Problem 37 - Calculus Early Transcendentals

Question

Find the mass of a thin wire shaped in the form of the cir- cular are \(y=\sqrt{9}-x^{2}(0 \leq x \leq 3)\) if the density function is \(\delta(x, y)=x \sqrt{y}\)

Step-by-Step Solution

Answer

The mass of the wire is \( \frac{27}{2} \).

Step 1: Determine the parameterization of the curve

The curve is given by the equation \( y = \sqrt{9} - x^2 \). This is a semi-circle centered at the origin with a radius of 3, and since \( 0 \leq x \leq 3 \), we're looking at the upper half. We parameterize this curve as \( x = t \) and \( y = \sqrt{9-t^2} \) for \( 0 \leq t \leq 3 \).

Step 2: Find the expression for arc length differential

To find the mass, we need to consider the arc length differential, \( ds \). For a parameterized curve \( x = t \) and \( y = \sqrt{9-t^2} \), the differential of arc length is \( ds = \sqrt{(dx/dt)^2 + (dy/dt)^2} dt \). Here, \( dx/dt = 1 \) and \( dy/dt = \frac{-t}{\sqrt{9-t^2}} \). Thus, \( ds = \sqrt{1 + \frac{t^2}{9-t^2}} dt = \frac{3}{\sqrt{9-t^2}} dt \).

Step 3: Set up the integral for mass

The mass \( M \) of the wire is calculated using the integral \( M = \int_{a}^{b} \delta(x(t), y(t)) \, ds \). Substituting the density function \( \delta(x, y) = x \sqrt{y} \) and the parameterization, we get \( M = \int_{0}^{3} t \sqrt{\sqrt{9-t^2}} \frac{3}{\sqrt{9-t^2}} dt \). Simplify this to \( M = \int_{0}^{3} 3t dt \).

Step 4: Evaluate the integral

Now, evaluate \( M = 3 \int_{0}^{3} t \, dt \). The antiderivative of \( t \) is \( \frac{t^2}{2} \), so the integral becomes \( 3 \left[\frac{t^2}{2}\right]_{0}^{3} = 3 \left[\frac{9}{2} - 0\right] = \frac{27}{2} \).

Step 5: Conclusion

The mass of the wire is thus \( \frac{27}{2} \) units.

Key Concepts

Parameterization

In calculus, parameterization is a technique used to express a curve using a parameter, often denoted as \( t \). This process is especially crucial when dealing with complex geometrical shapes, as it simplifies the analysis.
The curve given in the exercise, \( y = \sqrt{9} - x^2 \), is a semi-circle with a radius of 3, wherein \( x \) serves as the parameter. To parameterize, we set \( x = t \), which directly gives us the corresponding \( y \) as \( y = \sqrt{9 - t^2} \) for the range from 0 to 3. This parameterization allows us to handle the equation as a set of functions with respect to \( t \), making it easier to compute subsequent calculations like arc length and integrals.

Arc Length Differential

The arc length differential, denoted as \( ds \), represents an infinitesimal segment of a curve's length. It is derived from the parameterization of the curve and plays a crucial role in integral calculus, especially when calculating lengths and other properties involving curves.
For the parameterized curve \( x = t \) and \( y = \sqrt{9 - t^2} \), the arc length element \( ds \) is formulated as \( ds = \sqrt{(dx/dt)^2 + (dy/dt)^2} \, dt \). In this exercise, we find \( dx/dt = 1 \) since \( x = t \), and \( dy/dt = \frac{-t}{\sqrt{9-t^2}} \) by differentiating \( y \) with respect to \( t \). Combining these, the differential becomes \( ds = \frac{3}{\sqrt{9-t^2}} \, dt \), simplifying the calculation for the curve's length.

Integral Calculation

Integral calculation is a fundamental concept in calculus used to find areas, volumes, and in this exercise, the mass of a wire. We begin by setting up an integral expression based on the density function and arc length differential.
The mass \( M \) of the wire is determined using the integral \( M = \int_{0}^{3} \delta(x(t), y(t)) \, ds \), where the density function \( \delta(x, y) = x \sqrt{y} \) is defined based on the curve's parameterization. Substituting the parameterization and expression for \( ds \), the integral is \( M = \int_{0}^{3} t \sqrt{\sqrt{9-t^2}} \cdot \frac{3}{\sqrt{9-t^2}} \, dt \), which simplifies to \( M = \int_{0}^{3} 3t \, dt \). This integral reflects the cumulative mass of the wire across the defined arc.

Density Function

The density function in this context quantifies how mass is distributed along a curve. When given a function like \( \delta(x, y) = x \sqrt{y} \), it indicates how mass changes with respect to the position along the curve.
In the exercise, calculating the mass involves multiplying the density function by small arc length segments \( ds \), and integrating this product gives the total mass. Each element \( \delta(x, y) \cdot ds \) contributes to the integral that represents the wire's mass, providing a precise understanding of how mass density impacts integral evaluation and outcome. The function's nature requires us to consider both \( x \) and \( y \), which are related through the parameterization used, emphasizing the interconnectedness of parameterization and density in providing accurate results.