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Chapter 5: Problem 20

(a) If \(c\) is a positive constant, find the volume of the tetrahedron in the first octant bounded by the plane \(x+y+z=c\) and the three coordinate planes. (b) Consider the tetrahedron \(W\) in the first octant bounded by the plane \(x+y+z=1\) and the three coordinate planes. Suppose that you want to divide \(W\) into three pieces of equal volume by slicing it with two planes parallel to \(x+y+z=1\), i.e., with planes of the form \(x+y+z=c\). How should the slices be made?

Short Answer

Expert verified
(a) Volume is \(\frac{c^3}{6}\). (b) Use planes \(x+y+z=\sqrt[3]{\frac{1}{3}}\) and \(x+y+z=\sqrt[3]{\frac{2}{3}}\) for equal slices.

Step by step solution

01

Understanding Tetrahedron Boundaries

We are given the equation of a plane, \(x+y+z=c\), and we need to find the volume of the tetrahedron formed in the first octant. The tetrahedron is bounded by this plane and the coordinate planes \(x=0\), \(y=0\), and \(z=0\).
02

Identifying Intersection with Axes

The plane \(x+y+z=c\) intersects the coordinate axes at points \((c, 0, 0)\), \((0, c, 0)\), and \((0, 0, c)\). These points are the vertices of the tetrahedron, along with the origin \((0, 0, 0)\).
03

Volume of Tetrahedron in First Octant

The volume \(V\) of a tetrahedron with vertices at \((0, 0, 0)\), \((c, 0, 0)\), \((0, c, 0)\), and \((0, 0, c)\) is given by the formula: \[ V = \frac{1}{6} \times \text{Base Area} \times \text{Height} = \frac{1}{6} \times c \times c \times c = \frac{c^3}{6}.\]
04

Determine Volume for Second Part

We are given a specific case where \(c=1\). The volume of the entire tetrahedron \(W\) is \(\frac{1}{6}\).
05

Equal Volume Slicing Requirement

To divide \(W\) into three equal volumes, each piece must have a volume of \(\frac{1}{18}\) (since \(\frac{1}{6} \div 3 = \frac{1}{18}\)).
06

Find Values of \(c_1\) and \(c_2\) for Equal Slices

The volume of a smaller tetrahedron with a plane \(x+y+z=c_1\) is \(\frac{c_1^3}{6} = \frac{1}{18}\). Solving \(c_1^3 = \frac{1}{3}\), we get \(c_1 = \sqrt[3]{\frac{1}{3}}\). For the second slice, \(\frac{c_2^3}{6} - \frac{c_1^3}{6} = \frac{1}{18}\), leading to \(c_2^3 = \frac{2}{3}\). Thus, \(c_2 = \sqrt[3]{\frac{2}{3}}\).
07

