Volume of a Combination of Solids: Definition, Examples

A combination of solids can be seen in two hemispheres joined by a cylinder. For example, a capsule comprises two hemispheres and a cylinder. If we add the volume of two or more solid shapes, we will get the volume of the combination of solids.

Different shapes can have different volumes, and we use different formulae to calculate the volume of different shapes such as a cube, cuboid, sphere, cylinder, cone, etc. Combining two or more solids, we could get a new solid shape. From this article, we will learn about the volume of solids and the combination of solids.

Volume

The volume of a three-dimensional solid is the amount of space it occupies or space enclosed by a boundary or occupied by an object or the capacity to hold something.

The volume of the solids is measured in cubic units. For instance, units of volume are \(\mathrm{cm}^{3}\), \(\mathrm{mm}^{3}, \mathrm{~m}^{3}\), \(\text {inch} ^{3}\), etc.

Volume of Different Shapes

Let us learn about the volume of some important three-dimensional shapes,

Volume of a Cuboid

A cuboid is solid with six rectangular faces, and the adjacent faces are perpendicular to each other and opposite faces parallel to each other.

We can find the volume of a cuboid when the sides or the edges are given. The volume of the cuboid \(= \text {length} \times \text {breadth} \times \text {height}\)

Volume of a Cube

A cube is a symmetrical solid containing six equal square faces, and the adjacent faces are perpendicular to each other.

We can find the volume of a cube when the side or edge is given. The volume of a cube \(=\text {side} \times\text {side} \times \text {side} =(\text { side })^{3}[\because \text {length} = \text {breadth} = \text {height}]\)

Volume of a Cylinder

A cylinder is solid with a curved side and two identical circular faces. We can calculate the volume of a cylinder by finding the product of the area of the circular base and the height of the cylinder.

The volume of a cylinder \(=\text {area of the circular base} \times \text {height}\)

Area of the circular face \(=\pi r^{2}\)

The height of the cylinder is \(h\).

The volume of a cylinder \(=\pi r^{2} h\)

Volume of a Cone

A cone is like a pyramid with a circular base.
The volume of a cone \(=\frac{1}{3} \times \pi r^{2} \times h=\frac{1}{3} \pi r^{2} h\) where \(r\) is the radius of the circular base and \(h\) is the height of the cone.

Volume of a Sphere

A sphere is a round geometrical figure, with every point on its surface equidistant from its centre.

The volume of a sphere is \(V=\frac{4}{3} \pi r^{3}\) cubic units

Where \(r\) is the radius of the sphere.

Volume of Hemisphere

We know the hemisphere is a three-dimensional shape that is the half of a sphere. So, the volume of the solid hemisphere will be accurately half of the volume of a solid sphere.

So, the volume of a solid hemisphere is \(V=\frac{1}{2} \times \frac{4}{3} \pi r^{3}=\frac{2}{3} \pi r^{3}\)

Where \(r\) is the radius of the hemisphere.

Combination of Solids

We discussed the simple solid shapes and the formulas to calculate their volume. There are numerous cases where these basic shapes are combined to make a new shape, and that shape would have its unique criteria. 

For instance, if we combine two cubes, a cuboid is formed. An ice cream scoop with a cone is the combination of a cone and a hemisphere. The shape of a hut combines a cone and cylinder, two hemispheres, and a cylinder form a capsule; similarly, there are so many real-life examples that exist.

Volume of Combination of Solids

The volume of a combination of solids is the addition of the volumes of the separate solids that are combined to form that solid shape. For example, if we join two cubes by their ends, we will get a cuboid. Let us have a look at the below image.

Let us learn about some different shapes that are formed by basic solid shapes.
A capsule is made up of hemispheres and a cylinder.
The volume of a capsule \(=\) Volume of two identical hemispheres \(+\) volume of a cylinder

Thus, \(V=\frac{2}{3} \pi r^{3}+\frac{2}{3} \pi r^{3}+\pi r^{2} h\), when, \(h\) is the height of the cylinder, \(r\) is the radius of the hemisphere and the radius of the circular base of the cylinder.
Let us see another similar example.

