Question

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 `39` and the height of the cylinder is `17`. If the volume of the solid is `66 pi`, what is the area of the base of the cylinder?

Answer 1

`BA_(cyl) = pi(sqrt(11/5))^2 = 11/5pi`

Explanation:

Given: Volume cone and cylinder combined
`V_(Tot) = 66pi`
`Heights: H_(cyl) = 17; H_("cone")=39`
Required: Base Area of Cylinder, `BA_(cyl)`?
Solution:
`V_(Tot) = 66pi = pir^2*H_(cyl)+1/3pir^2*H_("cone")`
`66cancel(pi) = cancel(pi)r^2*17+1/3cancel(pi)r^2*39`
`66 = r^2(17+13) = 30r^2; r=sqrt(11/5)`

Now the base arear of the cylinder, `BA_(cyl) = pir^2`
`BA_(cyl) = pi(sqrt(11/5))^2 = 11/5pi`