12.2-Volume of a Combination of Solids

12.2-Volume of a Combination of Solids Important Formulae

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Grade 10 → Math → Surface Areas and Volumes → 12.2-Volume of a Combination of Solids

After successful completion of this topic, you should be able to:

  • Apply formulas of volume of 3D solids in order to derive the volume of a new solid.

The volume of a combination of solids refers to the total space occupied by two or more solids when they are combined. This calculation is important in various fields, such as architecture, engineering, and manufacturing, where understanding the capacity of different shapes is crucial.

To calculate the volume of combined solids, follow these steps:

  1. Identify the solids: Determine the types of solids involved in the combination, such as cubes, cylinders, cones, spheres, etc.
  2. Find the volume of individual solids: Calculate the volume of each solid using their respective formulas.

Common formulas for volumes include:

  • Cube: Volume $= a^3$, where $a$ is the length of a side.
  • Cylinder: Volume $= \pi r^2 h$, where $r$ is the radius and $h$ is the height.
  • Cone: Volume $= \frac{1}{3} \pi r^2 h$, where $r$ is the radius and $h$ is the height.
  • Sphere: Volume $= \frac{4}{3} \pi r^3$, where $r$ is the radius.
  1. Determine the volumes of combined solids: Add the volumes of the individual solids to find the total volume. Unlike surface area, the volume is additive, meaning we do not subtract any overlapping volumes when solids are combined.

Let’s consider a practical example:

Suppose we have a cylinder with a radius of $r = 3 \, \text{cm}$ and a height of $h = 5 \, \text{cm}$, with a cone of the same base radius placed on top of it, with a height of $h = 4 \, \text{cm}$. We will calculate the total volume of this combination.

First, we calculate the volumes:

  • Cylinder:
    • Volume of Cylinder $= \pi r^2 h = \pi \cdot 3^2 \cdot 5 = 45\pi \, \text{cm}^3$
  • Cone:
    • Volume of Cone $= \frac{1}{3} \pi r^2 h = \frac{1}{3} \cdot \pi \cdot 3^2 \cdot 4 = 12\pi \, \text{cm}^3$

Now, to find the total volume, we simply add the volumes of the cylinder and the cone:

Total Volume $= \text{Volume of Cylinder} + \text{Volume of Cone}$

Total Volume $= 45\pi + 12\pi = 57\pi \, \text{cm}^3$

This result shows that the total volume of the combination of the cylinder and the cone is $57\pi \, \text{cm}^3$. This method of calculating volumes allows for straightforward summation without the need for adjustments for overlapping or shared spaces.

In cases where solids are combined in more complex arrangements, such as a sphere placed inside a cylinder, the same principles apply: calculate individual volumes and sum them, taking care to understand the spatial relationships between the solids.


Oshkosh Clayton Concrete mixer 638 Moorestown, New Jersey.
Paul, CC BY-SA 2.0, via Wikimedia Commons

A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of $\pi$.

Solution:

Volume of the Solid

Given that the radius (r) of both the cone and the hemisphere is 1 cm, and the height (h) of the cone is equal to its radius, we have:

Height of the cone (h) = 1 cm.

The volume (V) of the hemisphere is given by the formula:

Vhemisphere = (2/3)πr3

Vhemisphere = (2/3)π(1)3 = (2/3)π cm3.

The volume (V) of the cone is given by the formula:

Vcone = (1/3)πr2h

Vcone = (1/3)π(1)2(1) = (1/3)π cm3.

The total volume (Vtotal) of the solid is the sum of the volumes of the hemisphere and the cone:

Vtotal = Vhemisphere + Vcone

Vtotal = (2/3)π + (1/3)π

Vtotal = (3/3)π = π cm3.

Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model that Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same.)

Solution:

Volume of Air Contained in the Model

Given the diameter of the model is 3 cm, the radius (r) is:

r = Diameter / 2 = 3 cm / 2 = 1.5 cm.

The length of the cylindrical part (hcylinder) is 12 cm, and the height of each cone (hcone) is 2 cm.

The volume (V) of the cylinder is given by the formula:

Vcylinder = πr2hcylinder

Vcylinder = π(1.5)2(12).

The volume (V) of one cone is given by the formula:

Vcone = (1/3)πr2hcone

Vcone = (1/3)π(1.5)2(2).

Since there are two cones, the total volume of the cones is:

Total Vcones = 2 * Vcone = 2 * (1/3)π(1.5)2(2).

The total volume of the model (Vtotal) is the sum of the volumes of the cylinder and the cones:

Vtotal = Vcylinder + Total Vcones.

Substituting the values:

Vcylinder = π(1.5)2(12) = π(2.25)(12) = 27π cm3.

Total Vcones = 2 * (1/3)π(1.5)2(2) = 2 * (1/3)π(2.25)(2) = 3π cm3.

Thus, Vtotal = 27π + 3π = 30π cm3.

A gulab jamun, contains sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm (see Fig. 12.15)

Solution:

Volume of a Gulab Jamun

Each gulab jamun is shaped like a cylinder with two hemispherical ends. First, we calculate the volume of the cylinder and the two hemispheres.

