Question

In Exercises \(11-14,\) a planar curve \(C\) is given along with a surface \(f\) that is defined over \(C\). Set up the line integral \(\int_{C} f(s) d s,\) then approximate its value using technology. $$ \begin{aligned} &C \text { is the portion of } y=x^{3} \text { on }[-1,1] ; \text { the surface is } f(x, y)=\\\ &2 x+3 y+5 \end{aligned} $$

Short answer

The approximate value of the line integral is 12.26.

Step-by-step solution

Understand the Problem

We need to evaluate the line integral \( \int_{C} f(s) \, ds \), where \( C \) is defined by \( y = x^3 \) for \( -1 \leq x \leq 1 \), and the surface function is \( f(x, y) = 2x + 3y + 5 \).

Parameterize the Curve

Since the curve is defined by \( y = x^3 \), we can parameterize \( C \) using \( x = t \) and \( y = t^3 \) where \( -1 \leq t \leq 1 \).

Write the Function in terms of Parameter

Substitute the parameterization into \( f(x, y) \). Thus, \( f(t, t^3) = 2t + 3(t^3) + 5 = 2t + 3t^3 + 5 \).

Determine the Differential \( ds \)

Using the parameterization, the differential of arc length is \( ds = \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt = \sqrt{1 + (3t^2)^2} \, dt = \sqrt{1 + 9t^4} \, dt \).

Set Up the Line Integral

The line integral becomes \( \int_{-1}^{1} (2t + 3t^3 + 5)\sqrt{1 + 9t^4} \, dt \).

Approximate the Integral Using Technology

Use a numerical integration tool like a calculator or software (e.g., Python, MATLAB) to approximate the value of the integral \( \int_{-1}^{1} (2t + 3t^3 + 5)\sqrt{1 + 9t^4} \, dt \). The approximate value is around 12.26.

Key concepts

Parameterization of Curves

Parameterization of curves is a technique used to express geometric curves using one or more parameters, usually denoted by \( t \). This allows us to describe a curve more easily, particularly for line integration, as it provides a single variable representation instead of multiple variables. In the context of the given exercise, we are dealing with a curve defined by \( y = x^3 \). To parameterize this curve, we set \( x = t \) and consequently \( y = t^3 \), where \( t \) spans from \(-1\) to \(1\).
This parameterization simplifies the process of replacing all occurrences of \( x \) and \( y \) with \( t \) in functions or equations, making it straightforward to handle in integration processes. This way, rather than working with \( y \) as a dependent variable, \( t \) acts as the single parameter, helping streamline computations in the line integration setup.

Numerical Integration

Numerical integration is the method used to approximate the value of integrals when an analytical solution is difficult or impossible to determine. This is particularly useful for calculating line integrals over complex curves or surfaces where an explicit solution isn't apparent. It employs techniques such as the trapezoidal rule, Simpson's rule, or more sophisticated algorithms available in computational tools.

• The goal is to break down the integration range into small sub-intervals.
• Then, the integral is approximated by summing values of the function at these intervals.

In our exercise, after setting up the line integral using parameterization and finding the arc length differential, we use numerical integration to find \( \int_{-1}^{1} (2t + 3t^3 + 5)\sqrt{1 + 9t^4} \, dt \), which roughly evaluates to 12.26. Technologies like Python or MATLAB can automate these calculations, providing a practical approach for solving such integrals.

Arc Length Differential

The arc length differential is a crucial concept when dealing with line integrals over a curve, capturing the influence of the curve's length on the integration process. It represents a small segment of the curve's arc length and is denoted by \( ds \). The formula for \( ds \) is derived using calculus and involves taking the derivative of the curve's parameterization with respect to the parameter.
For our curve \( y = x^3 \) parameterized by \( x = t \), and \( y = t^3 \):

• The differential \( dx/dt = 1 \) since \( x = t \).
• The differential \( dy/dt = 3t^2 \) since \( y = t^3 \).

Thus, the differential arc length is \( ds = \sqrt{(1)^2 + (3t^2)^2} \dt = \sqrt{1 + 9t^4} \, dt \). This formula for \( ds \) takes into account the curve's slope and horizontal differential, ensuring that the line integral correctly accounts for the true geometry of the curve.

Planar Curves

Planar curves are the most simple forms of curves that exist within a two-dimensional plane. They are called so because they lie entirely on a single plane, making them easier to analyze and compute integrals over in comparison to curves that exist in three-dimensional space.
Planar curves can be represented in various ways such as algebraic equations or parameterized forms, as we see in our exercise with \( y = x^3 \). This curve is planar because it can be plotted on a standard Cartesian plane with ease.
When performing line integrations over such curves, like \( C \) in the given exercise, it becomes essential to understand both its structure in the plane and how it interacts with other mathematical entities like surfaces or functions overlaid on these curves. Understanding planar curves helps us perceive the trajectory of the curve on the plane which is fundamental for determining the path over which the integration occurs.