Problem 57 - Calculus Early Transcendentals

Question

Mass from density Find the mass of the following solids with the given density functions. Note that density is described by the function \(f\) to avoid confusion with the radial spherical coordinate \(\rho\). The ball of radius 8 centered at the origin with a density \(f(\rho, \varphi, \theta)=2 e^{-\rho^{3}}\)

Step-by-Step Solution

Answer

Answer: The mass of the ball is approximately \(\frac{4\pi}{3}\).

Step 1: Identify the given density function and describe it in spherical coordinates

The ball has a density function \(f(\rho, \varphi, \theta)=2 e^{-\rho^{3}}\) where \(\rho\) is the radial spherical coordinate. This function describes the distribution of mass throughout the ball.

Step 2: Set up the integral for mass

The mass of the ball can be found by integrating the density function over the entire volume of the ball. In spherical coordinates, the volume element is given by \(dV=\rho^2\sin(\varphi)d\rho d\varphi d\theta\). The mass integral is: $$ m = \int\limits_{0}^{2\pi}\int\limits_{0}^{\pi}\int\limits_{0}^{8}2e^{-\rho^{3}}\rho^{2}\sin(\varphi) d\rho d\varphi d\theta $$

Step 3: Integrate over theta

First, we integrate the mass integral over the \(\theta\)-component: $$ m_{1} = \int_{0}^{8}\int_{0}^{\pi}2e^{-\rho^{3}}\rho^{2}\sin(\varphi) d\varphi d\rho \left[\int_{0}^{2\pi}d\theta\right] $$ The integral over \(\theta\) is simply \(2\pi\). Hence, $$ m_{1} = 2\pi \int_{0}^{8}\int_{0}^{\pi}e^{-\rho^{3}}\rho^{2}\sin(\varphi) d\varphi d\rho $$

Step 4: Integrate over varphi

Next, we integrate the mass integral over the \(\varphi\)-component: $$ m_{2} = 2\pi \int_0^8 e^{-\rho^3}\rho^2 \left[ -\cos(\varphi) \right]_0^{\pi} d\rho $$ Evaluating the integral over \(\varphi\) results in: $$ m_{2} = 2\pi \int_0^8 e^{-\rho^3}\rho^2(2) d\rho = 4\pi \int_0^8 e^{-\rho^3}\rho^2 d\rho $$

Step 5: Integrate over rho

Finally, we integrate the mass integral over the \(\rho\)-component using the substitution method. Let \(u = \rho^3\), then \(du = 3\rho^2 d\rho\): $$ m_{3} = 4\pi \int_0^{8^3} e^{-u}\frac{1}{3} du $$ Evaluating the integral over \(u\) results in: $$ m_{3} = \frac{4\pi}{3} \left[-e^{-u}\right]_0^{8^3} $$

Step 6: Calculate the mass

Now, we can substitute the limits of integration to find the mass of the ball: $$ m = \frac{4\pi}{3} \left[-e^{-(8^3)} + e^0\right] = \frac{4\pi}{3} \left(1 - e^{-512}\right) $$ Approximately, the mass of the ball is equal to: $$ m \approx \frac{4\pi}{3}(1 - 0) = \frac{4\pi}{3} $$

Key Concepts

Density Function

When calculating the mass of a solid object, understanding the concept of a density function is crucial. In mathematics and physics, a density function describes how mass is distributed within a given volume. For this particular exercise, the density function is given by \( f(\rho, \varphi, \theta)=2 e^{-\rho^{3}} \). Here, \( \rho \), \( \varphi \), and \( \theta \) are spherical coordinates representing the radial distance, polar angle, and azimuthal angle, respectively.

• The density function describes how dense the matter is within the ball: higher values of the function mean more mass at that point.
• The density function changes with varying \( \rho \) values, indicating that mass decreases exponentially as \( \rho \) increases due to the function \( e^{-\rho^3} \).

Understanding how the density function varies with the position helps in setting up integrals needed for calculating mass over varying volumes.

Spherical Coordinates

Spherical coordinates are a system of three-dimensional coordinates that are especially handy for calculations involving spheres and partial spheres. The coordinates \( \rho \), \( \varphi \), and \( \theta \) represent:

• \( \rho \): the radial distance from the origin
• \( \varphi \): the polar angle (angle from the positive z-axis)
• \( \theta \): the azimuthal angle in the xy-plane (angle from the positive x-axis)

The use of spherical coordinates in this problem simplifies the task of integrating over a spherical volume, such as a ball. The volume element in spherical coordinates is expressed as \( dV = \rho^2 \sin(\varphi) \, d\rho \, d\varphi \, d\theta \).

• This influence of \( \rho \), \( \varphi \), and \( \theta \) combines to provide a compact way to express regions within a sphere.
• The volume element accounts for the spherical nature of the space, ensuring that all points within a given spherical volume are considered in calculations.

Using spherical coordinates makes the problem of finding volumes or masses within spheres much more manageable compared to using Cartesian coordinates.

Volume Integration

Volume integration is the process of integrating a function over a three-dimensional region. In this exercise, the goal is to integrate the density function over the entire volume of the spherical ball with a radius of 8, using spherical coordinates.

Setting up the integral starts with the expression for mass \( m \) in spherical coordinates:\[m = \int\limits_{0}^{2\pi}\int\limits_{0}^{\pi}\int\limits_{0}^{8}2e^{-\rho^{3}}\rho^{2}\sin(\varphi) \, d\rho \, d\varphi \, d\theta\]This integral is evaluated by sequentially integrating over \( \theta \), \( \varphi \), and \( \rho \):

• First, integrate over \( \theta \), yielding a factor of \( 2\pi \).
• Second, integrate over \( \varphi \), which adjusts the expression based on the limits of the polar angle.
• Lastly, integrate over \( \rho \), using substitution if necessary to handle complex expressions.

The integral evaluations result in finding the total accumulated mass throughout the volume, providing crucial insight into how physical properties distribute across spherical regions. Using volume integration with spherical coordinates is essential for correctly accounting for spherical symmetry in such problems.