Topics
10. Modeling in Two & Three Dimensions
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United States of AmericaTennessee
Geometry

10.05 Spheres

Lesson
Practice
Lesson

Concept summary

Another common solid type of solid is the sphere.

Sphere

A solid consisting of all points at a fixed distance from a central point.

A three dimensional figure made from one curved surface. The distance from a central point to the surface is shown to be r.

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 centred 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

Worked examples

Example 1

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

Solution

We can substitute the given radius into the formula for the surface area of a sphere:

\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 snow ball with radius 4 \text{ in} and mass 64 \text{ lb}. Round your answer to two decimal places.

Approach

We can first find the volume of the snow ball, 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.

Solution

Finding the volume of the snow ball, 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}Simplify

So the volume of the snow ball 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{64}{\left(\dfrac{256\pi}{3}\right)}Substitute known values
\displaystyle =\displaystyle \dfrac{3}{4\pi}Simplify
\displaystyle {}\displaystyle \approx\displaystyle 0.23873\ldotsEvalute with a calculator

So the density of the snow ball is 0.24 \, \text{lb}/\text{in}^3, to two decimal places.

Outcomes

G.N.Q.A.1

Use units as a way to understand real-world problems.*

G.N.Q.A.1.A

Use appropriate quantities in formulas, converting units as necessary.

G.N.Q.A.1.C

Choose an appropriate level of accuracy when reporting quantities.

G.GMD.A.2

Use volume and surface area formulas for cylinders, cones, prisms, pyramids, and spheres to solve problems in a real-world context.*

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.*

G.MP1

Make sense of problems and persevere in solving them.

G.MP2

Reason abstractly and quantitatively.

G.MP3

Construct viable arguments and critique the reasoning of others.

G.MP4

Model with mathematics.

G.MP5

Use appropriate tools strategically.

G.MP6

Attend to precision.

G.MP7

Look for and make use of structure.

G.MP8

Look for and express regularity in repeated reasoning.

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