Volume of Combinations of Solids

The volume of combinations of solids refers to calculating the total space occupied by two or more three-dimensional shapes joined together, or by a single shape with a portion removed. This involves understanding the individual volumes of the basic solids (cubes, spheres, cylinders, cones, etc.) and then applying appropriate operations (addition or subtraction) based on how the shapes are combined.

When shapes are joined (e.g., a cube stacked on top of a cylinder), we typically add their individual volumes. When a portion is removed (e.g., a hole drilled through a sphere), we subtract the volume of the removed portion from the original shape’s volume. Careful visualization and understanding of the problem are crucial to identify the shapes involved and how they are combined.

Formulae

The core principle relies on knowing the volume formulae for basic solids. Here are some common ones:

• Cube: $V = s^3$ (where $s$ is the side length)
• Cuboid: $V = lwh$ (where $l$ is length, $w$ is width, and $h$ is height)
• Sphere: $V = \frac{4}{3}\pi r^3$ (where $r$ is the radius)
• Cylinder: $V = \pi r^2 h$ (where $r$ is the radius and $h$ is the height)
• Cone: $V = \frac{1}{3}\pi r^2 h$ (where $r$ is the radius and $h$ is the height)
• Pyramid: $V = \frac{1}{3}Bh$ (where $B$ is the area of the base and $h$ is the height)

Combining Volumes:

• Addition: If solids are joined, $V_{total} = V_1 + V_2 + … + V_n$
• Subtraction: If a portion is removed, $V_{remaining} = V_{original} – V_{removed}$

Examples

Example-1: A composite solid is formed by placing a cone on top of a cylinder. The cylinder has a radius of 3 cm and a height of 10 cm. The cone has the same radius as the cylinder and a height of 6 cm. Find the total volume.

Solution:

Volume of Cylinder: $V_{cylinder} = \pi r^2 h = \pi (3^2)(10) = 90\pi$ cm$^3$

Volume of Cone: $V_{cone} = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (3^2)(6) = 18\pi$ cm$^3$

Total Volume: $V_{total} = V_{cylinder} + V_{cone} = 90\pi + 18\pi = 108\pi$ cm$^3$ (approximately 339.29 cm$^3$)

Example-2: A solid wooden cube with a side length of 10 cm has a cylindrical hole drilled through it. The hole has a radius of 2 cm and passes completely through the cube. Find the volume of the remaining wood.

Solution:

Volume of Cube: $V_{cube} = s^3 = 10^3 = 1000$ cm$^3$

Volume of Cylindrical Hole: $V_{cylinder} = \pi r^2 h = \pi (2^2)(10) = 40\pi$ cm$^3$ (the height is equal to the cube’s side length)

Volume of Remaining Wood: $V_{remaining} = V_{cube} – V_{cylinder} = 1000 – 40\pi$ cm$^3$ (approximately 874.34 cm$^3$)

Common mistakes by students

• Incorrectly identifying the shapes involved: Failing to correctly recognize the different shapes (e.g., mistaking a frustum of a cone for a complete cone).
• Using the wrong formula: Applying the formula for the wrong solid.
• Forgetting units: Omitting units or mixing different units (e.g., cm and mm). Always provide the units in the answer.
• Incorrectly measuring the radius or height: Not paying attention to how the dimensions relate.
• Incorrect operations: Adding when you should subtract and vice versa. Make sure you understand whether volumes are combined or have a part removed from them.
• Not considering overlapping regions In the case of removing a hole from a cube or sphere, students sometimes forget that the volume of the hole is the volume of the cylinder that has been removed.

Real Life Application

Understanding the volume of combinations of solids is crucial in many real-world scenarios:

• Construction: Calculating the amount of materials needed (concrete, bricks, etc.) for building structures, like buildings or swimming pools.
• Manufacturing: Determining the volume of materials used in products, such as packaging (boxes, cans), or in designing parts of machines.
• Packaging: Designing efficient packaging that minimizes wasted space and costs.
• Engineering: Calculating the capacity of tanks, reservoirs, or other containers.
• Architecture and Design: Creating aesthetically pleasing and structurally sound designs, taking into account space and material use.

Fun Fact

Did you know that Archimedes, a famous Greek mathematician, used the concept of volumes to determine if a crown was made of pure gold? He compared the volume of the crown with the volume of an equal weight of pure gold, revealing that the crown was not pure, by using the principal of displacement!

Recommended YouTube Videos for Deeper Understanding

Q.1 A solid is formed by a hemisphere of radius 3 cm placed on top of a cone with the same radius and height. If the height of the cone is 4 cm, what is the total volume of the solid in cubic centimeters?

Check Solution

Ans: B

Volume of hemisphere = $\frac{2}{3}\pi r^3 = \frac{2}{3}\pi(3^3) = 18\pi$. Volume of cone = $\frac{1}{3}\pi r^2 h = \frac{1}{3}\pi(3^2)(4) = 12\pi$. Total volume = $18\pi + 12\pi = 30\pi$.

Q.2 A composite solid is made of a cylinder and a hemisphere. The cylinder has a radius of 2 cm and a height of 5 cm. The hemisphere has the same radius. What is the total volume of the solid in cubic centimeters?

Check Solution

Ans: C

Volume of cylinder = $\pi r^2 h = \pi(2^2)(5) = 20\pi$. Volume of hemisphere = $\frac{2}{3}\pi r^3 = \frac{2}{3}\pi(2^3) = \frac{16}{3}\pi$. Total volume = $20\pi + \frac{16}{3}\pi = \frac{76}{3}\pi$.

Q.3 A rectangular prism has dimensions 6 cm, 8 cm, and 10 cm. A cylindrical hole with radius 2 cm is drilled through the prism, passing through the center of two opposite faces with dimensions 6 cm by 8 cm. What is the volume of the solid remaining in cubic centimeters?

Check Solution

Ans: A

Volume of prism = $6*8*10 = 480$. Volume of cylindrical hole = $\pi r^2 h = \pi(2^2)(10) = 40\pi$. Volume of the remaining solid: $480 – 40\pi$.

Q.4 A solid is formed by attaching a cone to a cube. The cube has sides of length 5 cm, and the cone has a base radius of 2 cm and a height of 6 cm. What is the total volume of the solid in cubic centimeters?

Check Solution

Ans: A

Volume of cube = $5^3 = 125$. Volume of cone = $\frac{1}{3}\pi r^2 h = \frac{1}{3}\pi(2^2)(6) = 8\pi$. Total volume = $125 + 8\pi$.

Q.5 A toy is composed of a hemisphere and a cylinder. The hemisphere has a radius of 3 cm and the cylinder has the same radius and a height of 10 cm. What is the total volume of the toy in cubic centimeters?

Check Solution

Ans: B

Volume of hemisphere = $\frac{2}{3}\pi r^3 = \frac{2}{3}\pi(3^3) = 18\pi$. Volume of cylinder = $\pi r^2 h = \pi(3^2)(10) = 90\pi$. Total volume = $18\pi + 90\pi = 108\pi$.

Next Topic: Conversion of Solids: Volume Conservation