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G-GMD, G-MG Tennis Balls in a Can
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### Alignments to Content Standards

• Geometry
Domain
HSG-GMD: Geometric Measure and Dimension
Cluster
Visualize relationships between two-dimensional and three-dimensional objects
Standard
Identify the shapes of two-dimensional cross-sections of three-dimensional objects, and identify three-dimensional objects generated by rotations of two-dimensional objects.
• Geometry
Domain
HSG-MG: Modeling with Geometry
Cluster
Apply geometric concepts in modeling situations
Standard
Use geometric shapes, their measures, and their properties to describe objects (e.g., modeling a tree trunk or a human torso as a cylinder).

### Tags

Tags: MP 4

The official diameter of a tennis ball, as defined by the International Tennis Federation, is at least 2.575 inches and at most 2.700 inches. Tennis balls are sold in cylindrical containers that contain three balls each. To model the container and the balls in it, we will assume that the balls are 2.7 inches in diameter and that the container is a cylinder the interior of which measures 2.7 inches in diameter and $3 \times 2.7 = 8.1$ inches high.

1. Lying on its side, the container passes through an X-ray scanner in an airport. If the material of the container is opaque to X-rays, what outline will appear? With what dimensinos?

2. If the material of the container is partially opaque to X-rays and the material of the balls is completely opaque to X-rays, what will the outline look like (still assuming the can is lying on its side)?

3. The central axis of the container is a line that passes through the centers of the top and bottom. If one cuts the container and balls by a plane passing through the central axis, what does the intersection of the plane with the container and balls look like? (The intersection is also called a cross section. Imagine putting the cut surface on an ink pad and then stamping a piece of paper. The stamped image is a picture of the intersection.)

4. If the can is cut by a plane parallel to the central axis, but at a distance of 1 inch from the axis, what will the intersection of this plane with the container and balls look like?

5. If the can is cut by a plane parallel to one end of the can—a horizontal plane—what are the possible appearances of the intersections?

6. A cross-section by a horizontal plane at a height of $1.35 + w$ inches from the bottom is made, with $0 \lt w \lt 1.35$ (so the bottom ball is cut). What is the area of the portion of the cross section inside the container but outside the tennis ball?

7. Suppose the can is cut by a plane parallel to the central axis but at a distance of $w$ inches from the axis ($0\lt w\lt 1.35$). What fractional part of the cross section of the container is inside of a tennis ball?

### Commentary

This task is inspired by the derivation of the volume formula for the sphere. If a sphere of radius 1 is enclosed in a cylinder of radius 1 and height 2, then the volume not occupied by the sphere is equal to the volume of a “double-naped cone” with vertex at the center of the sphere and bases equal to the bases of the cylinder. This can be seen by slicing the figure parallel to the base of the cylinder and noting the areas of the annular slices consisting of portions of the volume that are inside the cylinder but outside the sphere are the same as the areas of the slices of the double-naped cone (and applying Cavalieri’s Principle). This almost magical fact about slices is a manifestation of Pythagorean Theorem. We see it at work in Part 6 of this task. The other parts of the task are exercises in 3D-visualization, which build up the spatial sense necessary to work on Part 6 with understanding. The visualization required here is used in calculus, in connection with procedures for calculating volumes by various slicing procedures.

Submitted by James Madden on behalf of the participants in the Louisiana Math and Science Teacher Institute On-Ramp.

#### Solutions

Solution: Solution
1. The shadow is a rectangle measuring 2.7 inches by 8.1 inches.

2. The shadow is a light rectangle (2.7 × 8.1 inches) with three disks inside. It looks like a traffic light:

3. The image is similar to the previous one, but now only the outlines are seen:

4. The intersection with the container is a narrower rectangle. The intersections with the balls are smaller circles. Because each ball touches the container along its whole “equator,” the circles must touch the long sides of the rectangle:

5. The intersections are two concentric circles, except when $w = 0, 2.7, 5.4, 8.1$ and when $w = 1.35, 4.05, 6.75$. In the former case, we see a circle (from the container) and a point (where the plane touches a sphere). In the latter case, we see a single circle corresponding to a place where the equator of a ball touches the container.

6. The intersection of the plane with the interior of the container is a disk of radius 1.35 inches. Its area is $\pi (1.35)^2 \text{ in}^2$. The intersection with the ball is a smaller disk that is contained in the first disk. The radius $r$ of the smaller disk is the square root of $(1.35)^2 - w^2$, as we see from the diagram below depicting the intersection of a plane through the central axis of the container with the bottom ball. Thus, the area of the smaller disk is $\pi ( (1.35)^2 - w^2 )$. Accordingly, the area inside the larger disk but outside the smaller is $\pi w^2$, provided that $0 \leq w \leq 1.35$. (It is notable that the radius of the ball does not appear explicitly in the expression for this annular area.)

7. Referring to Problem d), we see that we wish to find the ratio of the total area of three congruent disks to the area of a rectangle, one of whose dimensions is equal to the diameter of the disks. The same picture used in the previous problem, but interpreted as a view from one end of the container, gives us the radius of the small disks — namely, $\sqrt{(1.35)^2 - w^2}$, so the total area of the disks is $3 \pi ( (1.35)^2 - w^2 )$. The area of the rectangle is $(8.1) 2 \sqrt{(1.35)^2 - w^2}$. So, the ratio is

$$\frac{3 \pi ( (1.35)^2 - w^2 )}{(8.1) 2 \sqrt{(1.35)^2 – w^2}} = \frac{\pi \sqrt{ (1.35)^2 - w^2}}{5.4}$$