Problem 8 - Calculus Early Transcendentals: Pearson New International Edition

Question

\(\int_{C} y d x+x^{2} d y\), \(C\) is the right-angle curve from \((0,-1)\) to \((4,-1)\) to \((4,3)\).

Step-by-Step Solution

Answer

The integral evaluates to 60.

Step 1: Parameterize the First Segment

The curve \( C \) consists of two segments. The first segment goes from \((0, -1)\) to \((4, -1)\). Since \( y = -1 \) is constant we can parametrize this segment as \( x = t \), \( y = -1 \) where \( t \) ranges from 0 to 4.

Step 2: Evaluate the Line Integral Over the First Segment

Now we evaluate \( \int y \, dx + x^2 \, dy \) over the first segment. Since \( dy = 0 \) (as \( y \) is constant), the integral simplifies to \( \int y \, dx = \int_{0}^{4} -1 \, dt = [-t]_{0}^{4} = -4 \).

Step 3: Parameterize the Second Segment

The second segment goes from \((4, -1)\) to \((4, 3)\). Here we can keep \( x = 4 \) constant and let \( y = t \), where \( t \) ranges from -1 to 3.

Step 4: Evaluate the Line Integral Over the Second Segment

For this segment, \( dx = 0 \), so the integral becomes \( \int x^2 \, dy = \int_{-1}^{3} 4^2 \, dt = 16 \int_{-1}^{3} dt = 16[t]_{-1}^{3} = 16(3 - (-1)) = 64 \).

Step 5: Combine the Results

Add the results from both segments to find the total value of the integral over the entire curve. Therefore, the total value is \(-4 + 64 = 60 \).

Key Concepts

Parametrization of Curves

Parametrization is a technique used to represent curves and paths in a simplified manner by expressing them as functions of a single variable, usually denoted as \( t \). This variable, often referred to as the 'parameter', allows us to describe the path of a curve in terms of simple equations for each coordinate.

For line integrals, parametrization is a crucial first step because it translates the problem into a domain where we can apply calculus techniques. Imagine tracing the path of the curve with your finger. The parametrization turns this path into a runway along which our calculations can proceed smoothly.

In our exercise, the parametric equations for the segments of curve \( C \) are given:

• The first segment from \((0, -1)\) to \((4, -1)\) has \( x = t \) and \( y = -1 \), where \( t \) ranges from 0 to 4.
• The second segment from \((4, -1)\) to \((4, 3)\) has \( x = 4 \) and \( y = t \), where \( t \) ranges from -1 to 3.

These equations directly depict the movement along segments of the path and enable the computation of the integral over \( C \). Understanding how to parametrize effectively helps you to tackle complex curves with ease.

Definite Integrals

Definite integrals are a central concept in calculus that measure the accumulation of quantities. In the context of line integrals, they help us determine the total effect exerted along a curve by integrating the function over it.

A line integral computes the sum of values of a function along a curve. It is particularly useful in physics to calculate work done by a force field or flow of fluids along a path. Imagine integrating a small increment along every point in the path — the definite integral generalizes this process to the entire segment.

In this exercise, we see two separate definite integrals; one for each segment of the curve \( C \):

• For the first segment (where \( y = -1 \)): \[ \int_{0}^{4} -1 \, dt = -4 \]
• For the second segment (where \( x = 4 \)): \[ 16 \int_{-1}^{3} dt = 64 \]

The results from both integrals are then added to get a total of 60. This shows how the definite integral accumulates effects, represented by \( y \, dx + x^2 \, dy \), over each part of the path.

Piecewise Functions

Piecewise functions are instrumental when dealing with complex paths in line integrals as they allow us to consider different behavior in the different segments of a curve. They ensure that segments with varying functions or conditions can be evaluated in parts, and their results combined for a comprehensive solution.

In our example, the curve \( C \) was broken into two piecewise segments:

• The horizontal segment from \((0, -1)\) to \((4, -1)\)
• The vertical segment from \((4, -1)\) to \((4, 3)\)

By treating each of these segments as distinct functions, it becomes straightforward to compute the line integrals separately. This piecewise evaluation is a powerful tool since it simplifies complex problems by addressing one segment at a time.

The seamless addition of the results (\(-4\) from the first segment,\(64\) from the second segment) gives the total integral value 60. Dividing the task into piecewise functions makes handling intricate integral problems more manageable and clear.