A solid consists of a cone on top of a cylinder with a radius equal to that of the cone. The height of the cone is 66 and the height of the cylinder is 5 . If the volume of the solid is 64 pi, what is the area of the base of the cylinder?

2 Answers
Jun 6, 2018

A=64/27pi u^2

Explanation:

The volume of the cone is given by: v=1/3pir^2h
Since the height of the cone is 66, then h=66
So, v=1/3pir^2times66=22pir^2

The volume of a cylinder is given by: v=pir^2h
Since the height of the cylinder is 5, then h=5
So, v=pir^2h = pir^2times5=5pir^2

The total volume of the solid is 64pi
Therefore, 22pir^2+5pir^2=64pi

27pir^2=64pi
r^2=(64pi)/(27pi)
r^2=64/27
r=+-8/(3sqrt3)
Since r is the radius, it must be have the restriction: r>0
Therefore, r=8/(3sqrt3)units

To find the base of the cylinder, we need to know that the base is a circle. The area of a circle is given by A=pir^2 = pitimes(8/(3sqrt3))^2=64/27pi u^2

Jun 6, 2018

The area of the base of the cylinder is: A=pir^2=(64pi)/27

Explanation:

The area of the base we need to find is: A=pir^2, where r is the radius of the cylinder.

The volume of the cylinder is: pir^2*h_1
where h_1 is the height of the cylinder.
The volume of the cone is pir^2*h_2/3
where h_2 is the height of the cone.

The volume of the solid is the sum of those two volumes: V=pir^2*h_1 + pir^2*h_2/3
Factoring pir^2:
V= pir^2(h_1+h_2/3)
64pi = pir^2(5+66/3) = pir^2(5+22) = pir^2(27)

64pi = 27pir^2

pir^2 = (64pi)/27
And that is the area of the base: A=pir^2=(64pi)/27