Lesson 19
Estimating a Hemisphere
19.1: Notice and Wonder: Two Shapes (5 minutes)
Warm-up
The purpose of this warm-up is for students to review how to manipulate the formulas for volume of a cylinder and cone and consider what they look like when the height and radius are the same. Students will encounter these shapes again later in the lesson.
Launch
Arrange students in groups of 2. Tell students that they will look at an image, and their job is to think of at least one thing they notice and at least one thing they wonder. Display the image for all to see. Ask students to give a signal when they have noticed or wondered about something. Give students 1 minute of quiet think time, and then 1 minute to discuss the things they notice with their partner, followed by a whole-class discussion.
Student Facing
Here are two shapes.
![Two figures. A cone with unknown height and radius. Formula given, V = fraction one over three, pi, r cubed. A cylinder with unknown height and radius. Formula given V = pi, r cubed.](https://cms-im.s3.amazonaws.com/qkCq1YkHx6KtN1BckkYf9rEM?response-content-disposition=inline%3B%20filename%3D%228-8.5.CylinderAndCone.png%22%3B%20filename%2A%3DUTF-8%27%278-8.5.CylinderAndCone.png&response-content-type=image%2Fpng&X-Amz-Algorithm=AWS4-HMAC-SHA256&X-Amz-Credential=AKIAXQCCIHWF3XOEFOW4%2F20240907%2Fus-east-1%2Fs3%2Faws4_request&X-Amz-Date=20240907T233633Z&X-Amz-Expires=604800&X-Amz-SignedHeaders=host&X-Amz-Signature=e47bb32a09e707e4cca09ca2c028d11cae79fcd3a653e887c2393a1e5c322e60)
What do you notice? What do you wonder?
Student Response
For access, consult one of our IM Certified Partners.
Activity Synthesis
Select students to share things they have noticed and things they have wondered. Encourage students to think about the relationship with the dimension \(r\) and the volume \(V\) of each of these shapes. Ensure students understand that the two equations have no variable \(h\) for height since the \(h\) was replaced by \(r\) due to the height and radius being the same for both shapes.
19.2: Hemispheres in Boxes (15 minutes)
Activity
In this activity, students think about a hemisphere fitting inside the smallest possible box (or rectangular prism). Students reason that the smallest box in which a hemisphere can fit has a square base with edge length that is the same as the diameter of the hemisphere, and its height will be the radius of the hemisphere. Further, if we calculate the volume of the box, we have an upper bound for the volume of the hemisphere. This activity prepares students for the next, where they will calculate the upper and lower bounds for the volume of a hemisphere by considering cylinders and cones that fit outside or inside the hemisphere.
Launch
Ask students if they are familiar with the word sphere, and if they can think of examples of spheres. Some examples they might come up with are: ping pong balls, soap bubbles, baseballs, volleyballs, a globe. If students don’t come up with many examples, offer your own, or perhaps display an Internet image search for “sphere.” If you brought in any examples of physical spheres, display these or allow students to hold them. Explain to students that the radius of a sphere is the distance from the center of the sphere (the point in the exact middle) to any point on the sphere. If you brought in physical examples of spheres, ask students to rank them from smallest diameter to largest diameter.
Ask students if they are familiar with the word hemisphere, and if they can think of examples of hemispheres. Some examples are hemispherical (or ‘half’) balance boards, half of Earth, a dome or planetarium, and convex security mirrors. Display an example, illustration, or diagram of a hemisphere for all to see, pointing out how the radius is the distance from the center of the flat side of the sphere to any point on the curved surface of the sphere.
Tell students that in this activity, they will think about building a box (or rectangular prism) around a hemisphere. Arrange students in groups of 2. Ask students to quietly work through the first question and then share their reasoning with their partner. Pause for a whole-class discussion, and invite students to share their responses. The purpose of this problem is to see how the measurements of a hemisphere determine the dimensions of the prism. Ensure everyone understands the box must have edge lengths 6, 6, and 3 inches, and that the hemisphere must have a volume that is less than \(108\) cubic inches (since \(6^2\boldcdot 3=108\)). Discuss how much less students think the volume of the sphere is and their reasoning, then have students move onto the second problem with their partner.
