Problem 51
Question
Find the mass and center of mass of the region. The hemisphere \(z=\sqrt{1-x^{2}-y^{2}}, \rho(x, y, z)=1+x\)
Step-by-Step Solution
Verified Answer
The mass of the hemisphere is \(M = 5/8\pi\) and the center of mass of the hemisphere is at the point (0,0,3/8).
1Step 1: Find the mass of the hemisphere
To find the mass, integrate the density function \(\rho(x, y, z)\) over the entire volume of the hemisphere. This results in: \[ M = \int\int\int_V (1+x) dz dy dx \] where V represents the volume of the hemisphere. The hemispheric coordinates are \(x=\rho \cos \theta \sin \phi\), \(y=\rho \sin \theta \sin \phi\), and \(z=\rho \cos \phi\). So replace \(x\) in terms of \(\rho\). Now, the limits of integral fit a unit hemisphere. The azimuthal angle \(\phi\) ranges from \(0\) to \(\pi/2\), \(\rho\) ranges from \(0\) to \(1\), and \(\theta\) varies from \(0\) to \(2\pi\).
2Step 2: Evaluate the integral to obtain mass
We now evaluate the integral to obtain the mass of the hemisphere: \[ M = \int\int\int_V (1+ \rho \cos \theta \sin \phi) \rho^{2} \sin \phi\, d\phi\, d\theta\, d\rho \] This integral yields \(M = 5/8\pi\)
3Step 3: Find the coordinates of the center of mass
The coordinates of the center of mass (\(\bar{x}, \bar{y}, \bar{z}\)) can be obtained by using the formula: \[ \bar{x} = \frac{1}{M} \int\int\int_V x \rho(x,y,z) dz dy dx \] and similarly for \(\bar{y}\) and \(\bar{z}\). Substituting in the given mass and calculating these integrals yield the center of mass (\(\bar{x},\bar{y},\bar{z}\)).
4Step 4: Evaluate the integrals to obtain the center of mass
By evaluating the integrals, we find that \(\bar{x}=\bar{y}=0\) and \(\bar{z}=3/8\). Therefore, the center of mass is at the point (0,0,3/8).
Key Concepts
Mass of Hemisphere CalculusTriple Integral Mass CalculationSpherical Coordinates Integration
Mass of Hemisphere Calculus
Understanding the mass of a hemisphere requires integrating the given density function over the object's volume. In calculus, we often deal with non-uniform mass distributions. In the provided exercise, the density function is \(\rho(x, y, z) = 1+x\), which indicates that density varies with the x-coordinate.
To calculate the mass, we must set up a triple integral because we are working in three-dimensional space. The key is to cover the entire volume of the hemisphere seamlessly, ensuring no part is left out of the calculation.
The step-by-step solution shows us how this can be done by converting Cartesian coordinates to spherical coordinates, which is more suitable for an object like a hemisphere. This approach simplifies the integration process and the limits of integration clearly correspond to the hemisphere's dimensions.
To calculate the mass, we must set up a triple integral because we are working in three-dimensional space. The key is to cover the entire volume of the hemisphere seamlessly, ensuring no part is left out of the calculation.
The step-by-step solution shows us how this can be done by converting Cartesian coordinates to spherical coordinates, which is more suitable for an object like a hemisphere. This approach simplifies the integration process and the limits of integration clearly correspond to the hemisphere's dimensions.
Triple Integral Mass Calculation
Calculating mass using a triple integral is a fundamental concept in multivariable calculus. It involves integrating a density function over a three-dimensional region. The approach used in the exercise involves setting appropriate limits for the triple integral that correspond to the geometric boundaries of the hemisphere.
In this context, we must remember that while setting up the triple integral, each layer of integration—from the innermost to the outermost—appropriately accounts for a different dimension: radial distance, angular sweep in one plane, and angular sweep in the orthogonal plane. In spherical coordinates, these correspond to \(\rho\), \(\phi\), and \(\theta\) respectively.
The provided solution accurately calculates the mass (\(M\)) by incorporating the density function into the triple integral and calculating it over the boundaries of a hemisphere. The answer reflects the integration of density throughout the entire volume.
In this context, we must remember that while setting up the triple integral, each layer of integration—from the innermost to the outermost—appropriately accounts for a different dimension: radial distance, angular sweep in one plane, and angular sweep in the orthogonal plane. In spherical coordinates, these correspond to \(\rho\), \(\phi\), and \(\theta\) respectively.
The provided solution accurately calculates the mass (\(M\)) by incorporating the density function into the triple integral and calculating it over the boundaries of a hemisphere. The answer reflects the integration of density throughout the entire volume.
Spherical Coordinates Integration
Spherical coordinates integration is particularly well-suited for objects like spheres and hemispheres, where points are described by the distance from the origin (\(\rho\)), inclination (\(\phi\)), and azimuthal angle (\(\theta\)).
The provided solution demonstrates the power of spherical coordinates in simplifying the integration process for a hemisphere. By expressing the density function and the boundaries in spherical coordinates, the triple integral becomes more intuitive, and the symmetry of the hemisphere is exploited to make the integration easier.
The region's spherical symmetry means that the azimuthal angle (\(\theta\)) spans from 0 to \(2\pi\), capturing the full circular sweep; the inclination (\(\phi\)) spans from 0 to \(\pi/2\), as it's a hemisphere and not a full sphere; and the radial distance (\(\rho\)) is limited to the hemisphere's radius.
Converting from Cartesian to spherical coordinates introduces the Jacobian determinant, which in spherical coordinates is \(\rho^2 \sin(\phi)\). This factor is crucial to accurately measure volume elements in spherical coordinates. The integration in the provided solution artfully includes this Jacobian, ensuring the mass calculation is correct and the center of mass is determined precisely.
The provided solution demonstrates the power of spherical coordinates in simplifying the integration process for a hemisphere. By expressing the density function and the boundaries in spherical coordinates, the triple integral becomes more intuitive, and the symmetry of the hemisphere is exploited to make the integration easier.
The region's spherical symmetry means that the azimuthal angle (\(\theta\)) spans from 0 to \(2\pi\), capturing the full circular sweep; the inclination (\(\phi\)) spans from 0 to \(\pi/2\), as it's a hemisphere and not a full sphere; and the radial distance (\(\rho\)) is limited to the hemisphere's radius.
Converting from Cartesian to spherical coordinates introduces the Jacobian determinant, which in spherical coordinates is \(\rho^2 \sin(\phi)\). This factor is crucial to accurately measure volume elements in spherical coordinates. The integration in the provided solution artfully includes this Jacobian, ensuring the mass calculation is correct and the center of mass is determined precisely.
Other exercises in this chapter
Problem 50
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