Topics
11. Modeling in Two & Three Dimensions
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United States of AmericaFL
Grade 912

11.06 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

MA.912.GR.4.5

Solve mathematical and real-world problems involving the volume of three-dimensional figures limited to cylinders, pyramids, prisms, cones and spheres.

MA.912.GR.4.6

Solve mathematical and real-world problems involving the surface area of three-dimensional figures limited to cylinders, pyramids, prisms, cones and spheres.

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