Supports accessibility for: Memory; Language
Design Principle(s): Support sense-making
Student Facing
![A hemisphere with radius r sits snugly inside a box shaped like a rectangular prism.](https://cms-im.s3.amazonaws.com/LhKSsTi8kRc5LyiLiipEZmMm?response-content-disposition=inline%3B%20filename%3D%228-8.9.SphereRadiusR.png%22%3B%20filename%2A%3DUTF-8%27%278-8.9.SphereRadiusR.png&response-content-type=image%2Fpng&X-Amz-Algorithm=AWS4-HMAC-SHA256&X-Amz-Credential=AKIAXQCCIHWF3XOEFOW4%2F20240907%2Fus-east-1%2Fs3%2Faws4_request&X-Amz-Date=20240907T233633Z&X-Amz-Expires=604800&X-Amz-SignedHeaders=host&X-Amz-Signature=4e2af4ce52d5388a0531ba387b14fb9fb45df4a25b164bfd22706f901200a24f)
- Mai has a dome paperweight that she can use as a magnifier. The paperweight is shaped like a hemisphere made of solid glass, so she wants to design a box to keep it in so it won't get broken. Her paperweight has a radius of 3 cm.
- What should the dimensions of the inside of box be so the box is as small as possible?
- What is the volume of the box?
- What is a reasonable estimate for the volume of the paperweight?
- Tyler has a different box with side lengths that are twice as long as the sides of Mai's box. Tyler's box is just large enough to hold a different glass paperweight.
- What is the volume of the new box?
- What is a reasonable estimate for the volume of this glass paperweight?
- How many times bigger do you think the volume of the paperweight in this box is than the volume of Mai's paperweight? Explain your thinking.
Student Response
For access, consult one of our IM Certified Partners.
Anticipated Misconceptions
Students may believe that since this box is not a cube, but a rectangular prism, the volume will not increase by 8 when the side lengths are doubled. Since volume of a rectangular prism is the product of length, width, and height, the side lengths do not need to be equal; if every side length is doubled, then the volume will be increased by \(2 \boldcdot 2 \boldcdot 2 = 8\).
Activity Synthesis
Invite students to share their answers to the second question. The purpose of this discussion is for students to see that since a rectangular prism’s volume gets larger by a factor of \(2^3\) when edge lengths are doubled then it would make sense for a hemisphere’s volume to do the same. Ensure that students understand the volume calculated for the box holding the hemisphere is greater than the actual volume of the hemisphere because of the space left around the paperweight in the box and that this will be the upper bound of the volume of a sphere.
Tell students that in the next activity they are going to investigate a better way to estimate the volume of a hemisphere. If time allows, ask students to suggest shapes they are already familiar with that they could use to find the volume of a hemisphere.
19.3: Estimating Hemispheres (15 minutes)
Activity
In this activity, students use different solid figures to estimate an upper and lower bound for the volume of a hemisphere. For the upper bound, the hemisphere fits snugly inside a cylinder whose height and radius are equal to the radius of the hemisphere. For the lower bound, the cone fits snugly inside the hemisphere, and its radius and height also equal the radius and height of the hemisphere.
Select students who use the reasoning from the first problem to assist them in answering the second problem to share during the Activity Synthesis. For example, since the cylinder and cone have the same dimensions, the volume of the cone must be \(\frac13\) that of the cylinder.
Launch
Keep students in the same groups. Display the image of a hemisphere of radius 5 that fits snugly inside a cylinder of the same radius and height.