We can divide the shape of ice cream into two different parts. These are a hemisphere and a cone.
Therefore, the volume of the shape \(=\) Volume of a hemisphere \(+\) volume of a cone
\(V=\frac{2}{3} \pi r^{3}+\frac{1}{3} \pi r^{2} h\)
We can form a hut using a cylinder and a cone. So, the volume of the hut \(=\) volume of a cylinder \(+\) volume of a cone.

\(V=\frac{1}{3} \pi r^{2} h+\pi r^{2} h\)
We can form a unique shape using three different basic solid shapes. A sharpened pencil contains a cone, a cylinder and a hemisphere. Let us have a look at the following image.

Thus, the formula of the volume of the pencil \(=\frac{2}{3} \pi r^{3}+\pi r^{2} h_{1}+\frac{1}{3} \pi r^{2} h_{2}\) when, \(h_{1}\) is the height of the cylinder, \(h_{2}\) is the height of the cone, \(r\) is the radius of the hemisphere and the radius of the circular bases of the cylinder and the cone.

Solved Examples

Q.1. Three cubes each of \(15 \,\text {cm}\) side lengths are joined end to end. Find the volume of the resulting figure.
Ans:

In the above image, it is seen that three cubes form a cuboid when it is joined end to end.
We know,
The volume of a cuboid \(=\text {length} \times \text {breadth} \times \text {height} = \text {area of the base} \times \text {height}\)
The length of the new cuboid \(=15+15+15=45 \mathrm{~cm}\)
The height of the cuboid \(=45 \mathrm{~cm}\) and the breadth \(=15 \mathrm{~cm}\)
Therefore, the volume of the cuboid \(=45 \times 15 \times 15=10125 \,\text {cubic} \,\mathrm{cm}\)

Q.2. A solid combination of a hemisphere and a cone and that have a common base. The height of the cone is \(24 \mathrm{~cm}\), and the diameter of the base is \(14 \mathrm{~cm}\).
Ans:

In the above image, we can see that a cone and a hemisphere share a common base having a diameter \(14 \mathrm{~cm}\).
The volume of the given shape \(= \text {Volume of a cone} + \text {volume of a hemisphere}\)
Thus, the volume \(=\frac{2}{3} \pi r^{3}+\frac{1}{3} \pi r^{2} h\)
Radius \((r)=\frac{\text { Diameter }}{2}=\frac{14}{2}=7 \mathrm{~cm}\) and height \((h)=24 \mathrm{~cm}\).
Therefore, the volume of the given shape \(=\frac{2}{3} \pi(7)^{3}+\frac{1}{3} \pi(7)^{2} \times 24=1232+718.67=1950.67 \,\mathrm{cubic} \,\mathrm{cm}\)

Q.3. A globe of diameter \(8 \mathrm{~cm}\) fits inside a closed transparent cube box. The side length of the cube box is \(8 \mathrm{~cm}\) each. Find the amount of space in the box that is not occupied by the globe.
Ans:

The amount of space left in the box \(=\) Volume of the box \(-\) volume of the globe
Now, the volume of the box \(=(\text { side })^{3}=(8)^{3}=512 \,\text {cubic} \,\mathrm{cm}\).
Radius of the globe is \(\frac{8}{2}=4 \mathrm{~cm}\)
The volume of the globe \(=\frac{4}{3} \pi(r)^{3}=\frac{4}{3} \times \frac{22}{7} \times(4)^{3}=268.19\) (approx) \(\mathrm{cubic} \,\mathrm{cm}\).
Therefore, the amount of space in the box that is not occupied by the globe \(=512-268.19=243.81\) (approx) \(\mathrm{cubic} \,\mathrm{cm}\).