Cylinder Volume

Radius (r) = diameter/2 = 2.8 cm / 2 = 1.4 cm

Height (h) = 5 cm - (2 * radius) = 5 cm - (2 * 1.4 cm) = 5 cm - 2.8 cm = 2.2 cm

Volume of the cylinder (Vc) = πr²h = π(1.4)²(2.2) cm³

Hemispherical Volume

Volume of one hemisphere (Vh) = (2/3)πr³

Total volume of two hemispheres = 2 * Vh = 2 * (2/3)π(1.4)³ cm³

Total Volume of One Gulab Jamun

Total Volume (Vg) = Vc + 2 * Vh

Syrup Volume in One Gulab Jamun

Syrup volume = 30% of Vg = 0.3 * Vg

Total Syrup Volume in 45 Gulab Jamuns

Total Syrup = 45 * (0.3 * Vg)

A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand (see Fig. 12.16).

Solution:

Volume of the Pen Stand

The pen stand is shaped like a cuboid with conical depressions.

Cuboid Volume

Dimensions of the cuboid: length = 15 cm, breadth = 10 cm, height = 3.5 cm

Volume of the cuboid (Vc) = length × breadth × height = 15 cm × 10 cm × 3.5 cm

Volume of One Conical Depression

Radius (r) = 0.5 cm

Depth (h) = 1.4 cm

Volume of one cone (Vcone) = (1/3)πr²h = (1/3)π(0.5)²(1.4) cm³

Total Volume of Four Conical Depressions

Total volume of cones (Vcon) = 4 × Vcone

Volume of Wood in the Stand

Volume of wood = Volume of the cuboid - Total volume of cones

A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.

Solution:

Volume of the Inverted Cone Vessel

Height (h) = 8 cm, Radius (R) = 5 cm

Volume of the cone (Vcone) = (1/3)πR²h = (1/3)π(5)²(8) cm³

Volume of Water Displaced

When one-fourth of the water flows out, Volume of water displaced (Vdisplaced) = (1/4) × Vcone

Volume of One Lead Shot

Radius of each lead shot (r) = 0.5 cm

Volume of one sphere (Vsphere) = (4/3)πr³ = (4/3)π(0.5)³ cm³

Number of Lead Shots

Let n be the number of lead shots. Then, n × Vsphere = Vdisplaced

n = Vdisplaced / Vsphere

A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm3 of iron has approximately 8g mass. (Use $\pi$ = 3.14)

Solution:

Volume of the Solid Iron Pole

The pole consists of two cylinders: a larger cylinder and a smaller cylinder.

Volume of the Larger Cylinder

Height (h1) = 220 cm, Diameter = 24 cm

Radius (R1) = Diameter / 2 = 24 cm / 2 = 12 cm

Volume of the larger cylinder (V1) = πR1²h1 = 3.14 × (12)² × 220 cm³

Volume of the Smaller Cylinder

Height (h2) = 60 cm, Radius (R2) = 8 cm

Volume of the smaller cylinder (V2) = πR2²h2 = 3.14 × (8)² × 60 cm³

Total Volume of the Pole

Total Volume (Vtotal) = V1 + V2

Mass of the Pole

Density of iron = 8 g/cm³

Mass of the pole = Vtotal × Density

A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm.

Solution:

Volume of the Right Circular Cone

Height (h) = 120 cm, Radius (R) = 60 cm

Volume of the cone (Vcone) = (1/3)πR²h = (1/3) × 3.14 × (60)² × 120 cm³

Volume of the Hemisphere

Radius (r) = 60 cm

Volume of the hemisphere (Vhemisphere) = (2/3)πr³ = (2/3) × 3.14 × (60)³ cm³

Total Volume of the Solid

Total Volume (Vtotal) = Vcone + Vhemisphere

Volume of the Right Circular Cylinder

Radius of the cylinder (Rcylinder) = 60 cm, Height (H) = 180 cm

Volume of the cylinder (Vcylinder) = πRcylinder²H = 3.14 × (60)² × 180 cm³

Volume of Water Left in the Cylinder

Volume of water left = Vcylinder - Vtotal

A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm$^3$. Check whether she is correct, taking the above as the inside measurements, and $\pi$ = 3.14.

Solution:

Volume of the Spherical Part

Diameter of the sphere = 8.5 cm, Radius (r) = 8.5 cm / 2 = 4.25 cm

Volume of the sphere (Vsphere) = (4/3)πr³ = (4/3) × 3.14 × (4.25)³ cm³

Volume of the Cylindrical Neck

Diameter of the neck = 2 cm, Radius (R) = 2 cm / 2 = 1 cm, Height (h) = 8 cm

Volume of the cylinder (Vcylinder) = πR²h = 3.14 × (1)² × 8 cm³

Total Volume of the Glass Vessel

Total Volume (Vtotal) = Vsphere + Vcylinder

Comparison with Child's Measurement

Check if Vtotal = 345 cm³