How To Find The Volume Of A Truncated Cone (Conical Frustum) When The Slant Height Is Given

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Finding the volume of a truncated cone (or conical frustum) when the slant height is given requires additional steps to convert the slant height to the perpendicular height before using the standard volume formula. This guide will show you how to make that conversion and calculate the volume.


Volume Formula for a Truncated Cone

The volume (\( V \)) of a truncated cone is generally calculated using the perpendicular height \( h \), but when the slant height \( l \) is given, we first need to derive \( h \). The volume is given by:


\[ V = \dfrac{1}{3} \cdot \pi \cdot (r^2 + r \cdot R + R^2) \cdot h \]


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 perpendicular height.

- \( \pi \) is approximately equal to 3.14159.


Step-by-Step Process to Convert Slant Height to Perpendicular Height

To convert the slant height \( l \) to the perpendicular height \( h \), we use the Pythagorean theorem. Given the radii \( r \), \( R \), and the slant height \( l \):


\[ h = \sqrt{l^2 - (R - r)^2} \]


Explanation

1. Slant Height: \( l \) is the hypotenuse of the right triangle formed by the height \( h \), and the difference in radii (\( R - r \)).

2. Pythagorean Theorem: Use the relationship \( h = \sqrt{l^2 - (R - r)^2} \) to find the perpendicular height.


Example Calculation

Let's demonstrate the calculation with an example.


Given:

- \( r = 3 \) units (radius of the top base)

- \( R = 6 \) units (radius of the bottom base)

- \( l = 7 \) units (slant height of the truncated cone)


Step 1: Identify the Given Values

Given:

- \( r = 3 \) units

- \( R = 6 \) units

- \( l = 7 \) units


Step 2: Calculate the Perpendicular Height \( h \)

Using the formula:

\[ h = \sqrt{l^2 - (R - r)^2} \]


Substitute \( l = 7 \), \( R = 6 \), and \( r = 3 \):


\[ h = \sqrt{7^2 - (6 - 3)^2} \]


Simplify inside the square root:


\[ h = \sqrt{49 - 3^2} \]


\[ h = \sqrt{49 - 9} \]


\[ h = \sqrt{40} \]


\[ h \approx 6.325 \text{ units} \]


Step 3: Substitute \( h \) into the Volume Formula

Using the formula for volume:

\[ V = \dfrac{1}{3} \cdot \pi \cdot (r^2 + r \cdot R + R^2) \cdot h \]


Substitute \( r = 3 \), \( R = 6 \), and \( h = 6.325 \):


\[ V = \dfrac{1}{3} \cdot \pi \cdot (3^2 + 3 \cdot 6 + 6^2) \cdot 6.325 \]


Step 4: 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 5: Substitute and Simplify

Now substitute back into the volume formula:

\[ V = \dfrac{1}{3} \cdot \pi \cdot 63 \cdot 6.325 \]


\[ V = \dfrac{1}{3} \cdot 398.475 \cdot \pi \]


Step 6: Calculate the Final Value

Using \( \pi \approx 3.14159 \):

\[ V \approx \dfrac{398.475 \cdot 3.14159}{3} \]


\[ V \approx 417.703 \text{ cubic units} \]


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 slant height of 7 units is approximately \( 417.703 \) cubic units.


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