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10. Modeling in 2 & 3 Dimensions
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Geometry

10.05 Spheres

Lesson
Practice

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.

A sphere. The radius is labeled.

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

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.

Unlike the solids we have seen so far, we cannot unwrap a sphere to get a 2D net to calculate its area, so the surface area of a sphere is approached in a different way.

A cylinder showing radius r and height h which is equal to 2 r.

Archimedes discovered that the area of the curved face of a cylinder is equal to the surface area of the sphere.

We saw that the curved surface of a cylinder flattens out into a rectangle. Since the height of this cylinder is twice the radius of the inscribed sphere, which is also the radius of the circular ends, the area of the resulting rectangle is 2 \pi r \cdot 2 r = 4 \pi r^{2}.

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

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

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^{2}Surface area of a sphere
\displaystyle =\displaystyle 4 \pi \left(3\right)^{2}Substitute the radius
\displaystyle =\displaystyle 36 \piSimplify

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

Reflect and check

We found the exact surface area of the sphere in terms of pi. If we multiply 36 by \pi and round the result, we would have an approximation of the surface area which is less exact. The approximate surface of the sphere is about 113.1\text{ cm}^2.

Example 2

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

A cone with a hemisphere on top that resembles an icecream. The height of the cone is 15cm and the diameter is 8cm.
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 \frac{1}{2} \left(\frac{4}{3} \pi r^{3} \right)Volume of a hemisphere
\displaystyle =\displaystyle \frac{1}{2} \left(\frac{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 \frac{1}{3}BhVolume of a cone
\displaystyle =\displaystyle \frac{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 \frac{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[\frac{1}{2} \left(\frac{4}{3} \pi \left(4 \right)^{3} \right) \right] + \left[ \frac{1}{3} \left( \left(\pi \cdot 4^{2} \right) 15 \right) \right] + \left[\frac{1}{2} \left(\frac{4}{3} \pi \left(4 \right)^{3} \right) \right] \approx 519.40\,999

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.

Example 3

A sphere has a radius that is r\text{ cm} long and a volume of \dfrac{512\pi}{3}\text{ cm}^{3}.

Find the radius of the sphere. Round your answer to two decimal places.

Worked Solution
Create a strategy

Substitute the given volume into the volume formula V=\dfrac{4}{3}\pi r^{3} and solve for the radius.

Apply the idea
\displaystyle V\displaystyle =\displaystyle \frac{4}{3}\pi r^{3}Start with the formula for the volume of a sphere
\displaystyle \frac{512\pi}{3}\displaystyle =\displaystyle \frac{4}{3}\pi r^{3}Substitute the volume
\displaystyle 512\pi\displaystyle =\displaystyle 4\pi r^{3}Multiply both sides by 3
\displaystyle \frac{512\pi}{4\pi}\displaystyle =\displaystyle r^{3}Divide both sides by 4\pi
\displaystyle r^{3} \displaystyle =\displaystyle 128Simplify
\displaystyle r\displaystyle =\displaystyle \sqrt[3]{128}Take the cube root of both sides
\displaystyle \approx\displaystyle 5.04\text{ cm}Evaluate and round

Example 4

A spherical lamp base, initially designed with a diameter of 10 inches, needs to be resized to fit a new space. Its diameter needs to be reduced by 20 percent.

a

What is the diameter of the lamp's base after the redesign?

Worked Solution
Create a strategy

To find the new diameter, we will calculate 20 percent of the original diameter and subtract it from the original diameter.

Apply the idea

The original diameter is 10 inches. Reducing this by 20 percent, we calculate the reduction as 10 \cdot 20\% = 2Thus, the new diameter is 10 - 2 = 8 inches.

Reflect and check

An alternative way to find the 20\% decrease is by multiplying the original diameter by 80\%.100\% \cdot 10 - 20\% \cdot 10 = 80\% \cdot 10This allows us to find the new diameter in a single step:0.8\left(10\right)=8\text{ in}

b

By what percentage does the surface area decrease due to the design change?

