CK-12 Geometry Concepts

http://www.ck12.org Chapter 11. Surface Area and Volume

The circumference is referring to the circumference of a great circle. UseC= 2 πr.

2 πr= 26 π

r= 13 f t.

Example B

Find the surface area of a sphere with a radius of 14 feet.

Use the formula,r= 14 f t.

SA= 4 π( 14 )^2

= 784 πf t^2

Example C

Find the volume of a sphere with a radius of 6 m.

Use the formula for the volume of a sphere.

V=

4

3

π 63

=

4

3

π( 216 )

= 288 π

Watch this video for help with the Examples above.

MEDIA

Click image to the left for more content.

CK-12 Foundation: Chapter11SpheresB

Concept Problem Revisited

If the maximum circumference of a bowling ball is 27 inches, then the maximum radius would be 27= 2 πr, or
r= 4 .30 inches. Therefore, the surface area would be 4π 4. 32 ≈ 232. 35 in^2 , and the volume would be^43 π 4. 33 ≈
333. 04 in^3. The weight of the bowling ball refers to its density, how heavy something is. The volume of the ball tells
us how much it can hold.

Vocabulary

Asphereis the set of all points, in three-dimensional space, which are equidistant from a point. Theradiushas one
endpoint on the sphere and the other endpoint at the center of that sphere. Thediameterof a sphere must contain
the center. Thegreat circleis a plane that contains the diameter. Ahemisphereis half of a sphere.