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Section Integrals of Mass
Worksheet IC-3 Reading
Objectives
Things you should understand and know:
What is meant by center of mass or center of gravity
What a density function refers to
What moment of inertia tells about a two-dimensional object
Things you should be able to do:
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
You will be assessed on the following criteria:
Problem structure: Solution should be organized and easy to follow, with clear, complete steps that follow progressively from one to the next.
Integrals for Mass Calculations: Use multiple integrals to solve physical problems, such as finding the mass of a lamina or solid.
All of the readings for this course come from the OpenStax Calculus Volume 3 textbook, which is available for free online at
https://openstax.org/details/books/calculus-volume-3 . You can read the textbook online or download a PDF version. The textbook is also available in print from various retailers if you prefer a physical copy.
For this lesson, you should read sections
5.6 Calculating Centers of Mass and Moments of Inertia . You should read through the sections carefully, making sure to understand the definitions, examples, and key ideas. You should also work through the exercises at the end of each section to test your understanding and practice applying the concepts. Some of the exercises at the end of the section are assigned in the WP assignments.
Worksheet IC-3 Videos
Calculating Mass and Moments of Inertia
Worksheet IC-3 Written Practice
INFORMATION ABOUT ALL WRITTEN PRACTICE (WP) ASSIGNMENTS:
THE WP ASSIGNMENTS ARE DESIGNED TO PREPARE YOU FOR THE QUIZZES AND ASSESSMENTS
Work on every problem on every assignment.
If you get stuck or do not understand a solution, ask questions.
For each problem, try it once or twice before looking at solutions or asking for help.
Mark the problems you did correctly on the first try.
Highlight problems where you got help (using solutions, videos, a tutor, etc) -These are the problem types that you will need more practice on to be ready for the assessments.
Understand that just copying the solutions instead of working through the problems will greatly reduce your chances of success in this class.
Calculating Centers of Mass and Moments of Inertia: Complete problems 297, 302, 309, 314, 321, 326
Worksheet IC-3 Sample Outcomes
For the quiz on this outcome you would be given 40 minutes to complete 1 problem with parts. For these you will be asked to set up all three problems but only asked to fully evaluate one. Here are samples of the types of problems you will encounter:
1. Let the lamina
\(R\) be bounded by the
\(x=1-y^2\) and the y-axis. Let
\(\rho(x,y)=12x\) \(mg/cm^2\)
(a) Sketch the lamina
\(R\text{.}\)
Answer .
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(b) Compute the mass of the lamina
\(R\text{.}\)
Hint .
To find the mass, integrate the density function over the region
\(R\text{.}\)
Answer .
\(mass=\frac{32}{5}\) grams
Solution .
Based on the graph of the region
\(R\text{,}\) we can see that the limits on the integral will be
\(0 \leq x \leq 1-y^2\) and
\(-1 \leq y \leq 1\text{.}\)
\(\iint_R 12x dA = \int_{-1}^{1} \int_{0}^{1-y^2} 12x dx dy\)
\(\iint_R 12x dA = \int_{-1}^{1} \left[6x^2\right]_0^{1-y^2} dy\)
\(\iint_R 12x dA = \int_{-1}^{1} 6(1-y^2)^2 dy\)
\(\iint_R 12x dA = \int_{-1}^{1} 6(1-2y^2+y^4) dy\)
\(\iint_R 12x dA = \int_{-1}^{1} 6-12y^2+6y^4 dy\)
\(\iint_R 12x dA = \left[6y-4y^3+\frac{6}{5}y^5\right]_{-1}^{1} dy\)
\(\iint_R 12x dA = 6(1)-4(1)+\frac{6}{5}(1) - (6(-1)-4(-1)+\frac{6}{5}(-1)))\)
\(\iint_R 12x dA = \frac{32}{5}\) grams
(c) Without calculation, explain why
\(M_x = 0\text{.}\)
Hint .
Think about the symmetry of the region
\(R\) with respect to the
\(x\) -axis.
Answer .
The region
\(R\) is symmetric about the
\(x\) -axis, so the first moment about the
\(x\) -axis is zero.
Solution .
\(M_x\) measures the tendency of the lamina to rotate about the
\(x\) -axis. Here the density function depends only on
\(x\text{,}\) so the mass is distributed symmetrically above and below the
\(x\) -axis. Since the region
\(R\) is symmetric about the
\(x\) -axis, the first moment about the
\(x\) -axis is zero.
(d) Compute the center of mass of the lamina
\(R\text{.}\)
Hint .
Use the formulas for the center of mass coordinates. Calculate the moment about the
\(y\) -axis.
Answer .
\(M_y=\frac{128}{5}\) grams
Solution .
