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Cavalieri’s Principle
Mathematical Goals: Teachers will be able to
· Derive the formulas for the circumference and area of a circle.
· Derive the formula for a sphere using Cavalieri’s Principle.
Pedagogical Goals: Teachers will be able to
· Discuss what supports students might need in deriving formulas.
· Consider how students might approach the tasks presented.
Mathematical Practices:
· Make sense of problems and persevere in solving them.
· Reason abstractly and quantitatively.
· Construct viable arguments and critique the reasoning of others.
· Use appropriate tools strategically.
· Attend to precision.
· Look for and make use of structure.
Length of session: 90 minutes
Materials needed: String, scissors, copies of segmented circles for cutting, pennies, rulers, can/cup, Cavalieri’s Principle Participant Handout
Overview:
In this session participants will derive the formulas for the circumference and area of a circle using dissection methods. They will also explore Cavalieri’s Principle and derive the formula for the volume of a sphere using Cavalieri’s Principle.
Estimated # of Minutes / Activity40 minutes / Deriving Circumference and Area of Circle Formulas
· To start this session, we will derive the circumference and area of circle formulas.
· Providing just a can or cup, plenty of string, rulers and scissors, have participants determine how these materials can be used to derive the formula for the circumference of a circle. To do this, participants should take a string and wrap it around a can or cup to record its circumference. They should cut this piece of string to the length of the circumference and lay it flat on the table. Then, participants should measure the diameter of the can or cup using another piece of string, cutting it to the appropriate length. Now, all the participants have to do is use the diameter to estimate the length of the string representing the circumference. They should find that the circumference is approximately 3.14 times the diameter. An example is shown here below. Here you can see that copying the diameter three times almost gets you the length of the circumference. But, to estimate the leftover portion if you take another piece of string the length of the diameter, half it and then half the half, you can see that the left over piece of the circumference is still smaller than that. Thus, if you half the ¼ piece, you can see that the leftover piece of the circumference is just larger than 1/8 of the diameter. Thus, the remaining piece of the circumference is between 1/8 = .125 and 1/4 = .25, but closer to the 1/8 mark. Therefore, the remaining piece is approximately .14 of the diameter. This gives the circumference equal to approximately 3.14 times the diameter, yielding the formula we know and use: C=πd.
· Next, provide the participants with the segmented circles and scissors and ask them how these materials can be used to derive the area of a circle. To do this, participants must cut the segments of the circle out and arrange them so that they approximate the shape of a parallelogram as shown below. Then, recalling the area of a parallelogram is base times height, the participants can estimate the area of a circle to be the height of their approximated parallelogram, the length of the radius, times the base of the approximated parallelogram, the length of one half the circumference. Thus, you have A=1/2Cr=1/2(πd)r=1/2(π*2r)r=πr2.
· Questions to consider:
1. Often by the time students get to high school, they know these formulas. Do you think they would be able to derive these formulas as you did with just the materials provided? What additional supports might they need? If solving problems and persevering in mathematics are norms of the classroom, students should be able to come up with how these materials provide the formulas. It is important that the teacher be aware of what students must know before the activity (e.g., area of parallelogram). After some time struggling, students may need some additional supports if they are getting frustrated. This might come in the form of questioning from the teacher. For example, the teacher might ask for the circumference problem, how are the diameter and circumference related? Such questions, however, should only be used as a last resort. So while it may take students some time to get there, they can!
2. Discuss the advantages to having students provide informal arguments for these formulas. By having students provide informal arguments for these formulas, they will have a better conceptual understanding of them. They will also understand the relationship between the diameter and circumference of a circle and how the area of a circle is similar to that of a parallelogram.
50 minutes / Cavalieri’s Principle
· Provide participants with pennies. Ask them to build a model of the Leaning Tower of Pisa.
· Questions to consider:
1. Create a model of the Leaning Tower of Pisa using your pennies. How many pennies did you use to create your model of the Leaning Tower of Pisa? How could you represent the volume of your leaning tower? Answers will vary. The volume is equivalent to the area of a penny times the height of a penny times the number of pennies used.
2. Now straighten your tower. Does the volume of your model change by making it straight? Why or why not? The volume of the model does not change. It still uses the same number of pennies. More specifically, the height of either tower is the height of a penny times the number of pennies used. This doesn’t change in either scenario, nor does the area of the base (a penny) thus allowing the volume to remain the same.
3. What would happen if we made a straight tower of 15 quarters and compared it to a straight tower of 15 pennies? Would these two towers have the same volume? These two towers would not have the same volume since they have different sized bases. Further, their heights will only be equivalent if the height of a penny and quarter are the same.
4. Cavalieri’s Principle states: If two solids have the same height and the same cross-sectional area at every level, then they have the same volume. Discuss how the Leaning Tower of Pisa task illustrates this principle. Essentially, the leaning tower and the straight tower both have the same height and each cross section has the same area since each cross section is a penny. Thus, the towers have the same volume.
5. What important terminology could be discussed with students as a result of completing this task? This task could introduce the difference between oblique and right cylinders. The leaning tower would be considered an oblique cylinder since its top and bottom are not aligned with one right above the other, while the straight tower would be considered a right cylinder since its top and bottom are aligned one directly above the other.
· Pause here for a discussion of the leaning tower task. Then, present the following problem to participants asking them to pretend they only know the formulas for the volume of a cone and a cylinder as well as Cavalieri’s Principle. In other words, they have to pretend they do not know the volume for a sphere, as this activity would be used to introduce that.
