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5. Modeling in Two & Three Dimensions
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United States of AmericaTennessee
Integrated Math 3

5.06 Modeling with 3D figures

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Lesson

Concept summary

When solving real-world problems involving volume or surface area, it is useful to approximate the real-world object with some appropriate geometric solids. Usually, we will need to combine a few different basic solids together to properly approximate a real-world object.

An image of a water bottle and its approximation as a composition of basic solids: a small cylinder stacked on a truncated cone stacked on a large cylinder.

A solid formed by combining basic solids is called a composite solid.

Composite solid

A solid that is formed by adding (or subtracting) basic solids to each other.

Basic solids can include:

  • Cube
  • Rectangular prism
  • Triangular prism
  • Pyramid - base could be square, rectangular, triangular, etc
  • Cylinder
  • Cone
  • Sphere

By approximating real-world objects as composite solids, we can estimate their volume by adding or subtracting the volumes of the basic solids that make up the composite solid.

Similarly, we can estimate surface area by approximating real-world objects as composite solids and then adding or subtracting the area of 2D surfaces on the basic solids.

There are also cases where a real-world object is similar to two different basic solids, somewhere in between the two. In these cases, we can average the two approximations to better estimate the object.

Worked examples

Example 1

Describe how we could approximate the jack-in-a-box as a composite solid.

An image of a jack-in-a-box. Talk to your teacher for more information.

Approach

When approximating a real-world object as a composite solid, we want to keep the composite solid as simple as possible while representing the object properly. This is to balance accuracy and complexity.

We can see that the main parts of the jack-in-a-box are the box, the body of the jack, and the head of the jack. We could also consider the lid of the box and the arms and hat of the jack.

Solution

Approximating only the main parts of the jack-in-a-box, we can use a composite solid composed of a sphere, a cylinder, and a cube.

A composite solid made up of a sphere on top of a cylinder on top of a cube.

Reflection

If we wanted to include the other parts of the jack-in-a-box in our approximation, we could use a rectangular prism for the lid, cylinders and spheres for the arms, and cones and spheres for the hat.

However, while adding all these small parts increases the accuracy of the approximation, it also makes any calculations with the composite solid more complex.

When approximating real-world objects with composite solids, we can choose how accurate we want our approximation to match the problem that needs to be solved and how closely we want to estimate the solution.

Example 2

Nhat has a glass jar of gumballs. They measure that the jar has a height of 12 centimeters and a base diameter of 8 centimeters.

An image of a jar of gumballs. The jar is 12 centimeters high and has a base diameter of 8 centimeters. Talk to your teacher for more information.
a

If the glass has a thickness of 0.5 centimeters, estimate the volume of glass needed to make the jar.

Approach

We can approximate the volume of glass as a product of the surface area of the jar (minus the lid) and the thickness of the glass.

We can approximate the jar as a cylinder to find the required surface area. The cylinder will have a height of 12\text{ cm} and a diameter of 8\text{ cm} (which is a radius of 4\text{ cm}).

Solution

The surface area of a cylinder (minus its lid) is made up of the curved surface of the cylinder and the circular base.

\displaystyle SA\displaystyle =\displaystyle 2\pi rh+\pi r^2Formulas
\displaystyle =\displaystyle 2\pi \left(4\right)\left(12\right)+\pi \left(4\right)^2Substitute
\displaystyle =\displaystyle 96\pi + 16\piEvaluate
\displaystyle =\displaystyle 112\piSimplify

The surface area of the glass jar (minus the lid) is 112\pi\text{ cm}^2. We can multiply this by the thickness of the glass to find the volume.

\displaystyle \text{Volume}_{\text{glass}}\displaystyle =\displaystyle 112\pi \cdot 0.5
\displaystyle =\displaystyle 56\pi
\displaystyle \approx\displaystyle 175.93

The volume of glass needed to make the jar is approximately 175.93\text{ cm}^3.

Reflection

Another way to estimate the volume of glass is to consider subtracting one cylinder from another in order to get only the glass part of the jar.

This would involve finding the volume of the jar using the given dimensions, then subtracting the volume of a smaller cylinder with dimensions based on the thickness of the glass. The smaller cylinder would have a height of 11.5 centimeters and a base diameter of 7 centimeters to account for the 0.5 centimeters we are removing on each side, excluding the lid.

This approximation will not give the same exact answer as the approximation used in the example, but is just as valid for estimating the volume of the glass.

b

Nhat measures the gumballs to have diameters of 2 centimeters. If there are 30 gumballs in the jar, estimate the volume of air in the jar.

Approach

We can approximate the volume of air in the jar as a cylinder with lots of spherical holes in it.

Solution

We can approximate the volume of the jar using a cylinder, making sure to take into account the 0.5\text{cm} thickness of the glass from part

a
.

\displaystyle \text{Volume}_{\text{jar}}\displaystyle \approx\displaystyle \pi r^2 hFormula
\displaystyle =\displaystyle \pi\left(3.5\right)^2\left(11.5\right)Substitute
\displaystyle =\displaystyle 140.875\piEvaluate

We can approximate the volume of a gumball using a sphere.

\displaystyle \text{Volume}_{\text{gumball}}\displaystyle \approx\displaystyle \frac{4}{3}\pi r^3Formula
\displaystyle =\displaystyle \frac{4}{3}\pi \left(1\right)^3Substitute
\displaystyle =\displaystyle \frac{4}{3}\piEvaluate

We can consider the volume of the air in the jar as a composite solid composed by subtracting the gumballs from the jar.

\displaystyle \text{Volume}_{\text{air}}\displaystyle \approx\displaystyle \text{Volume}_{\text{jar}}-30\cdot\text{Volume}_{\text{gumball}}
\displaystyle =\displaystyle 140.875\pi-40\piSubstitute
\displaystyle =\displaystyle 100.875\piSimplify
\displaystyle \approx\displaystyle 316.91Evaluate

The volume of air in the jar is approximately 316.91\text{ cm}^3.

Outcomes

M3.N.Q.A.1.A

Choose and interpret the scale and the origin in graphs and data displays.

M3.N.Q.A.1.B

Use appropriate quantities in formulas, converting units as necessary.

M3.N.Q.A.1.C

Define and justify appropriate quantities within a context for the purpose of modeling.

M3.G.MG.A.1

Use geometric shapes, their measures, and their properties to model objects found in a real-world context for the purpose of approximating solutions to problems.*

M3.MP1

Make sense of problems and persevere in solving them.

M3.MP2

Reason abstractly and quantitatively.

M3.MP3

Construct viable arguments and critique the reasoning of others.

M3.MP4

Model with mathematics.

M3.MP5

Use appropriate tools strategically.

M3.MP6

Attend to precision.

M3.MP7

Look for and make use of structure.

M3.MP8

Look for and express regularity in repeated reasoning.

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