To find the moment about the
\(y\) -axis, we integrate
\(x\) times the density function over the region
\(R\text{.}\) Using the same limits (region) as before we get:
\(\iint_R 12x^2 dA = \int_{-1}^{1} \int_{0}^{1-y^2} 12x^2 dx dy\)
\(\iint_R 12x^2 dA = \int_{-1}^{1} \left[4x^3\right]_0^{1-y^2} dy\)
\(\iint_R 12x^2 dA = \int_{-1}^{1} 4(1-y^2)^3 dy\)
\(\iint_R 12x^2 dA = \int_{-1}^{1} 4(1-3y^2+3y^4-y^6) dy\)
\(\iint_R 12x^2 dA = \int_{-1}^{1} 4-12y^2+12y^4-4y^6 dy\)
\(\iint_R 12x^2 dA = \left[4y-4y^3+\frac{12}{5}y^5-\frac{4}{7}y^7\right]_{-1}^{1} dy\)
\(\iint_R 12x^2 dA = 4(1)-4(1)+\frac{12}{5}(1) - (4(-1)-4(-1)+\frac{12}{5}(-1)))\)
\(\iint_R 12x^2 dA = \frac{128}{5}\) grams
2. Given the thin plate of the constant density
\(\rho=5\) modeled by the plane region
\(R:\) The rectangle
\([-2,1]]\times\left[1,3]\)
(a) Find the center of mass of the lamina
\(R\text{.}\)
Hint .
Set up the integrals for the moments and the mass using the limits of the rectangle.
Answer .
\(\left(\frac{M_y}{M}, \frac{M_x}{M}\right) = \left(0, 2\right)\)
Solution .
\(M = \iint_R \rho dA = 5 \iint_R dA\)
\(M = 5 \int_{-2}^1 \int_1^3 dy dx\)
\(M = 5 \int_{-2}^1 \left[y\right]_1^3 dx\)
\(M = 5 \int_{-2}^1 2 dx\)
\(M = 10 \left[x\right]_{-2}^1\)
\(M = 10 (3) = 30\) grams
Next, find the moment about the
\(y\) -axis:
\(M_y = \iint_R x \rho dA = 5 \iint_R x dA\)
\(M_y = 5 \int_{-2}^1 \int_1^3 x dy dx\)
\(M_y = 5 \int_{-2}^1 x \left[y\right]_1^3 dx\)
\(M_y = 5 \int_{-2}^1 2x dx\)
\(M_y = 5 \left[x^2\right]_{-2}^1\)
\(M_y = 5 (1 - 4) = -15\)
Finally, find the moment about the
\(x\) -axis:
\(M_x = \iint_R y \rho dA = 5 \iint_R y dA\)
\(M_x = 5 \int_{-2}^1 \int_1^3 y dx dy\)
\(M_x = 5 \int_{-2}^1 \left[\frac{y^2}{2}\right]_1^3 dx\)
\(M_x = 5 \int_{-2}^1 \frac{1}{2} (9 - 1) dx\)
\(M_x = 5 \int_{-2}^1 4 dx\)
\(M_x = 20 \left[x\right]_{-2}^1\)
Thus, the center of mass is at
\(\left(\frac{M_y}{M}, \frac{M_x}{M}\right) = \left(0, \frac{60}{30}\right) = \left(0, 2\right)\text{.}\)
(b) Find the moments of inertia.
Hint .
Set up the integrals for the moments of inertia using the limits of the rectangle.
Answer .
\(I_x = 130\text{,}\) \(I_y = 30\text{,}\) and
\(I_0 = 160\)
Solution .
To find the moments of inertia, we use the formulas:
\(I_x = \iint_R (y^2) \rho dA\)
\(I_y = \iint_R (x^2) \rho dA\)
Using the same limits as before:
\(I_x = 5 \int_{-2}^1 \int_1^3 y^2 dy dx\)
\(I_x = 5 \int_{-2}^1 \left[\frac{y^3}{3}\right]_1^3 dx\)
\(I_x = 5 \int_{-2}^1 \frac{1}{3} (27 - 1) dx\)
\(I_x = 5 \int_{-2}^1 \frac{26}{3} dx\)
\(I_x = \frac{130}{3} \left[x\right]_{-2}^1\)
\(I_x = \frac{130}{3} (1 - (-2)) = \frac{130}{3} (3) = 130\)
\(I_y = 5 \int_{-2}^1 \int_1^3 x^2 dx dy\)
\(I_y = 5 \int_{-2}^1 x^2 \left[y\right]_1^3 dy\)
\(I_y = 5 \int_{-2}^1 2x^2 dy\)
\(I_y = 10 \int_{-2}^1 x^2 dx\)
\(I_y = 10 \left[\frac{x^3}{3}\right]_{-2}^1\)
\(I_y = 10 \frac{1}{3} (1 - (-8)) = 10 \frac{9}{3} = 30\)
\(I_0 = I_x + I_y = 130 + 30 = 160\)
