Chapter 13: Problem 36
Find the coordinates of the center of mass of the following solids with variable density. The interior of the cube in the first octant formed by the planes \(x=1, y=1,\) and \(z=1\) with \(\rho(x, y, z)=2+x+y+z\)
Short Answer
Expert verified
The coordinates of the center of mass are (0.5, 0.5, 0.5).
Step by step solution
01
Calculate the mass of the solid
To find the mass of the solid, we have to find the integral of the density function over the entire solid. Since it's the interior of a cube in the first octant, we can express the limits of integration as \(0 \le x \le 1, 0 \le y \le 1, 0 \le z \le 1\). Thus, $$M = \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} (2+x+y+z)dxdydz$$
02
Calculate the expressions for \(M_x, M_y,\) and \(M_z\)
Now, we have to find the expressions for \(M_x, M_y\), and \(M_z\). We have to find the integral of \(x \rho(x, y, z)\), \(y \rho(x, y, z)\), and \(z \rho(x, y, z)\). \(M_x = \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} x(2+x+y+z)dxdydz\), \(M_y = \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} y(2+x+y+z)dxdydz\), and \(M_z = \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} z(2+x+y+z)dxdydz\)
03
Evaluate the integrals for mass and \(M_x, M_y,\) and \(M_z\)
Now, we have to evaluate the integrals we found above:
$$M = \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} (2+x+y+z)dxdydz = 4$$
$$M_x = \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} x(2+x+y+z)dxdydz = 2$$
$$M_y = \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} y(2+x+y+z)dxdydz = 2$$
$$M_z = \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} z(2+x+y+z)dxdydz = 2$$
04
Calculate the coordinates of the center of mass
Finally, we need to use the results we obtained in the previous step to find the coordinates of the center of mass:
$$x_{cm} = \frac{M_x}{M} = \frac{2}{4} = 0.5$$
$$y_{cm} = \frac{M_y}{M} = \frac{2}{4} = 0.5$$
$$z_{cm} = \frac{M_z}{M} = \frac{2}{4} = 0.5$$
The coordinates of the center of mass are \((0.5, 0.5, 0.5)\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Variable Density
When dealing with objects that have variable density, the density changes across different points within the object. In mathematical terms, this is described by a density function, often denoted as \( \rho(x, y, z) \). In our specific problem, the density varies as \( \rho(x, y, z) = 2 + x + y + z \). This means the density at any point
- increases by 1 as you move along the x-axis,
- increases by 1 as you move along the y-axis,
- and again increases by 1 as you move along the z-axis.
Triple Integration
Triple integration is an extension of single integration into three dimensions. It is used when you need to integrate over a region in three-dimensional space. In our problem, the triple integral is expressed as:\[\int_{0}^{1} \int_{0}^{1} \int_{0}^{1} f(x, y, z) \, dx \, dy \, dz\]Where the function \( f(x, y, z) \) is the density function \( \rho(x, y, z) \). This triple integral calculates volume attributes, such as the mass, of a three-dimensional object.
The process follows these steps:
The process follows these steps:
- Evaluate the innermost integral: Typically, the integrals are computed from the innermost to the outermost. For the innermost integral (with respect to \(x\)), treat \(y\) and \(z\) as constants.
- Evaluate the middle integral: Once the innermost integration is done, move onto the next integral (with respect to \(y\)), treating \(z\) as constant.
- Evaluate the outermost integral: Finally, integrate with respect to \(z\).
Mass Calculation
Calculating the mass of an object with variable density involves setting up and evaluating an integral that takes into account the density function over the object's volume. The formula to do so is:\[M = \int \int \int \rho(x, y, z) \, dV\]Where \( \rho(x, y, z) \) is the density function and \( dV \) represents the infinitesimal volume element, often written as \( dx \, dy \, dz \) in Cartesian coordinates.
Here's how the steps unfold in practice:
Here's how the steps unfold in practice:
- Define the limits of integration: Based on the volume the object occupies. In our example, it’s a cube within the first octant, ranging from 0 to 1 for \(x\), \(y\), and \(z\).
- Integrate the density function: Over the defined limits. In our case, we found that the mass \( M = 4 \).