Confirm Slice Positions

We have found the two planes \(x+y+z=\sqrt[3]{\frac{1}{3}}\) and \(x+y+z=\sqrt[3]{\frac{2}{3}}\) that divide the initial tetrahedron into three smaller tetrahedrons of equal volume.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Volume of a Tetrahedron
The volume of a tetrahedron is a fascinating concept in multivariable calculus. A tetrahedron is a 3-dimensional geometric shape with four triangular faces. It resembles a pyramid but with a triangular base.
To calculate the volume of a tetrahedron, especially one in the first octant, like our original problem, we need to identify its vertices first:
  • The plane equation is given as \(x+y+z=c\).
  • Its intersections with the coordinate axes are \((c, 0, 0)\), \((0, c, 0)\), and \((0, 0, c)\).
  • With the origin \((0, 0, 0)\) forming the fourth vertex.
The formula used to calculate the volume of a tetrahedron with these vertices is:
\[ V = \frac{1}{6} \times \text{Base Area} \times \text{Height} \]
For a base parallel to the coordinate plane, both the base area and height are \(c\), yielding the formula:
\[ V = \frac{c^3}{6} \]
Coordinate Planes
Coordinate planes are fundamental to understanding the geometry of a 3D space.
These planes are essentially the surfaces defined by holding one of the three dimensions constant.
In our problem, we are dealing with:
  • The x-coordinate plane: \(x=0\)
  • The y-coordinate plane: \(y=0\)
  • The z-coordinate plane: \(z=0\)
Each plane divides the space into the opposite 3D sections known as octants. In the first octant, all x, y, and z coordinates are positive. Our tetrahedron resides in this first octant, being naturally bounded by these planes. They essentially form the three edges of the tetrahedron and are crucial for defining its shape and volume.
Plane Intersection
The concept of plane intersection is crucial for formulating the problem of the tetrahedron volume.
When a plane like \(x+y+z=c\) intersects with the coordinate planes, vertices of the tetrahedron are formed at these intersection points.
Understanding these points is essential as they are the limits of the tetrahedron in each axis.
  • The intersection of \(x+y+z=c\) with the x-axis is at \((c, 0, 0)\).
  • With the y-axis, it's at \((0, c, 0)\).
  • And with the z-axis, it's at \((0, 0, c)\).
  • Additionally, the origin \((0, 0, 0)\) acts as a pivotal point.
These intersections help define the tetrahedron's boundaries and are integral for its volumetric calculations.
Volume Slicing
Volume slicing is an advanced yet straightforward approach to dividing a geometric shape into smaller sections with equal volume.
For our problem, slicing is done by utilizing planes parallel to an initial bounding plane, \(x+y+z=1\). By doing this, each slicer or cross-section retains a consistent geometric proportionality to the whole.
Here's what happens during volume slicing:
  • Identify the total volume - here it's \(\frac{1}{6}\) for \(c=1\).
  • Set the goal to divide this volume into three equal parts - each being \(\frac{1}{18}\).
  • Use planes of form \(x+y+z=c_1\) and \(x+y+z=c_2\) to slice.
These slice planes are carefully calculated to:
  • First obtain \(c_1 = \sqrt[3]{\frac{1}{3}}\).
  • Then, achieve \(c_2 = \sqrt[3]{\frac{2}{3}}\) beyond the first slice.
This method explains how to create smaller tetrahedrons, each with an equal volume, from the original shape.

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Most popular questions from this chapter

Find the volume of the pyramid-shaped solid in the first octant bounded by the three coordinate planes and the planes \(x+z=1\) and \(y+2 z=2\).

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Let \(R=[0,1] \times[0,1],\) and let \(f: R \rightarrow \mathbb{R}\) be the function given by: $$ f(x, y)=\left\\{\begin{array}{ll} 1 & \text { if }(x, y)=\left(\frac{1}{2}, \frac{1}{2}\right), \\ 0 & \text { otherwise } \end{array}\right. $$ (a) Let \(R\) be subdivided into a 3 by 3 grid of subrectangles of equal size (Figure 5.36 ). What are the possible values of Riemann sums based on this subdivision? (Different choices of sample points may give different values.) (b) Repeat for a 4 by 4 grid of subrectangles of equal size. (c) How about for an \(n\) by \(n\) grid of subrectangles of equal size? (d) Let \(\delta\) be a positive real number. If \(R\) is subdivided into a grid of subrectangles of dimensions \(\triangle x_{i}\) by \(\triangle y_{j},\) where \(\triangle x_{i}<\delta\) and \(\triangle y_{j}<\delta\) for all \(i, j,\) show that any Riemann sum based on the subdivision lies in the range \(0 \leq \sum_{i, j} f\left(\mathbf{p}_{i j}\right) \Delta x_{i} \Delta y_{j}<4 \delta^{2}\). (e) Show that \(\iint_{R} f(x, y) d A=0,\) i.e., that \(\lim _{\triangle x_{i} \rightarrow 0 \atop \triangle y_{j} \rightarrow 0} \sum_{i, j} f\left(\mathbf{p}_{i j}\right) \Delta x_{i} \Delta y_{j}=0 .\) Technically, this means you need to show that, given any \(\epsilon>0,\) there exists a \(\delta>0\) such that, for any subdivision of \(R\) into subrectangles with \(\triangle x_{i}<\delta\) and \(\Delta y_{j}<\delta\) for all \(i, j\), every Riemann sum based on the subdivision satisfies \(\left|\sum_{i, j} f\left(\mathbf{p}_{i j}\right) \Delta x_{i} \Delta y_{j}-0\right|<\epsilon\).

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