![A hemisphere fits snugly in a cylinder. The hemisphere has radius 5 and the cylinder has height 5.](https://cms-im.s3.amazonaws.com/kWHTKhhRpZUHN3MYmuht7nz5?response-content-disposition=inline%3B%20filename%3D%228-8.5.HemisphereInCylinder.png%22%3B%20filename%2A%3DUTF-8%27%278-8.5.HemisphereInCylinder.png&response-content-type=image%2Fpng&X-Amz-Algorithm=AWS4-HMAC-SHA256&X-Amz-Credential=AKIAXQCCIHWF3XOEFOW4%2F20240907%2Fus-east-1%2Fs3%2Faws4_request&X-Amz-Date=20240907T233633Z&X-Amz-Expires=604800&X-Amz-SignedHeaders=host&X-Amz-Signature=5ad9994c1508e785bcbfa4d3dea74401da71611250e8c3bab4407b2b3acf200d)
Ask “How does the hemisphere affect the height of the cylinder?” and then give students one minute of quiet think time, then one minute to discuss their response with a partner. Ask partners to share their responses. If not mentioned by students, point out that the height of the cylinder is equal to the radius of the hemisphere.
Give students work time for the activity followed by a whole-class discussion.
Supports accessibility for: Visual-spatial processing; Conceptual processing
Student Facing
- A hemisphere with radius 5 units fits snugly into a cylinder of the same radius and height.
- Calculate the volume of the cylinder.
- Estimate the volume of the hemisphere. Explain your reasoning.
- A cone fits snugly inside a hemisphere, and they share a radius of 5.
- What is the volume of the cone?
- Estimate the volume of the hemisphere. Explain your reasoning.
- Compare your estimate for the hemisphere with the cone inside to your estimate of the hemisphere inside the cylinder. How do they compare to the volumes of the cylinder and the cone?
Student Response
For access, consult one of our IM Certified Partners.
Student Facing
Are you ready for more?
Estimate what fraction of the volume of the cube is occupied by the pyramid that shares the base and a top vertex with the cube, as in the figure.
![A cube. A slice is made from a top corner to two bottom opposite corners forming a pyramid with a square base.](https://cms-im.s3.amazonaws.com/1W5LQ9EBVRQmC7QoBoq57WyX?response-content-disposition=inline%3B%20filename%3D%228-8.5.E19.Image.Revision.19.png%22%3B%20filename%2A%3DUTF-8%27%278-8.5.E19.Image.Revision.19.png&response-content-type=image%2Fpng&X-Amz-Algorithm=AWS4-HMAC-SHA256&X-Amz-Credential=AKIAXQCCIHWF3XOEFOW4%2F20240907%2Fus-east-1%2Fs3%2Faws4_request&X-Amz-Date=20240907T233633Z&X-Amz-Expires=604800&X-Amz-SignedHeaders=host&X-Amz-Signature=a5f0dcdfab89b73f3440d99c04e793c4b051420e30725d59b06786696509d401)
Student Response
For access, consult one of our IM Certified Partners.
Anticipated Misconceptions
Students may not realize that the radius of the hemisphere determines the height of the cone and cylinder. If students struggle with identifying the needed information, remind them of the previous activity in which the radius of the cylinder was also the height of the box (rectangular prism). Ask students how that can help them determine the dimensions of the cone or cylinder.
Activity Synthesis
Ask previously identified students to share their responses to the first two problems. Draw attention to any connections made between the two problems.
The purpose of this discussion is for students to recognize how the upper and lower bounds for the volume of a hemisphere are established by the cylinder and cone. Consider asking the following question:
- “What do the volumes of the cone and cylinder tell us about the volume of the hemisphere?” (The volume of the hemisphere has to be between the two volumes.)
- “Did anyone revise their original estimate for the hemisphere based on the calculation of the volume of the cone?” (Answers vary. If students estimated low values after calculating the volume of a cylinder, some may have needed to adjust their estimates.)
- “Compare the equations for volume of a cylinder and cone where radius and height are equal. If the volume of the hemisphere has to be between these two, what might an equation for the volume of a hemisphere look like?” (Cylinder volume: \(V=\pi r^3\); Cone volume: \(V= \frac 13 \pi r^3\). A possible hemisphere volume might be the average of these two, or \(V= \frac 23 \pi r^3\).)
