Question

Find the mass of the following objects with the given density functions. The ball of radius 8 centered at the origin with a density \(f(\rho, \varphi, \theta)=2 e^{-\rho^{3}}\)

Short answer

Answer: The mass of the object is \(M = \frac{4\pi}{3}(1 - e^{-512})\).

Step-by-step solution

Set up the triple integral

We will convert the given integration from Cartesian to spherical coordinates using the conversion formula: $$ dV = \rho^2 \sin(\varphi) \, d\rho \, d\varphi \, d\theta $$ Then the triple integral representing the mass of the object is: $$ M = \int\int\int f(\rho, \varphi, \theta) \rho^2 \sin(\varphi) \, d\rho \, d\varphi \, d\theta $$ where \(f(\rho,\varphi,\theta) = 2e^{-\rho^3}\).

Determine the bounds

With the ball centered at the origin and radius \(8\), we need to determine the appropriate bounds for each coordinate. For \(\rho\), the distance from the origin, it ranges from \(0\) to \(8\). Thus, $$ 0 \le \rho \le 8 $$ For \(\varphi\), the polar angle, it ranges from \(0\) to \(\pi\). Hence, $$ 0 \le \varphi \le \pi $$ For \(\theta\), the azimuthal angle, it ranges from \(0\) to \(2\pi\). Therefore, $$ 0 \le \theta \le 2\pi $$

Evaluate the triple integral

Now, substitute the density function and bounds into the integral, and solve: $$ M = \int_{0}^{8} \int_{0}^{\pi} \int_{0}^{2\pi} 2e^{-\rho^3} \rho^2 \sin(\varphi) \, d\rho \, d\varphi \, d\theta $$ We can split the integral into three separate integrals, as the variables are independent: $$ M = \left(\int_{0}^{8} 2e^{-\rho^3} \rho^2 \, d\rho\right) \left(\int_{0}^{\pi} \sin(\varphi) \, d\varphi\right) \left(\int_{0}^{2\pi} d\theta\right) $$ Evaluate each integral separately: $$ \int_{0}^{8} 2e^{-\rho^3} \rho^2 \, d\rho = -\frac{2}{3}e^{-\rho^3}\Bigg|_0^8 = -\frac{2}{3}(e^{-8^3} - 1) = \frac{2}{3}(1 - e^{-512}) $$ $$ \int_{0}^{\pi} \sin(\varphi) \, d\varphi = \left[-\cos(\varphi)\right]_0^\pi = -[\cos(\pi) - \cos(0)] = 2 $$ $$ \int_{0}^{2\pi} d\theta = \theta \Bigg|_0^{2\pi} = 2\pi $$ Now, multiply the results of the three integrals together to find the mass: $$ M = \left(\frac{2}{3}(1 - e^{-512})\right)(2)(2\pi) = \frac{4\pi}{3}(1 - e^{-512}) $$ So, the mass of the ball with the given density function is \(M = \frac{4\pi}{3}(1 - e^{-512})\).

Key concepts

Triple Integral

A triple integral is used to compute volumes or other properties, like mass, in three-dimensional spaces. Specifically, it's a way to integrate over a volume. For spherical coordinates, the infinitesimally small volume element is given by \[dV = \rho^2 \sin(\varphi) \, d\rho \, d\varphi \, d\theta\]This formula comes from transforming the integration from Cartesian to spherical coordinates. In spherical coordinates, a point in space is defined by its distance from the origin (\(\rho\)), the polar angle (\(\varphi\)), and the azimuthal angle (\(\theta\)). By using the triple integral, we can sum up infinitesimal volumes, weighted by the density function, to find the total mass. This concept is particularly useful in physics for regular shapes like spheres.

Density Function

The density function describes how mass is distributed within the object. For this particular problem, the density function given is \(f(\rho, \varphi, \theta) = 2e^{-\rho^3}\). It indicates that as you move away from the origin, the density decreases exponentially based on \(\rho\), the radial distance.

• The function \(2e^{-\rho^3}\) tells us the concentration of mass at any point within the ball.
• The exponential term \(e^{-\rho^3}\) ensures that the density diminishes more sharply the further we get from the center.

Understanding the density function is crucial, as it directly affects how much mass is present at each infinitesimal volume in the sphere.

Mass Calculation

The mass of an object can be calculated with a triple integral that incorporates the density function. The triple integral takes the form \[M = \int\int\int f(\rho, \varphi, \theta) \rho^2 \sin(\varphi) \, d\rho \, d\varphi \, d\theta\]This involves:

• The density function \(f(\rho, \varphi, \theta)\) within the integral, representing mass per unit volume.
• The infinitesimal volume element \(\rho^2 \sin(\varphi) \, d\rho \, d\varphi \, d\theta\).

In our example, upon evaluating the separate integrals based on the object's symmetry and density distribution in spherical coordinates, we determined that the mass is \(M = \frac{4\pi}{3}(1 - e^{-512})\). This means that the total mass is spread throughout the volume of the sphere, diminishing with distance from the center.

Bounds of Integration

The bounds of integration define the limits on each of the spherical coordinates, dictating the region over which the integration is conducted. In this problem:

• \(\rho\), representing the radial distance, ranges from \(0\) to \(8\), as it cannot exceed the sphere's radius.
• \(\varphi\), the polar angle, ranges from \(0\) to \(\pi\), covering the full range from the "north pole" to the "south pole" of the sphere.
• \(\theta\), the azimuthal angle, ranges from \(0\) to \(2\pi\), completing a full rotation around the vertical axis.

These bounds ensure that the integral evaluates over the entire volume of the sphere, including all points from its surface down to its center.