CK-12 Geometry - Second Edition

11.6. Surface Area and Volume of Spheres http://www.ck12.org

V=

4

3

π 63

=

4

3

π( 216 )

= 288 π

Example 7:A sphere has a volume of 14137. 167 f t^3 , what is the radius?

Solution:Because we have a decimal, our radius might be an approximation. Plug in what you know to the formula
and solve forr.

14137. 167 =

4

3

πr^3

3

4 π

· 14137. 167 =r^3

3375 =r^3

At this point, you will need to take thecubed rootof 3375. Depending on your calculator, you can use the^3

√

x
function or∧

( 1

3

)

. The cubed root is the inverse of cubing a number, just like the square root is the inverse, or how
you undo, the square of a number.

√ 3

3375 = 15 =r The radius is 15f t.

Example 8:Find the volume of the following solid.

Solution:To find the volume of this solid, we need the volume of a cylinder and the volume of the hemisphere.

Vcylinder=π 62 ( 13 ) = 78 π

Vhemis phere=

1

2

(

4

3

π 63

)

= 36 π

Vtotal= 78 π+ 36 π= 114 πin^3

Know What? RevisitedIf the maximum circumference of a bowling ball is 27 inches, then the maximum radius
would be 27= 2 πr, orr= 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.