Design Principle(s): Optimize output (for explanation)
Lesson Synthesis
Lesson Synthesis
To have students describe some of the important highlights of the lesson, ask:
- “How did we use a cylinder to estimate a volume of a hemisphere that is an overestimate?”
- “How did we use a cone to estimate a volume of a hemisphere that is an underestimate?”
- “How did we get a closer estimate for the volume of a hemisphere?”
- “How did we use today’s work to estimate the volume of a sphere?”
Explain to students that we used figures we know how to find the volume of (a cone and cylinder) to try and estimate the volume of a figure we do not know how to find the volume of (a sphere). In the next lesson, we will do something similar to learn how to find the volume of a sphere and see how close our reasoning today was to the actual calculation.
19.4: Cool-down - A Mirror Box (5 minutes)
Cool-Down
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Student Lesson Summary
Student Facing
We can estimate the volume of a hemisphere by comparing it to other shapes for which we know the volume. For example, a hemisphere of radius 1 unit fits inside a cylinder with a radius of 1 unit and height of 1 unit.
Since the hemisphere is inside the cylinder, it must have a smaller volume than the cylinder making the cylinder's volume a reasonable over-estimate for the volume of the hemisphere.
The volume of this particular cylinder is about 3.14 units3 since \(\pi(1)^2(1)=\pi\), so we know the volume of the hemisphere is less than 3.14 cubic units.
![A hemisphere sits snugly inside a cylinder. Each has radius 1, the cylinder has height 1.](https://cms-im.s3.amazonaws.com/yQyJAhZzdcMsmE4JxXaXP9Eg?response-content-disposition=inline%3B%20filename%3D%228-8.5.Radius1InCylinder.png%22%3B%20filename%2A%3DUTF-8%27%278-8.5.Radius1InCylinder.png&response-content-type=image%2Fpng&X-Amz-Algorithm=AWS4-HMAC-SHA256&X-Amz-Credential=AKIAXQCCIHWF3XOEFOW4%2F20240907%2Fus-east-1%2Fs3%2Faws4_request&X-Amz-Date=20240907T233633Z&X-Amz-Expires=604800&X-Amz-SignedHeaders=host&X-Amz-Signature=06a4fd3c9ce209ac040ef017a7b596e70d8fa9d0ff51207723c7f28e74348579)
Using similar logic, a cone of radius 1 unit and height 1 unit fits inside of the hemisphere of radius 1 unit.
![A cone fits snugly inside a hemisphere. The base of the cone is the flat surface of the hemisphere. The height of the cone is the radius of the hemisphere, which is 1 unit.](https://cms-im.s3.amazonaws.com/iKZ45b4Tuu52prgnkMULMRw6?response-content-disposition=inline%3B%20filename%3D%228-8.5.ConeInHemisphere4.png%22%3B%20filename%2A%3DUTF-8%27%278-8.5.ConeInHemisphere4.png&response-content-type=image%2Fpng&X-Amz-Algorithm=AWS4-HMAC-SHA256&X-Amz-Credential=AKIAXQCCIHWF3XOEFOW4%2F20240907%2Fus-east-1%2Fs3%2Faws4_request&X-Amz-Date=20240907T233633Z&X-Amz-Expires=604800&X-Amz-SignedHeaders=host&X-Amz-Signature=4c440a2d12a0959a5a74d542f7d94a151d99c8f6c46b4773ac04432d4e48ae48)
Since the cone is inside the hemisphere, the cone must have a smaller volume than the hemisphere making the cone's volume a reasonable under-estimate for the volume of the hemisphere.
The volume of this particular cone is about 1.05 units3 since \(\frac13 \pi(1)^2(1)=\frac13 \pi \approx 1.05\), so we know the volume of the hemisphere is more than 1.05 cubic units.
Averaging the volumes of the cylinder and the cone, we can estimate the volume of the hemisphere to be about 2.10 units3 since \(\frac {3.14+1.05}2 \approx 2.10\). And, since a hemisphere is half of a sphere, we can also estimate that a sphere with radius of 1 would be double this volume, or about 4.20 units3.