Worked Solution
Create a strategy

We need to calculate the surface area of the spherical lamp base before and after the redesign using the formula SA=4\pi r^{2}, where r is the radius of the spherical lamp.

Since we are given the diameter, we need to find the radius by dividing the diameter by two.

Dividing the difference between the surface areas by the surface area before the redesign, and then multiplying by 100\% would give the percentage decrease.

\text{Percentage decrease}=\left(\frac{SA_{\text{before}}-SA_{\text{after}} }{SA_{\text{before}}}\right)\cdot 100\%

Apply the idea

Let's calculate the surface area before the redesign:

Given the diameter of 10 \text{ in}, the radius is 10 \cdot \dfrac{1}{2}=5.

\displaystyle SA_{\text{before}}\displaystyle =\displaystyle 4\pi \left(5\right)^{2}Substitute r=5
\displaystyle =\displaystyle 314.16 \text{ in}^{2}Evaluate

Let's calculate the surface area after the redesign:

Given the diameter after the redesign in part (a), the radius is 8\cdot \dfrac{1}{2}=4.

\displaystyle SA_{\text{after}}\displaystyle =\displaystyle 4\pi \left(4\right)^{2}Substitute r=4
\displaystyle =\displaystyle 201.06 \text{ in}^{2}Evaluate

Now, let's calculate the percentage decrease:

\displaystyle \text{Percentage decrease}\displaystyle =\displaystyle \left(\frac{SA_{\text{before}}-SA_{\text{after}} }{SA_{\text{before}}}\right)\cdot 100\%
\displaystyle =\displaystyle \left(\frac{314.16 \text{ in}^{2}-201.06\text{ in}^{2}}{314.16\text{ in}^{2}}\right)\cdot 100\%
\displaystyle =\displaystyle \left(\frac{113.1\text{ in}^{2}}{314.16\text{ in}^{2}}\right)\cdot 100\%
\displaystyle =\displaystyle 0.36\cdot 100\%
\displaystyle =\displaystyle 36\%

This shows that decreasing the diameter by 20\% will lead to a 36\% decrease in the surface area.

c

By what percentage does the volume decrease due to the design change?

Worked Solution
Create a strategy

We need to calculate the volume of the spherical lamp base before and after the redesign using the formula V=\dfrac{4}{3}\pi r^{3}, where r is the radius of the spherical lamp.

Dividing the difference between the volumes by the volume before the redesign, and then multiplying by 100\% would give the percentage decrease.

\text{Percentage decrease}=\left(\frac{V_{\text{before}}-V_{\text{after}} }{V_{\text{before}}}\right)\cdot 100\%

Apply the idea

Let's calculate the volume before the redesign:

The radius is 5\text{ in}, as calculated in part (b).

\displaystyle VA_{\text{before}}\displaystyle =\displaystyle \frac{4}{3}\pi \left(5\right)^{3}Substitute r=5
\displaystyle =\displaystyle 523.6 \text{ in}^{3}Evaluate

Let's calculate the volume after the redesign:

The radius of the new design is 4\text{ in}, as calculated in part (b).

\displaystyle V_{\text{after}}\displaystyle =\displaystyle \frac{4}{3}\pi \left(4\right)^{3}Substitute r=4
\displaystyle =\displaystyle 268.08 \text{ in}^{3}Evaluate

Now, let's calculate the percentage decrease:

\displaystyle \text{Percentage decrease}\displaystyle =\displaystyle \left(\frac{V_{\text{before}}-V_{\text{after}} }{V_{\text{before}}}\right)\cdot 100\%
\displaystyle =\displaystyle \left(\frac{523.6 \text{ in}^{3}-268.08\text{ in}^{3}}{523.6\text{ in}^{3}}\right)\cdot 100\%
\displaystyle =\displaystyle \left(\frac{255.52\text{ in}^{3}}{523.6\text{ in}^{3}}\right)\cdot 100\%
\displaystyle =\displaystyle 0.4880\cdot 100\%
\displaystyle =\displaystyle 48.80\%

This means a 20\% decrease in the diameter leads to a 48.8\% decrease in the volume.