Q.4. The height of a capsule-shaped figure is \(16 \mathrm{~cm}\) and radius is \(8 \mathrm{~cm}\). Find the volume of the solid.
Ans:

The volume of a capsule \(=\) Volume of two identical hemispheres \(+\) volume of a cylinder

Thus, \(V=\frac{2}{3} \pi r^{3}+\frac{2}{3} \pi r^{3}+\pi r^{2} h\), when, \(h\) is the height of the cylinder, \(r\) is the radius of the hemisphere and the radius of the circular base of the cylinder
Therefore, the volume:
\(=\frac{2}{3} \pi \times 8^{3}+\frac{2}{3} \pi \times 8^{3}+\pi \times 8^{2} \times 16=1072.76+1072.76+3218.28\)
\(=5363.8 \,\text {cubic cm}\) (approx).

Q.5. Madhura has three wax balls of radii \(3 \,\text {cm}\),\(4 \,\text {cm}\) and \(5 \,\text {cm}\). She melted all three balls and recast them into a single solid ball. Find the radius of the new single solid?
Ans: The balls are nothing but spheres. The volume of the new solid \(=\) Volume of the ball of \(3 \mathrm{~cm}\,+\) Volume of the ball of \(4 \mathrm{~cm}\,+\) Volume of the ball of \(5 \mathrm{~cm}\)
We know that the volume of a sphere is calculated as:
\(V=\frac{4}{3} \pi r^{3}\)
Considering the radius of the single solid ball as \(R\), we get the volume as:
\(\frac{4}{3} \pi R^{3}=\frac{4}{3} \pi \times 3^{3}+\frac{4}{3} \pi \times 4^{3}+\frac{4}{3} \pi \times 5^{3}\)
\(\Rightarrow R^{3}=3^{3}+4^{3}+5^{3}\)
\(\Rightarrow R^{3}=27+64+125=216\)
\(\Rightarrow R=6\)
Hence, the radius of the new single solid is \(6 \mathrm{~cm}\).

Summary

This article tells about different types of solids, the combination of solids, volume formula of basic solids, volume of a combination of solids, examples of the volume of the combination of solids. We solved examples of the volume of the combination of solids.

FAQs

Q.1. What is the combination of solid?
Ans: There are numerous cases where these basic shapes are combined to make a new shape, and that shape would have its unique properties. For instance, If we combine two cubes, a cuboid is formed.

Q.2. What is the volume in simple words?
Ans: The volume describes the capacity of any three-dimensional object. The volume of a cuboidal box indicates the amount of water or any substance contained in it.

Q.3. How do you find the volume of a combined solid?
Ans: The volume of the combination of solids is the addition of the volumes of the separate solids that are combined to form that solid shape.
For example, the volume of a capsule \(=\) Volume of two identical hemispheres \(+\) volume of a cylinder.

Q.4. What is the formula of volume of solids?
Ans: The formula of volume of the cuboid:
\(V=(l \times b \times h)\) cubic units
The formula of volume of a cube, \(V=a^{3} \,\text {cubic units}\)
The formula of volume of a cylinder, \(V=\pi r^{2}h \,\text {cubic units}\)
The formula of the volume of a cone, \(V=\frac{1}{3} \pi r^{2} h\,\text {cubic units}\)
The formula of the volume of a sphere, \(V=\frac{4}{3} \pi r^{3} \,\text {cubic units}\)

Q.5. What is the formula of the surface area of solids?
Ans: The surface area of:
1. Cuboid \(=2(l b+b h+h l)\),
2. Cube \(=6 a^{2}\)
3. Cylinder \(=2 \pi r(r+h)\)
4. Cone \(=\pi r(l+r)\)
5. Sphere \(=4 \pi r^{2}\)
6. Hemisphere \(=3 \pi r^{2}\)

We hope this detailed article on the volume of a combination of solids proves helpful to you. If you have any doubts or queries regarding this topic, feel to ask us in the comment section and we will help you at the earliest.