Problem 19 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

A substance has been put in a centrifuge. We now have a cylindrical sample (radius 3 centimeters, height 4 centimeters) in which density varies with \(x\), the distance (in centimeters) from the central axis. If the density is given by \(\rho(x) \mathrm{mg} / \mathrm{cm}^{3}\), write an integral that gives the total mass of the substance.

Step-by-Step Solution

Answer

The integral that gives the total mass of the substance is \[m = \int_0^4 \int_0^{2\pi} \int_0^3 \rho(x) \cdot r \, dr \, d\theta \, dz\].

Step 1: Understand the relationship between mass, density and volume

The density of an object is defined as its mass divided by its volume, or \(\rho = \frac{m}{V}\). Therefore, the mass of an object can be calculated as the product of its volume and density, or \(m = \rho \cdot V\). In this case, the density is given as \(\rho(x)\) and we are asked to find the total mass. That means we need to integrate the product of density and the differential volume over the whole cylinder.

Step 2: Set up the integral

To set up the integral, we must consider the volume element in cylindrical coordinates. The volume element, \(dV\), in cylindrical coordinates is \(rd \theta dx dz\). Hence, the integral will be the double integral of the product of \(\rho(x)\) and \(dV\), over the limits of \(r\) (0 to 3), \(\theta\) (0 to \(2\pi\)), and \(z\) (0 to 4).

Step 3: Write down the integral

By substituting the given density \(\rho(x)\) and the volume \(dV\) into the equation, we can get the integral form which gives the total mass, \(m\), of the substance. Hence, the integral is\[m = \int_0^4 \int_0^{2\pi} \int_0^3 \rho(x) \cdot r \, dr \, d\theta \, dz\].

Key Concepts

Mass and Density

When dealing with the physical properties of materials, it's essential to understand the relationship between mass and density. Density, symbolized as \( \rho \), is essentially how much 'stuff' is packed into a particular space. In formal terms, it's the mass per unit volume of a substance and is usually expressed as \( \rho = \frac{m}{V} \) where \( m \) is the mass and \( V \) is volume. This measure tells us how compact or spread out the material is within a given area or volume.

In contexts like the given exercise, the substance is often not uniform, meaning its density varies from one point to another. This variable density is represented as \( \rho(x) \) and indicates that density depends on the position (in this case, the distance from the central axis of a cylinder). Such a condition often requires integration to find the total mass, as we must sum up small bits of mass throughout the entire volume of the substance, representing the true spread of the material's density.

Volume Element

To compute mass in three dimensions when dealing with shapes like cylinders, spheres, or cones, it’s crucial to understand the concept of a volume element. A volume element is a small part of the total volume that can be described in a precise mathematical way.

In cylindrical coordinates, which are particularly useful for problems involving symmetry around an axis (like our cylindrical sample), the volume element is defined as \( dV = r dr d\theta dz \). Here, \( r \) represents the radius, \( \theta \) the angle around the central axis, and \( z \) the height along the axis. This represents an infinitesimally small cylinder within the larger one, making it the building block for our integral. We sum up these infinitesimally small volumes over the entire volume of the cylinder to determine properties like mass, when multiplied by the density function.

Cylindrical Sample Mass Calculation

The cylindrical sample mass calculation uses integration to add up small contributions to the overall mass from each volume element within the cylinder. Since density varies as a function of \( x \) within our sample, it's vital to account for the changes in density across different distances from the central axis. The mass \( m \) of the sample can be calculated by integrating the product of density \( \rho(x) \) and the volume element \( dV \) across all the coordinates that outline the shape of the cylindrical sample.

The integral to calculate the total mass, in this case, would be a triple integral over the cylindrical coordinates, written as \[ m = \int_0^4 \int_0^{2\pi} \int_0^3 \rho(x) \cdot r dr d\theta dz \]. This triple integral will accumulate the product of the variable density and the infinitesimal volume pieces across the height (0 to 4 cm), around the full circle (0 to 2\pi radians), and from the central axis to the outer edge of the cylinder (0 to 3 cm). Through this process, we obtain the total mass of the cylindrical sample, accounting for the variable density and cylindrical shape.