# Volume of a Combination of Solids

Unlike the surface area of the combination of solids, while calculating the volume of the combined solids we need to find the volume of each solid and add them together.

Given below are few example problems based on finding the volume of the combined solids.

Â

Circus tent has cylindrical shape surmounted by a conical roof. The radius of the cylindrical base is 20m. The heights of the cylindrical and conicalÂ portions are 4.2 m and 2.1m, respectively. Find the volume of the tent.

Radius of the cylindrical base = 20m

Height of the cylindrical portion = 4.2m

Height of the conical portion = 2.1m

Volume of the tent = Volume of the cylindrical portion + volume of the conical portion

**=Â +
=Â
=Â **

= 6160 m

^{3}

Volume of the tent = 6160 m

^{3}.

Â

Find the volume of a solid in the form of a right circular cylinder withÂ hemispherical ends whose total length is 2.7m and the diameter ofÂ each hemispherical end is 0.7 m.

Radius of the hemispherical end = 0.7m/2 = 0.35m

Length of the cylinder = 2.7 â€“ (0.35 + 0.35) = 2m

Volume of a hemisphere =Â = 0.0898 m^{3}

Volume of a right circular cylinder =Â = 0.77m^{3}

Volume of a solid = 2

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 2Ã—Â 0.0898 + 0.77

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 0.1796 + 0.77Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 0.9496 m

^{3}

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 0.95 m

^{3}

Volume of the solid = 0.95 m^{3}.

Â

Â

An iron pole consisting of a cylindrical portion 110 m high and ofÂ base diameter 12 m is surmounted by a cone 9 m high. Find theÂ mass of the pole, given that 1 m^{3}Â of iron has 8 g mass (approx).Â (UseÂ **Ï€Â =Â ).**

Radius of the cone =Â Â = 6 m

Height of the cone = 9 m

Height of the cylindrical portion = (110 - 9) m = 101 m

Volume of the cone =Â

Volume of the cylindrical portion =

Volume of the iron pole = Volume of the cone + Volume of the cylindrical portion

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â =**+Â **

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â =

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â =

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 12785.14 m^{3}

The mass of the pole = 12785.14Â Ã—Â 8 = 102281.14 g = 102.281 kg.

Â

A cylindrical vessel of diameter 14 cm and height 42 cm is fixedÂ symmetrically inside a similar vessel of diameter 16 cm and heightÂ 42 cm. The total space between the two vessels is filled with corkÂ dust for heat insulation purposes. How many cubic centimetres ofÂ cork dust will be required?

Radius of inner cylindrical vessel = 7cm

Radius of outer cylindrical vessel = 8cm

Height of both inner and outer cylindrical vessel = 42cm

The total space between the two vessels is filled with cork dust for heat insulation

= Volume of outer cylindrical vessel - Volume of inner cylindrical vessel

=Â -Â

=

=

= 1980cm^{3}

Â

A petrol tank is a cylinder of base diameter 21 cm and length 18 cm fitted withÂ conical ends each of axis length 9 cm. Determine the capacity of the tank.

Diameter of the base of the cylinder = 21 cmÂ âˆ´Â Radius of the base of the cylinder = 10.5 cm

Length of the cylinder = 18 cm

Radius of the conical end = 10.5 cm

Height of the conical end = 9 cmÂ âˆ´Â Volume of the cylindrical part =Â Ï€Â r^{2}h

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â =Â Ã—Â 10.5Â Ã—Â 10.5Â Ã—Â 18 cm^{3}

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 6237 cm^{3}**Â âˆ´Â **Volume of one conical end =Â Ï€Â r^{2}h

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â =Â Ã—Â Ã—Â 10.5Â Ã—Â 10.5Â Ã—Â 9 cm^{3
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â }= 1039.5 cm^{3}**Â âˆ´Â **Volume of two conical ends = 2Â Ã—Â 1039.5 cm^{3Â }= 2079 cm^{3}Â âˆ´Â Volume of the petrol tank = Volume of the cylindrical part + 2Â Ã—Â Volume of the conical portion

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = (6237 + 2079) cm^{3}

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 8316 cm^{3}.

Â