Example 5

Soledad has two spheres. The volume of the larger sphere is 4 times the volume of the smaller sphere. How much greater is the radius of the larger sphere than the smaller sphere?

Worked Solution
Create a strategy

We can set up an equation given the following:

  • The volume of a sphere is given by V=\dfrac{4}{3} \pi r^{3}.

  • The new sphere has 4 times the volume of the original sphere.

We will let the volume and radius of the larger sphere be V_{\text{L}} and r_{\text{L}} respectively. For the smaller sphere, the volume and radius will be V_{\text{S}} and r_{\text{S}}. Using this information, we can use {V_{\text{L}}=4\cdot V_{\text{S}}} or \dfrac{4}{3}\pi r_{\text{L}}^3 = 4 \cdot\left(\dfrac{4}{3}\pi r_{\text{S}}^3\right) Then, we can simplify it to find the relationship between their radii.

Apply the idea
\displaystyle V_{\text{L}}\displaystyle =\displaystyle 4\cdot V_{\text{S}}Set up the equation
\displaystyle \frac{4}{3}\pi r_{\text{L}}^{3}\displaystyle =\displaystyle 4\cdot \left(\frac{4}{3}\pi r_{\text{S}}^{3}\right)Substitute the volumes
\displaystyle r_{\text{L}}^{3}\displaystyle =\displaystyle 4\cdot r_{\text{S}}^{3}Divide both sides by \dfrac{4}{3} \pi
\displaystyle \sqrt[3]{r_{\text{L}}^{3}}\displaystyle =\displaystyle \sqrt[3]{4\cdot r_{\text{S}}^{3}}Take the cube root of both sides
\displaystyle r_{\text{L}}\displaystyle =\displaystyle r_{\text{S}}\sqrt[3]{4}Simplify

This means that the radius of the larger sphere is \sqrt[3]{4} times radius of the smaller sphere.

Reflect and check

To verify our answer, we can choose specific values for the radii of each sphere and see if the volume of the larger sphere is 4 times the volume of the smaller sphere. Let's set the radius of the smaller sphere to 2\text{ in}, which would make the radius of the larger sphere be 2\sqrt[3]{4}\text{ in}.

The volume of the smaller sphere is:

\displaystyle V_{\text{S}}\displaystyle =\displaystyle \frac{4}{3}\pi \left(2\right)^{3}Substitute r=2
\displaystyle =\displaystyle 33.51\text{ in}^{3}Simplify and round to two decimal places

The volume of the larger sphere is:

\displaystyle V_{\text{L}}\displaystyle =\displaystyle \frac{4}{3}\pi \left(2\sqrt[3]{4}\right)^{3}Substitute r=2\sqrt[3]{4}
\displaystyle =\displaystyle 134.04 \text{ in}^{3}Simplify and round to two decimal places

Then divide the volume of the larger sphere by volume of the smaller sphere:

\frac{134.04 \text{ in}^{3}}{33.51\text{ in}^{3}}=4

As shown, the volume of the larger sphere is 4 times the volume of the smaller sphere. This verifies our answer.

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.DF.1

The student will create models and solve problems, including those in context, involving surface area and volume of rectangular and triangular prisms, cylinders, cones, pyramids, and spheres.

G.DF.1b

Create models and solve problems, including those in context, involving surface area of threedimensional figures, as well as composite three-dimensional figures.

G.DF.1c

Solve multistep problems, including those in context, involving volume of three-dimensional figures, as well as composite three-dimensional figures.

G.DF.1d

Determine unknown measurements of three-dimensional figures using information such as length of a side, area of a face, or volume.

G.DF.2

The student will determine the effect of changing one or more dimensions of a three-dimensional geometric figure and describe the relationship between the original and changed figure.

G.DF.2a

Describe how changes in one or more dimensions of a figure affect other derived measures (perimeter, area, total surface area, and volume) of the figure.

G.DF.2b

Describe how changes in surface area and/or volume of a figure affect the measures of one or more dimensions of the figure.

G.DF.2c

Solve problems, including those in context, involving changing the dimensions or derived measures of a three-dimensional figure.

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