Calculus: Early Transcendentals
8th Edition
ISBN: 9781285741550
Author: James Stewart
Publisher: Cengage Learning
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![### Calculating the Volume of a Tetrahedron
**Problem Statement:**
1) Compute the volume of a tetrahedron.
**Step-by-Step Solution:**
(a) Illustrate the tetrahedron that has vertices at (0,0,0), (2,0,0), (0,4,0), and (0,0,6) in Cartesian coordinates.
This tetrahedron sits inside a rectangular box with side lengths 2, 4, and 6. The volume of this box is calculated using the formula for the volume of a rectangular prism:
\[ V_{\text{box}} = \text{length} \times \text{width} \times \text{height} \]
Substituting the given dimensions:
\[ V_{\text{box}} = 2 \times 4 \times 6 = 48 \text{ cubic units} \]
Since the tetrahedron occupies a portion of this box, the volume of the tetrahedron can be found by determining what fraction of the box it occupies. The vertices of the tetrahedron suggest it is a specific fraction of the box:
\[ V_{\text{tetrahedron}} = \frac{V_{\text{box}}}{6} \]
This is because a tetrahedron that fits perfectly inside a rectangular box defined by one vertex at the origin and other vertices at the axes will occupy 1/6th of the volume of the box.
Therefore, the volume of the tetrahedron is:
\[ V_{\text{tetrahedron}} = \frac{48}{6} = 8 \text{ cubic units} \]
**Summary:**
- The volume of the tetrahedron with the given vertices is **8 cubic units**.](https://content.bartleby.com/qna-images/question/23435939-1300-49d7-9ff3-84762f7d9052/7e8e5e72-296d-436d-a4ed-02dc6126ff58/qwswdf9_processed.jpeg)
Transcribed Image Text:### Calculating the Volume of a Tetrahedron
**Problem Statement:**
1) Compute the volume of a tetrahedron.
**Step-by-Step Solution:**
(a) Illustrate the tetrahedron that has vertices at (0,0,0), (2,0,0), (0,4,0), and (0,0,6) in Cartesian coordinates.
This tetrahedron sits inside a rectangular box with side lengths 2, 4, and 6. The volume of this box is calculated using the formula for the volume of a rectangular prism:
\[ V_{\text{box}} = \text{length} \times \text{width} \times \text{height} \]
Substituting the given dimensions:
\[ V_{\text{box}} = 2 \times 4 \times 6 = 48 \text{ cubic units} \]
Since the tetrahedron occupies a portion of this box, the volume of the tetrahedron can be found by determining what fraction of the box it occupies. The vertices of the tetrahedron suggest it is a specific fraction of the box:
\[ V_{\text{tetrahedron}} = \frac{V_{\text{box}}}{6} \]
This is because a tetrahedron that fits perfectly inside a rectangular box defined by one vertex at the origin and other vertices at the axes will occupy 1/6th of the volume of the box.
Therefore, the volume of the tetrahedron is:
\[ V_{\text{tetrahedron}} = \frac{48}{6} = 8 \text{ cubic units} \]
**Summary:**
- The volume of the tetrahedron with the given vertices is **8 cubic units**.
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