6. Consider the following problem:
The management of an ocean life museum needs to build an aquarium for its new exhibit. They have two options, Aquarium A or Aquarium B, and would like to choose the aquarium that utilizes the least amount of water. Aquarium A is a right cylinder with a diameter of 10 feet and a height of 5 feet. Covering the lower base of Aquarium A is an “underwater mountain” in the shape of a 5-foot-tall right cone. This aquarium would be built into a pillar in the center of the exhibit room. Aquarium B is half of a 10-foot-diameter sphere. This aquarium would protrude from the ceiling of the exhibit room. Does it matter which aquarium the management chooses? Hint: Use Cavalieri’s Principle to find the volume of Aquarium B.
Note: This task presents a context that leads to the discovery of the formula for calculating the volume of a sphere. To solve the problem, one must be familiar with the volume of a cylinder, the formula for the volume of a cone, and Cavalieri’s principle. So, for the participants, they should pretend they do not know the volume of a sphere; for students, this should be used to introduce the volume of a sphere.
The number of cubic feet of water that Aquarium A will hold is found by subtracting the volume of the cone from the volume of the cylinder: πr2h-13πr2h=23πr2h=23π*52*5=250π3≈262 cubic feet of water. Now, recall Cavalieri’s Principle states that if two solids have the same height and cross sectional area at every level, then they have the same volume. We already know our solids have the same height, so now let’s compare the cross sectional areas. For Aquarium A, the radius and height of the cylinder are both 5 feet, so the vertical cross-section through the vertex of the cone shown below shows that the cross section of the water forms an isosceles right triangle. This is because the vertical cross section of the cone is an isosceles triangle. Dropping an altitude from the vertex angle will divide this cross section of the cone into two right isosceles triangles; therefore, each of its base angles is 45 degrees, leaving the complementary angle of the cross section of the water to be 45 degrees also. Thus, since the top of the water forms a 90-degree angle with the side of the cylinder, the other remaining angle of the triangular cross section of water is also 45 degrees. Therefore, the triangles representing the cross sections of water are isosceles right triangles. This means that the horizontal distance between the cone and the aquarium wall at height h is also h.
Now if we consider the horizontal cross section at height h, we need to find the area of the water’s surface for that height. This can be done by taking the area of the outer circle and subtracting from it the area of the inner circle (which has radius 5-h). Thus we get: π*52-π5-h2=25π-π25-10h+h2=25π-25π+10πh-πh2=10πh-πh2.
Secondly, let’s consider the cross sections of Aquarium B. Momentarily, the horizontal distance between the curved wall and the sphere’s center at height h is denoted by x. But, x can be expressed in terms of h since x and 5-h both represent leg lengths of a right triangle with hypotenuse 5. In other words:
x2+5-h2=25
x2+25-10h+h2=25
x2=10h-h2
x=10h-h2
Therefore, the area of the water’s surface at height h for Aquarium B is: π(10h-h2)2=π10h-h2=10πh-πh2.
Therefore, each of these solids also has the same cross sectional area for the water’s surface. Thus, the volume of Aquarium A is equivalent to the volume of Aquarium B by Cavalieri’s Principle.
7. Use what you discovered in question 6 to derive the formula for the volume of a sphere. From question 6, we essentially learned that, for solids of the same height, the volume of a cylinder minus the volume of a cone is equivalent to half the volume of a sphere. In other words πr2h-13πr2h=23πr2h, which should be half the volume of a sphere. Thus, the volume of a sphere is 2*23πr2h=43πr2h.
Cavalieri’s Principle
Participant Handout
Circumference of a circle
With the materials provided for you, derive the formula for the circumference of a circle.
Area of a circle
With the materials provided for you, derive the formula for the circumference of a circle.
Leaning Tower of Pisa Model
Using the pennies provided, create a model of the Leaning Tower of Pisa.
Questions to consider:
1. Create a model of the Leaning Tower of Pisa using your pennies. How many pennies did you use to create your model of the Leaning Tower of Pisa? How could you represent the volume of your leaning tower?
2. Now straighten your tower. Does the volume of your model change by making it straight? Why or why not?
3. What would happen if we made a straight tower of 15 quarters and compared it to a straight tower of 15 pennies? Would these two towers have the same volume?
4. Cavalieri’s Principle states: If two solids have the same height and the same cross-sectional area at every level, then they have the same volume. Discuss how the Leaning Tower of Pisa task illustrates this principle.
5. What important terminology could be discussed with students as a result of completing this task?
For the next few questions, pretend you only know the formulas for the volume of a cone and a cylinder as well as Cavalieri’s Principle. In other words, pretend you do not know the volume for a sphere.
6. Consider the following problem:
The management of an ocean life museum needs to build an aquarium for its new exhibit. They have two options, Aquarium A or Aquarium B, and would like to choose the aquarium that utilizes the least amount of water. Aquarium A is a right cylinder with a diameter of 10 feet and a height of 5 feet. Covering the lower base of Aquarium A is an “underwater mountain” in the shape of a 5-foot-tall right cone. This aquarium would be built into a pillar in the center of the exhibit room. Aquarium B is half of a 10-foot-diameter sphere. This aquarium would protrude from the ceiling of the exhibit room. Does it matter which aquarium the management chooses? Hint: Use Cavalieri’s Principle to find the volume of Aquarium B.
7. Use what you discovered in question 6 to derive the formula for the volume of a sphere.
Adapted from: S. Conklin’s Geometry Unit 7: 3D Shapes and Volume (http://cc.betterlesson.com/lesson/442620/exploring-circumference-area-and-cavalieri-s-principle) and from Illustrative Mathematics (https://www.illustrativemathematics.org/illustrations/530)