Topics
11. Modeling in Two and Three dimensions
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United States of America
Grades 9-12

11.04 Spheres

Lesson
Practice

Introduction

We were introduced to the formula for calculating the volume of a sphere in 8th grade, and this lesson continues using that formula. We will also find the surface area of a sphere and apply these formulas to solve problems in this lesson.

Ideas

Spheres

Another common type of solid is the sphere. Recall that a sphere is a solid consisting of all points at a fixed distance from a central point.

The central point of a sphere is often just called its center. The distance of each point on the sphere from the center is called the radius, just like for a circle.

A sphere has many special properties resulting from its symmetry.

A diagram of a spherical object similar to a ball drawn with its cross section, a circle.

A cross section of a sphere will always be a circle, no matter where the slice is made. The only change will be the size of the circle.

A two quadrant coordinate plane with no scales on its axes showing only quadrant 1 and 4. A sphere is plotted as a result of rotation of a circle about its x-axis.

A sphere can be formed as a volume of revolution of a circle centered along the axis of revolution.

In the image shown, the circle is rotated about the x-axis to form a sphere.

We can calculate the volume of a sphere using the formula:

\displaystyle V = \dfrac{4}{3} \pi r^3
\bm{r}
the radius of the sphere

We can calculate the surface area of a sphere using the formula:

\displaystyle SA = 4\pi r^2
\bm{r}
the radius of the sphere

Examples

Example 1

Find the surface area of a sphere with radius 3 \text{ cm}.

Worked Solution
Create a strategy

Substitute the given radius into the formula for the surface area of a sphere.

Apply the idea
\displaystyle SA\displaystyle =\displaystyle 4 \pi r^2Surface area of a sphere
\displaystyle =\displaystyle 4 \pi (3)^2Substitute the radius
\displaystyle =\displaystyle 36 \piSimplify

So the surface area of the sphere is 36 \pi \text{ cm}^2.

Example 2

Find the density of a snowball with radius 4 \text{ in} and mass \frac{1}{2} \text{ lb}. Round your answer to three decimal places.

Worked Solution
Create a strategy

We can first find the volume of the snowball, using the formula V = \dfrac{4}{3} \pi r^3, where r is its radius.

Then we can calculate the density by dividing the given mass by the volume.

Apply the idea

Finding the volume of the snowball, we have:

\displaystyle V\displaystyle =\displaystyle \dfrac{4}{3}\pi r^3Volume of a sphere
\displaystyle =\displaystyle \dfrac{4}{3} \pi (4)^3Substitute the radius
\displaystyle =\displaystyle \dfrac{256 \pi}{3}Evaluate the exponent and multiplication

So the volume of the snowball is \dfrac{256 \pi}{3}\text{ in}^3.

We can now use this to calculate the density:

\displaystyle \text{Density}\displaystyle =\displaystyle \dfrac{\text{Mass}}{\text{Volume}}Density formula
\displaystyle =\displaystyle \dfrac{\frac{1}{2}}{\left(\dfrac{256\pi}{3}\right)}Substitute the mass and volume
\displaystyle =\displaystyle \dfrac{3}{512\pi}Evaluate the division
\displaystyle {}\displaystyle \approx\displaystyle 0.001865\ldots

So the density of the snowball is 0.002 \, \text{lb}/\text{in}^3, to three decimal places.

Example 3

The ice cream cones at an ice creamery have the dimensions indicated in the diagram:

a

Given that 1 cubic centimeter is equivalent to 1 milliliter, how many milliliters of ice cream can fit in each cone, including the hemisphere scoop on top? Round your answer to the nearest milliliter.

Worked Solution
Create a strategy

Calculate the volume of the hemisphere and the cone, then find their sum. The radius of the hemisphere and cone is 4 \text{ cm}.

Apply the idea

A hemisphere is half a sphere, so we can calculate half the volume of a sphere to find the amount of ice cream on top of the cone as follows:

\displaystyle V\displaystyle =\displaystyle \dfrac{1}{2} \left(\dfrac{4}{3} \pi r^3 \right)Volume of a hemisphere
\displaystyle =\displaystyle \dfrac{1}{2} \left(\dfrac{4}{3} \pi \left(4 \right)^3 \right)Substitute the radius
\displaystyle \approx\displaystyle 134.04Evaluate the exponent and multiplication

The volume of the ice cream scoop is 134.04 \text{ cm}^3, or 134.04 \text{ mL}.

