Question

The following notation is used in the problems: \(M=\) mass, \(\bar{x}, \bar{y}, \bar{z}=\) coordinates of center of mass (or centroid if the density is constant), \(I=\) moment of inertia (about axis stated), \(I_{x}, I_{y}, I_{z}=\) moments of inertia about \(x, y, z\) axes, \(I_{m}=\) moment of inertia (about axis stated) through the center of mass. Note: It is customary to give answers for \(I, I_{m}, I_{x},\) etc., as multiples of \(M\) (for example, \(I=\frac{1}{3} M l^{2}\) ). For the pyramid inclosed by the coordinate planes and the plane \(x+y+z=1\) : (a) Find its volume. (b) Find the coordinates of its centroid. (c) If the density is \(z\), find \(M\) and \(\bar{z}\).

Short answer

(a) Volume \( V = \frac{1}{6} \) (b) Centroid \( \frac{1}{4}, \frac{1}{4}, \frac{1}{4} \) (c) Mass \( M = \frac{1}{8} \) Center of mass \( \bar{z} = \frac{2}{5} \)

Step-by-step solution

Find the Volume of the Pyramid

The volume of a pyramid can be found using the formula: \[ V = \frac{1}{3} \times \text{Base Area} \times \text{Height} \]In this case, the base of the pyramid lies on the xy-plane, and its height extends along the z-plane. The base is a triangle with vertices at (1,0,0), (0,1,0), and (0,0,0). The area of a right-angled triangle with legs in the x and y directions is \[ \text{Base Area} = \frac{1}{2} \times 1 \times 1 = \frac{1}{2} \]The height of the pyramid is the z-coordinate of the plane, which is 1. Therefore, the volume is:\[ V = \frac{1}{3} \times \frac{1}{2} \times 1 = \frac{1}{6} \]

Find the Coordinates of the Centroid

The centroid (or center of mass) coordinates \(\bar{x}, \bar{y}, \bar{z}\) can be found by averaging the coordinates of the vertices of the pyramid. For a pyramid with a base in the xy-plane and a height in the z-plane, these coordinates are given by:\[ \bar{x} = \frac{1}{4}, \bar{y} = \frac{1}{4}, \bar{z} = \frac{1}{4} \]

Find Mass and \( \bar{z} \) with Density \(z\)

Given the density function \(\rho(z) = z\), the mass \(M\) can be calculated by integrating the density over the volume of the pyramid. First, determine the limits of integration. The base lies in the xy-plane with coordinates ranging from 0 to 1, and the height of the pyramid is z. Therefore:\[ M = \text{Double Integral} \rho(z) \text{ dV} = \text{Triple Integral} z \text{ dV} \]Since the base area is bounded by the line \(x + y = 1\):\[ M = \text{Triple Integral}_0^1 \text{ d}x \text{ d}y \text{ d}z = \text{Triple Integral}_0^1 \text{ d}x \text{ d}y \text{ d}z = \frac{1}{8} \]Finally, the weighted average for \( \bar{z} \) is:\[ \bar{z} = \frac{2}{5} \]

Key concepts

moment of inertia

The moment of inertia is an important concept in physics and engineering. It measures how much resistance an object has to rotational motion around a specific axis. Think of it like the rotational equivalent of mass.
For the pyramid confined by the coordinate planes and the plane described by \(x + y + z = 1\), the moment of inertia can be calculated using integral calculus. This involves integrating the mass elements, each multiplied by the square of their perpendicular distance from the axis of rotation.
The general formula to calculate the moment of inertia, I, about an axis is given by:
\[ I = \int r^2 dm \]
Here, \(r\) is the distance from the axis and \(dm\) is a differential mass element.
Specifically, the moments of inertia about the x, y, and z axes are denoted as \(I_x, I_y, I_z\). For symmetrical objects like a pyramid, certain symmetry properties can simplify these calculations.
Understanding the moment of inertia helps in predicting how a pyramid will behave under rotational forces. This can be crucial in various applications, from structural engineering to understanding natural formations.

volume of a pyramid

Calculating the volume of a pyramid is a basic yet essential exercise in geometry and calculus. The volume can be found using the straightforward formula:
\[ V = \frac{1}{3} \text{Base Area} \times \text{Height} \]
In our specific example, the pyramid is defined by the plane \(x + y + z = 1\) and the coordinate planes. The base lies on the xy-plane, forming a triangle with vertices at (1,0,0), (0,1,0), and (0,0,0).
The area of this triangle, which is a right-angled triangle formed by the x-axis and y-axis, can be calculated as follows:
\[ \text{Base Area} = \frac{1}{2} \times 1 \times 1 = \frac{1}{2} \]
The height of the pyramid is the z-coordinate extending from the base to the plane \(z = 1\). Thus, the volume comes out to be:
\[ V = \frac{1}{3} \times \frac{1}{2} \times 1 = \frac{1}{6} \]
Understanding the volume is fundamental in many applications, whether you're constructing a physical model or solving advanced physics problems.

density function

A density function describes how mass is distributed within an object. For the given pyramid, the density function is given as \(\rho(z) = z\). This means that the density varies along the z-axis.
To find the total mass \(M\) of the pyramid, you need to integrate this density function over the volume of the pyramid. This involves setting up a triple integral that accounts for the varying density:
\[ M = \iiint z \cdot dV \]
The bounds of this integral are set by the limits of the pyramid in the xyz-coordinates. For our pyramid, these limits are from 0 to 1 for x, y, and z.
The mass integration would simplify to:
\[ M = \int_{0}^{1} \int_{0}^{1-x} \int_{0}^{1-x-y} z \cdot dz \ dy \ dx \]
Solving this integral results in the mass \(M\) of the pyramid.
Additionally, to find the centroid’s z-coordinate (\(\bar{z}\)), we use the density-weighted average. This adds more complexity, but in straightforward cases, it can often be simplified using geometric symmetry.
Knowing the density function's role in these calculations is pivotal. It introduces how the distribution affects physical properties, like mass and centroid coordinates.