A truncated cone, also known as a conical frustum, is a cone with its top cut off parallel to the base. Calculating its volume requires knowing the radii of the top and bottom bases and the height of the frustum. This guide will show you how to find the volume using a straightforward formula.
Formula to Calculate the Volume of a Truncated Cone
The volume (\( V \)) of a truncated cone can be calculated using the following formula:
\[ V = \dfrac{1}{3} \cdot \pi \cdot (r^2 + r \cdot R + R^2) \cdot h \]
Where:
- \( V \) is the volume of the truncated cone.
- \( r \) is the radius of the top base.
- \( R \) is the radius of the bottom base.
- \( h \) is the height of the truncated cone.
- \( \pi \) is a mathematical constant approximately equal to 3.14159.
Explanation of the Formula
1. Base Radii Contribution: The formula combines the areas of three circles: the top base (\( r^2 \)), the bottom base (\( R^2 \)), and an average area (\( r \cdot R \)) to account for the slanted surface between the two bases.
2. Height Contribution: The height (\( h \)) stretches these areas into the volume, and dividing by 3 adjusts for the fact that it's a conical frustum, not a full cone.
Example Calculation
Let's walk through an example to demonstrate the calculation.
Given:
- \( r = 3 \) units (radius of the top base)
- \( R = 6 \) units (radius of the bottom base)
- \( h = 5 \) units (height of the truncated cone)
Step 1: Identify the Given Values
Given:
- \( r = 3 \) units
- \( R = 6 \) units
- \( h = 5 \) units
Step 2: Substitute the Values into the Volume Formula
\[ V = \dfrac{1}{3} \cdot \pi \cdot (r^2 + r \cdot R + R^2) \cdot h \]
Substitute \( r = 3 \), \( R = 6 \), and \( h = 5 \):
\[ V = \dfrac{1}{3} \cdot \pi \cdot (3^2 + 3 \cdot 6 + 6^2) \cdot 5 \]
Step 3: Simplify the Terms Inside the Parentheses
Calculate each term:
\[ 3^2 = 9 \]
\[ 3 \cdot 6 = 18 \]
\[ 6^2 = 36 \]
Add them together:
\[ 9 + 18 + 36 = 63 \]
Step 4: Substitute and Simplify
Now substitute back into the formula:
\[ V = \dfrac{1}{3} \cdot \pi \cdot 63 \cdot 5 \]
\[ V = \dfrac{1}{3} \cdot 315 \cdot \pi \]
Step 5: Calculate the Final Value
Using \( \pi \approx 3.14159 \):
\[ V \approx \dfrac{315 \cdot 3.14159}{3} \]
\[ V \approx 330.846 \]
Final Value
The volume of a truncated cone with a top base radius of 3 units, a bottom base radius of 6 units, and a height of 5 units is approximately \( 330.846 \) cubic units.
