delta h\)

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\(\displaystyle (15+\Delta h)\pi r_i^2\cdot \Delta h_i\)

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\(\displaystyle (15+\sqrt{h_i})\pi r^2\cdot \Delta h\)

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Activity 6.5.10.

Consider a solid where the cross section of the solid at \(x=x_i\) has area \(A(x_i)\text{,}\) and the density when \(x=x_i\) is \(\delta(x_i)\text{.}\)

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(a)

If we used prisms of width \(\Delta x\) to approximate this solid, what is the mass of the slice associated with \(x_i?\)

\(\displaystyle A(x)\delta(x)\Delta x\)

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\(\displaystyle \pi A(x)^2\delta(x_i)\Delta x\)

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\(\displaystyle A(x_i)\delta(x_i)\Delta x\)

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\(\displaystyle A(x_i)\delta(x_i)\Delta x_i\)

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Fact 6.5.11.

Consider a solid where the cross section of the solid at \(x=x_i\) has area \(A(x_i)\text{,}\) and the density when \(x=x_i\) is \(\delta(x_i)\text{.}\) Suppose the interval \([a,b]\) represents the \(x\) values of this solid. If one slices the solid into \(n\) pieces of width \(\Delta x=\frac{b-a}{n}\text{,}\) then one can approximate the mass of the solid by

\begin{equation*} \sum_{i=1}^n \delta(x_i)A(x_i)\Delta x. \end{equation*}

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We can then find actual mass by taking the limit as \(n\to\infty\text{:}\)

\begin{equation*} \lim_{n\to\infty} \left(\sum_{i=1}^n \delta(x_i)A(x_i)\Delta x\right)=\int_a^b \delta(x)A(x) dx. \end{equation*}

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Activity 6.5.12.

Consider that for the prism from Activity 6.5.1, a cross section of height \(h\) is \(A(h)=10^2=100\) m\(^2\text{.}\) Also recall that the density of the prism is \(\delta(h)=10+h\) kg/m\(^3\text{,}\) where \(h\) is the height in meters.

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Use Fact 6.5.11 to find the mass of the prism.

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Activity 6.5.13.

Consider that for the cone from Activity 6.5.6, a cross section of height \(h\) is \(A(h)=\pi r^2\) in\(^2\text{,}\) where \(r\) is the radius of the circular cross-section at height \(h\) inches. Also recall that the density of the cone is \(\delta(h)=15+\sqrt{h}\) oz/in\(^3\text{,}\) where \(h\) is the height in inches.

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(a)

When the height is \(h\) inches, what is \(r\text{?}\)

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Hint.

Use similar triangles:

a 15 by 60 triangle, with a similar r by 60-h triangle. — Figure 145. The right triangles in this figure are similar.
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(b)

Find \(A(h)\) as a function of \(h\) using this information.

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(c)

Use Fact 6.5.11 to find the mass of the cone.

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Activity 6.5.14.

\(8\times 8\)\(\delta(h)=10+\cos(\pi h)\)\(^3\)\(h\)

(a)

When the height is \(h\) feet, what is the area of the square cross section at that height, \(A(h)\text{?}\)

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Hint.

Use similar triangles:

a 8 by 60 triangle, with a similar r by 16-h triangle. — Figure 146. The triangles in this figure are similar.
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(b)

Use Fact 6.5.11 to find the mass of the pyramid.

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Activity 6.5.15.

Consider a board sitting atop the \(x\)-axis with six \(1\times 1\) blocks each weighing 1 kg placed upon it in the following way: two blocks are atop the 1, three blocks are atop the 2, and one block is atop the 6.

Six 1 by 1 blocks on the x-axis, each weighing a kilogram. Two blocks atop the 1, three blocks atop the 2, and one block atop the 6. — Figure 147. Six 1 kg blocks atop the \(x\)-axis.
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Which of the following describes the \(x\)-value of the center of gravity of the board with the blocks?

\(\displaystyle\frac{1+6}{2}=3.5\text{.}\)

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\(\displaystyle\frac{1+2+6}{3}=3\text{.}\)

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\(\displaystyle\frac{2\cdot1+3\cdot2+1\cdot6}{6}\approx 2.3333\text{.}\)

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Activity 6.5.16.

Six 1 by 1 blocks on the x-axis, each weighing a kilogram. Two blocks atop the 1, three blocks atop the 2, and one block atop the 8. — Figure 148. Six 1 kg blocks atop the \(x\)-axis.
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Find the \(x\)-value of the center of gravity of the board with the blocks.

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Fact 6.5.17.

Consider a solid where the cross section of the solid at \(x=x_i\) has area \(A(x_i)\text{,}\) and the density when \(x=x_i\) is \(\delta(x_i)\text{.}\) Suppose the interval \([a,b]\) represents the \(x\) values of this solid. Since each slice has approximate mass \(\delta(x_i)A(x_i)\delta(x_i)\text{,}\) we can approximate the center of mass by taking the weighted “average” of the \(x_i\)-values weighted by the associated mass:

\begin{equation*} \frac{\sum_{i=1}^n x_i\delta(x_i)A(x_i)\Delta x}{\sum_{i=1}^n \delta(x_i)A(x_i)\Delta x}. \end{equation*}

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We can then find actual center of mass by taking the limit as \(n\to\infty\text{:}\)

\begin{equation*} \lim_{n\to\infty} \left(\frac{\sum_{i=1}^n x_i\delta(x_i)A(x_i)\Delta x}{\sum_{i=1}^n \delta(x_i)A(x_i)\Delta x}\right)=\frac{\int_a^b x\delta(x)A(x)dx}{\int_a^b \delta(x)A(x)dx}=\frac{\int_a^b x\delta(x)A(x)dx}{\text{The Total Mass}}. \end{equation*}

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Activity 6.5.18.

Consider that for the prism from Activity 6.5.12, a cross section of height \(h\) is \(A(h)=10^2=100\) m\(^2\text{.}\) Also recall that the density of the prism is \(\delta(h)=10+h\) kg/m\(^3\text{,}\) where \(h\) is the height in meters, and that we found the total mass to be 40000 kg.

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Use Fact 6.5.17 to find the height where the center of mass occurs.

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Activity 6.5.19.

Consider that for the prism from Activity 6.5.13, a cross section of height \(h\) is \(A(h)=\pi\cdot \left( \frac{60-h}{4}\right)^2\) in\(^2\text{.}\) Also recall that the density of the cone is \(\delta(h)=15+\sqrt{h}\) oz/in\(^3\text{,}\) where \(h\) is the height in inches, and that we found the total mass to be about 142492.6 oz.

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Use Fact 6.5.17 to find the height where the center of mass occurs.

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Activity 6.5.20.

Consider that for the pyramid from Activity 6.5.14, a cross section of height \(h\) is \(A(h)=\pi\cdot \left( \frac{16-h}{2}\right)^2\) ft\(^2\text{.}\) Also recall that the density of the pyramid is \(\delta(h)=10+\cos{\pi h}\) lb/feet\(^3\text{,}\) where \(h\) is the height in feet, and that we found the total mass to be about 3414.14.6 lbs.

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Use Fact 6.5.17 to find the height where the center of mass occurs.

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Subsection 6.5.2 Videos

Figure 151. Video: Set up integrals to solve problems involving density, mass, and center of mass

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Subsection 6.5.3 Exercises

Exercises available at https://library.tbil.org/2025/calculus/exercises/#/bank/AI5/

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