#### Question

A solid toy is in the form of a hemisphere surmounted by a right circular cone. height of the cone is 2 cm and the diameter of the base is 4 cm. If a right circular cylinder circumscribes the toy, find how much more space it will cover.

#### Solution

Let r_{1} r_{2} and r_{3} cm denate the radii of the base of the cylinder, cone and hemisphere respectively. Then,

r_{1} = r_{2} = r_{3} = `4/3` =2 cm

Let h_{1} and l cm be the heights of the cylinder and cone respectively. Then,

h = 4cm and l= 2cm

s0, the remaining volume of the cylinder when the toy is inserted into it.

= Volume of the cylinder -(volume of the cone + volume of the hemisphere)

`= pir_1^2h-(1/3pir_2^2l+2/3pir_3^3)`

`=pixx(2)^2xx4-(1/3pixx(2)^2xx2+2/3pixx(2)^3)`

`=16pi-((8pi)/3+(16pi)/3)`

`=16pi-((24pi)/3)`

=16π - 8π

= 8π cm^{3}