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Magazine Chapter 10: Problem 20
Find the total mass of a mass distribution of density \(\rho\) in region \(\mathrm{V}\) of space: $$ \rho=r^{3} e^{-r} $$ V: all space
Short Answer
Expert verified
The total mass is \( 480\pi \).
Step by step solution
01
Understand the Density Function
The given density function is \( \rho = r^{3} e^{-r} \), where \( r \) is the radial distance from the origin in spherical coordinates. This type of function is radial and depends only on the distance \( r \), not on angles.
02
Set Up the Mass Integral
To find the total mass of a continuous density distribution, integrate the density \( \rho \) over the entire region \( V \). Since \( V \) is all space in spherical coordinates, we integrate \( \rho \) from \( r = 0 \) to \( \infty \), including the spherical volume element \( dV = r^2 \sin \theta \, dr \, d\theta \, d\phi \).
03
Write the Integral for Total Mass
The integral for total mass \( M \) is\[M = \int_{0}^{\infty} \int_{0}^{\pi} \int_{0}^{2\pi} \rho(r) \, r^2 \sin \theta \, dr \, d\theta \, d\phi.\]Substituting \( \rho(r) = r^{3} e^{-r} \) gives\[M = \int_{0}^{\infty} \int_{0}^{\pi} \int_{0}^{2\pi} r^{3} e^{-r} \cdot r^2 \sin \theta \, dr \, d\theta \, d\phi.\]
04
Evaluate the Angular Integrals
The integral over \( \theta \) is \( \int_{0}^{\pi} \sin \theta \, d\theta = 2 \) and the integral over \( \phi \) is \( \int_{0}^{2\pi} d\phi = 2\pi \). Hence the angular part equals \( 4\pi \).
05
Evaluate the Radial Integral
The radial integral becomes\[\int_{0}^{\infty} r^{5} e^{-r} \, dr.\]This integral resembles the gamma function \( \Gamma(n) = \int_{0}^{\infty} x^{n-1} e^{-x} \, dx \). Setting the substitution \( r \to x \), the integral becomes \( \Gamma(6) \).
06
Compute the Gamma Function
The value \( \Gamma(n) \) is \( (n-1)! \). Therefore, \( \Gamma(6) = 5! = 120 \).
07
Combine Results
Combining all parts, the total mass \( M \) is calculated as follows:\[M = 4\pi \times 120 = 480\pi.\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Density Function
In the context of mass distribution, a density function, such as \( \rho = r^{3} e^{-r} \), plays a crucial role. It describes how mass is distributed in a given region. This particular function depends on the radial distance \( r \), meaning it is radial. Understanding this can simplify complex problems, as the influence of angles is absent.
- By using spherical coordinates where \( r \) is the distance from the origin, the density function becomes \( \rho(r) \).
- This radial dependency makes the integrals easier to manage since they are functions of \( r \) only, not \( \theta \) or \( \phi \).
Mass Integral
The mass integral helps us find the total mass of an object by integrating the density over a volume. For a spherical distribution, we integrate over all space.
- The formula for the mass integral is given by \[ M = \int_{0}^{\infty} \int_{0}^{\pi} \int_{0}^{2\pi} \rho(r) \, r^2 \sin \theta \, dr \, d\theta \, d\phi. \]
- This accounts for the entire region by covering all possible values of \( r \), \( \theta \), and \( \phi \).
Gamma Function
In some cases, evaluating integrals can be tackled using the gamma function. This special function is particularly useful when the integral resembles a specific form.
- The gamma function \( \Gamma(n) = \int_{0}^{\infty} x^{n-1} e^{-x} \, dx \) can simplify the process.
- For our radial integral, \( \int_{0}^{\infty} r^{5} e^{-r} \, dr \) is recognized as \( \Gamma(6) \).
- Knowing that \( \Gamma(n) = (n-1)! \), helps in easily finding values without computing the integral manually.
Spherical Coordinates
Spherical coordinates offer a convenient way of expressing three-dimensional problems, especially where symmetry about a point is involved. They are defined by three parameters: radial distance \( r \), polar angle \( \theta \), and azimuthal angle \( \phi \).
- Radial distance \( r \) indicates the distance from the origin.
- The polar angle \( \theta \) measures the angle from the positive z-axis.
- The azimuthal angle \( \phi \) is the angle from the positive x-axis in the xy-plane.
Radial Integral
The radial integral operates over the radial component, \( r \), of spherical coordinates. It captures the mass contribution due to density as it extends radially.For our density function, the radial part of the integral is:\[ \int_{0}^{\infty} r^{5} e^{-r} \, dr. \]
- This function grows as \( r \) increases initially (due to \( r^5 \)) but eventually diminishes to zero with \( e^{-r} \).
- The evaluation of this integral is made straightforward by recognizing it as a scaled gamma function \( \Gamma(6) \).
- Computing this yields the factorial result, critical to finding the total mass.
