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 4 centered at the origin with a density \(f(\rho, \varphi, \theta)=1+\rho\)

Short answer

The mass of the given solid with the density function \(f(\rho, \varphi, \theta) = 1+\rho\) is \(384\pi\).

Step-by-step solution

Convert the density function to spherical coordinates

To convert the density function to spherical coordinates, we will use the transformation rule for Jacobian. The Jacobian for spherical coordinates is given by \(\rho^2\sin\varphi\). Therefore, our density function in spherical coordinates will be: \(f(\rho,\varphi,\theta) = (1+\rho)(\rho^2\sin\varphi)\).

Determine the limits of integration

Since the solid is a ball of radius 4 centered at the origin, the limits of integration for the spherical coordinates are: $$ 0 \le \rho \le 4 \\ 0 \le \varphi \le \pi \\ 0 \le \theta \le 2\pi $$

Write and evaluate the triple integral

Now, let's write down the triple integral representing the mass of the solid: $$ M = \int_{0}^{4} \int_{0}^{\pi} \int_{0}^{2\pi} (1+\rho)(\rho^2\sin\varphi)\,d\theta\,d\varphi\,d\rho $$ To evaluate this triple integral, we will need to first integrate with respect to \(\theta\), then \(\varphi\), and finally \(\rho\).

Step 3.1: Integrate with respect to \(\theta\)

Integrating with respect to \(\theta\), we get: $$ M = \int_{0}^{4} \int_{0}^{\pi} (1+\rho)(\rho^2\sin\varphi) [\theta]_0^{2\pi}\,d\varphi\,d\rho = 2\pi \int_{0}^{4} \int_{0}^{\pi}(1+\rho)(\rho^2\sin\varphi)\,d\varphi\,d\rho $$

Step 3.2: Integrate with respect to \(\varphi\)

Integrating with respect to \(\varphi\), we get: $$ M = 2\pi \int_{0}^{4} (\rho^2+\rho^3) [-\cos\varphi]_0^{\pi}\,d\rho = 2\pi \int_{0}^{4} 2(\rho^2+\rho^3)\,d\rho $$

Step 3.3: Integrate with respect to \(\rho\)

Integrating with respect to \(\rho\), we get: $$ M = 2\pi [\frac{2}{3}\rho^3 + \frac{1}{4}\rho^4]_0^{4} = 2\pi (\frac{2}{3}(4)^3 + \frac{1}{4}(4)^4) = 2\pi (128 + 64) = 384\pi $$ So, the mass of the given solid with the density function \(f(\rho, \varphi, \theta) = 1+\rho\) is \(384\pi\).

Key concepts

Density Function

In the realm of spherical integration, the density function is a pivotal concept. It's an expression that describes how mass is distributed throughout a solid object. In this exercise, we find that the density is described by the function \(f(\rho, \varphi, \theta)=1+\rho\). This function varies as a function of the radial distance \(\rho\) from the center of the sphere, impacting the overall mass calculation.

The density function can vary based on different parameters such as radius, angle, and height, depending on the context of the problem. When we express density functions in spherical coordinates, it's important to adjust for the transformation from Cartesian coordinates, which often involves a Jacobian transformation. This process accounts for the curvature and dimensionality differences when moving from one coordinate system to another.

Understanding the density function allows us to calculate important physical properties, like mass, by integrating it over the volume of the object. It's essentially how the mass per unit volume changes in space. This concept is crucial when solving physics problems involving mass distribution, such as this spherical ball.

Triple Integral

Triple integrals extend the concept of single or double integration to three dimensions, allowing us to compute volumes and properties like mass for 3D objects. In this exercise, we use the triple integral to find the mass of a solid sphere, considering its varying density.

A triple integral integrates a function over a three-dimensional region. In our scenario, we write the integral as:
\[M = \int_{0}^{4} \int_{0}^{\pi} \int_{0}^{2\pi} (1+\rho)(\rho^2\sin\varphi)\,d\theta\,d\varphi\,d\rho\]
This illustrates integrating first with respect to \(\theta\), then \(\varphi\), and finally \(\rho\). Each integration step simplifies the expression, ultimately scaling the volume according to density. The triple integral not only encompasses the entire volume of the sphere but also considers how the density varies within this volume. This step-by-step approach is essential for obtaining accurate mass and volumetric measurements.

The limits of integration are critical, as they define the boundaries of the region we are considering. In spherical coordinates, these are defined by the geometry of the sphere: \(\rho\) for radial distance, \(\varphi\) for polar angle, and \(\theta\) for azimuthal angle. The triple integral process is a powerful mathematical tool for solving complex physical problems involving three-dimensional objects.

Spherical Coordinates

The concept of spherical coordinates is key to simplifying complex geometry problems, such as the computation of mass in this exercise. Unlike Cartesian coordinates, which involve \(x, y, z\) distances, spherical coordinates use three variables: \(\rho\), \(\varphi\), and \(\theta\).

These coordinates are particularly useful for spherical objects or problems involving radial symmetry. Here's a brief breakdown:

• \(\rho\) denotes the radial distance from the origin to a point within the sphere.
• \(\varphi\) represents the polar angle, measured from the positive z-axis.
• \(\theta\) is the azimuthal angle, rotating around the z-axis, similar to the angle in polar coordinates on the xy-plane.

When a function is described in spherical coordinates, the integration and transformations require specific alterations, like multiplying by \(\rho^2 \sin \varphi\) for the Jacobian, ensuring we account for the geometry's natural curvature.
This system is indispensable for physics and engineering tasks that involve spherical objects, as it more accurately reflects the true nature of space and dimension in spherical shapes than traditional Cartesian coordinates do. Problems defined in this way often involve more intuitive limits and simpler integrals, making calculations more straightforward and manageable.