Now, we can calculate the volume of the ice cream that fills the cone as follows:

\displaystyle V\displaystyle =\displaystyle \dfrac{1}{3}BhVolume of a cone
\displaystyle =\displaystyle \dfrac{1}{3} \left( \left(\pi \cdot 4^2 \right) 15 \right)Substitute the area of the base and height
\displaystyle \approx\displaystyle 251.33Evaluate the exponent and multiplication

Since the volume of the ice cream in the cone is 251.33 \text{ cm}^3 or 251.33 \text{ mL} and the ice cream on top is 134.04 \text{ mL}, the amount of ice cream that can fit in each cone is 251.33 \text{ mL} + 134.04 \text{ mL} \approx 385 \text{ mL}.

b

The ice cream is bought in 10 \text{ L} tubs. How many whole cones can be made with a single tub of ice cream?

Worked Solution
Create a strategy

Convert 10 \text{ L} to milliliters by multiplying by \dfrac{1\,000 \text{ mL}}{1 \text{ L}}, then divide the amount by 385 \text{ mL}.

Apply the idea

Since 10 \text{ L} is equivalent to 10 \,000 \text{ mL}, we can find the number of ice cream cones that a tub of ice cream makes as follows:10\,000 \text{ mL} \div \dfrac{385 \text{ mL}}{1 \text{ cone}} \approx 25.97 \text{ cones}

A tub of ice cream can make 25 cones.

Reflect and check

While 25.97 could be rounded up in a non-contextual problem, the tub of ice cream will not actually have enough to make the next cone with exactly 385 \text{ mL}, so we state that there is enough ice cream to make 25 cones.

c

Double cones are served with a second hemispherical scoop of the same dimensions as the first scoop. How many double cones can be made with from a 10 \text{ L} tub?

Worked Solution
Create a strategy

We will need to calculate the amount of ice cream needed for a double cone, then divide the amount of ice cream in a tub by the amount of ice cream needed per double scoop cone.

Apply the idea

Since the amount of ice cream in an ice cream cone is approximately 385.37 \text{ mL}, by adding another scoop, the amount of ice cream needed for a double cone is 385.37 \text{ mL} + 134.04 \text{ mL}\approx 519 \text{ mL}.

We can find the number of double scoop ice cream cones that a tub of ice cream makes as follows: 10\,000 \text{ mL} \div \dfrac{519 \text{ mL}}{1 \text{ cone}} \approx 19.27 \text{ cones}

A tub of ice cream can make 19 double scoop ice cream cones.

Reflect and check

Note that as we calculated the math further, we used the milliliters of ice cream that were rounded to two decimal places. The most accurate way to add the volume of the next scoop would be to keep the calculations in their original forms for as long as possible before rounding, like this:\left[\dfrac{1}{2} \left(\dfrac{4}{3} \pi \left(4 \right)^3 \right) \right] + \left[ \dfrac{1}{3} \left( \left(\pi \cdot 4^2 \right) 15 \right) \right] + \left[\dfrac{1}{2} \left(\dfrac{4}{3} \pi \left(4 \right)^3 \right) \right] \approx 519.40999

Our calculations yielded the same amount of ice cream in a double cone, because we only needed accuracy to the nearest integer. While performing calculations, we need to keep in mind the degree of accuracy needed for a specific context.

Idea summary

We can calculate the volume of a sphere using the formula:

\displaystyle V = \dfrac{4}{3} \pi r^3
\bm{r}
the radius of the sphere

We can calculate the surface area of a sphere using the formula:

\displaystyle SA = 4\pi r^2
\bm{r}
the radius of the sphere

Outcomes

G.GMD.A.3

Use volume formulas for cylinders, pyramids, cones, and spheres to solve problems.

G.MG.A.1

Use geometric shapes, their measures, and their properties to describe objects.

G.MG.A.2

Apply concepts of density based on area and volume in modeling situations.

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