Problem 16 - Calculus Early Transcendentals

Question

Find the mass of the following thin bars with the given density function. $$\rho(x)=\left\\{\begin{array}{ll} x^{2} & \text { if } 0 \leq x \leq 1 \\ x(2-x) & \text { if } 1

Step-by-Step Solution

Answer

Answer: The mass of the thin bar is \(\frac{2}{3}\).

Step 1: Find the mass of the first region

To find the mass of the first region, we need to integrate the density function (\(x^2\)) from \(0\) to \(1\). This is done by the following integral: $$ m_1 = \int_{0}^{1} \rho(x) \, dx = \int_{0}^{1} x^{2} \, dx $$

Step 2: Calculate the integral for the first region

To calculate the integral, we will use the power rule for integration, which states: $$ \int x^n dx = \frac{x^{n+1}}{n+1} + C $$ Applying the power rule to our integral above, we get: $$ m_1 = \left[ \frac{x^3}{3} \right]_{0}^{1} = \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3} $$

Step 3: Find the mass of the second region

Now, we need to find the mass of the second region by integrating the density function (\(x(2-x)\)) from \(1\) to \(2\): $$ m_2 = \int_{1}^{2} \rho(x) \, dx = \int_{1}^{2} x(2-x) \, dx $$

Step 4: Calculate the integral for the second region

First, expand \(x(2-x)\) then apply the power rule for integration: $$ m_2 = \int_{1}^{2} (2x - x^2) \, dx = \left[ 2\frac{x^2}{2} - \frac{x^3}{3} \right]_{1}^{2} = \left(2 - \frac{8}{3}\right) - \left(1 - \frac{1}{3}\right) = \frac{1}{3} $$

Step 5: Add masses of both regions to find the total mass

Finally, add \(m_1\) and \(m_2\) to find the total mass of the thin bar: $$ m = m_1 + m_2 = \frac{1}{3} + \frac{1}{3} = \frac{2}{3} $$ So, the mass of the thin bar is \(\frac{2}{3}\).

Key Concepts

Density Function

The concept of the density function is fundamental in calculus, especially when dealing with objects of varying density. A density function, denoted typically by \( \rho(x) \), describes how much mass is concentrated at each point along an object. It allows us to understand and compute the distribution of mass across an object.

In the given problem, the density function \( \rho(x) \) is provided in a piecewise manner. This means the density changes depending on the range of \( x \).

• For \( 0 \leq x \leq 1 \), \( \rho(x) = x^2 \), indicating an increasing density as \( x \) increases.
• For \( 1 < x \leq 2 \), \( \rho(x) = x(2-x) \), which suggests the density first increases and then decreases within this interval.

By understanding the density function, we can better grasp how the mass of an object, like a bar in this example, changes across its length. This forms the foundation for using calculus to determine total mass or other related dynamic properties.

Integral Calculus

Integral calculus is a powerful mathematical tool that allows us to find quantities like area under a curve, total accumulation, or, as in this exercise, the mass of an object with a given density function. The integral of a function over a specified interval gives us a way to sum infinitely many small quantities effectively.

In the example provided, we use integrals to determine the mass of a bar with a varying density. To do this, one needs to:

• Set up an integral using the density function as the integrand.
• Define the limits of integration based on the region of interest.
• Calculate the integral, which involves finding an antiderivative and evaluating it at the boundaries.

The integration over the two regions with the given density functions is essential to capture the complete behavior of the mass across the bar. With integral calculus, we are piecing together the contributions of mass from infinitesimally small segments of the bar to find the total mass.

Integration by Parts

Integration by parts is an integration technique derived from the product rule of differentiation. However, in this particular problem, we didn't directly use integration by parts, but it's relevant since it can help solve more complex integral equations that can arise in density function problems.

When employing integration by parts, remember the key formula:\[\int u \, dv = uv - \int v \, du\]
This formula is particularly useful when dealing with products of functions where one function can be easily differentiated, and the other can be easily integrated.

• Choose \( u \) and \( dv \) carefully based on ease of differentiation and integration.
• Compute \( du \) (the derivative of \( u \)) and \( v \) (the integral of \( dv \)).
• Substitute into the formula and simplify.

While integration by parts wasn't used in this case due to the straightforward nature of the density functions, it's a valuable tool for handling more challenging integrals you might encounter in advanced density function problems.