Problem 98 - Calculus Volume 3

Question

David and Sandra plan to evaluate line integral \(\int_{C} \mathbf{F} \cdot d \mathbf{r}\) along a path in the \(x y\) -plane from (0,0) to \((1,\) 1). The force field is \(\mathbf{F}(x, y)=(x+2 y) \mathbf{i}+\left(-x+y^{2}\right) \mathbf{j}\). David chooses the path that runs along the \(x\) -axis from (0, 0) to (1,0) and then runs along the vertical line \(x=1\) from (1, 0 ) to the final point (1,1) . Sandra chooses the direct path along the diagonal line \(y=x\) from (0,0) to (1,1) . Whose line integral is larger and by how much?

Step-by-Step Solution

Answer

Sandra's integral is larger by \(\frac{1}{6}\).

Step 1: Understand the Problem

We need to evaluate the line integral \(\int_{C} \mathbf{F} \cdot d\mathbf{r}\), where \(\mathbf{F}(x, y) = (x + 2y) \mathbf{i} + (-x + y^2) \mathbf{j}\), along two different paths from \((0, 0)\) to \((1, 1)\). David chooses a piecewise path and Sandra a diagonal path. We have to calculate both integrals and compare them.

Step 2: Evaluate the Line Integral Along David's Path

David's path consists of two parts, \(C_1\) from \((0, 0)\) to \((1, 0)\) along the x-axis, and \(C_2\) from \((1, 0)\) to \((1, 1)\) along the line \(x = 1\).For \(C_1\):\[d\mathbf{r} = dx \mathbf{i}\]\(y = 0\) implies \(\mathbf{F} = (x) \mathbf{i}\).\[\int_{C_1} \mathbf{F} \cdot d\mathbf{r} = \int_{0}^{1} x \, dx = \frac{1}{2}\]For \(C_2\):\[d\mathbf{r} = dy \mathbf{j}\]\(x = 1\) implies \(\mathbf{F} = (1 + 2y) \mathbf{i} + (y^2) \mathbf{j}\).\[\int_{C_2} \mathbf{F} \cdot d\mathbf{r} = \int_{0}^{1} y^2 \, dy = \frac{1}{3}\]The total integral for David's path is:\[\int_C \mathbf{F} \cdot d\mathbf{r} = \frac{1}{2} + \frac{1}{3} = \frac{5}{6}\]

Step 3: Evaluate the Line Integral Along Sandra's Path

Sandra's path is along the diagonal \(y = x\), from \((0,0)\) to \((1,1)\).Along this path:\[d\mathbf{r} = (dx \mathbf{i} + dx \mathbf{j})\] leading to\[\mathbf{F} = (x + 2x) \mathbf{i} + (-x + x^2) \mathbf{j} = (3x) \mathbf{i} + (x^2 - x) \mathbf{j}\]The line integral becomes:\[\int_{C} \mathbf{F} \cdot d\mathbf{r} = \int_{0}^{1} (3x + (x^2 - x)) \, dx = \int_{0}^{1} (2x + x^2) \, dx\]Calculating the integral:\[= \int_{0}^{1} (x^2 + 2x) \, dx = \left[ \frac{x^3}{3} + x^2 \right]_{0}^{1} = \frac{1}{3} + 1 = \frac{4}{3}\]

Step 4: Compare the Results

David's line integral along his path is \(\frac{5}{6}\), while Sandra's is \(\frac{4}{3}\).Convert \(\frac{4}{3}\) to a common denominator with \(\frac{5}{6}\), which is \(\frac{8}{6}\). Therefore, Sandra's line integral is larger.

Key Concepts

Vector Fields

When we talk about vector fields in mathematics, we refer to a function that assigns a vector to each point in a subset of space. Imagine placing a tiny arrow at every point on a plane or in space; these arrows vary in direction and magnitude. In our given exercise, the vector field is described by the equation \(\mathbf{F}(x, y) = (x + 2y) \mathbf{i} + (-x + y^2) \mathbf{j}\). Here, \(\mathbf{i}\) and \(\mathbf{j}\) are the unit vectors along the x and y axes, respectively.
In a practical context, vector fields can represent various physical phenomena such as fluid flow, where each vector depicts the velocity of the fluid at that point, or magnetic and electric fields showing the force that a charged particle would experience.
Understanding vector fields is crucial for visualizing how they behave in space and for proceeding to compute line integrals. We use these fields to understand how a function changes along a path or curve.

Path Independence

Path independence is a very interesting concept in vector calculus. It tells us that if we are dealing with a specific type of vector field, the value of a line integral from one point to another is the same no matter what path we take. Specifically, it is independent of the path between the start and end points.
In our exercise, we compare the line integrals of David's path and Sandra's path from \((0, 0)\) to \((1, 1)\) in the vector field \(\mathbf{F}(x, y)\). However, as shown by the different results of these integrals, the vector field in question does not possess path independence. For a field to have this characteristic, it must be conservative, meaning that \(\mathbf{F}\) must be the gradient of some scalar potential function. If it were a conservative field, David and Sandra would get the same results, no matter their chosen paths.

Parameterized Curves

In calculus, parameterizing a curve means expressing a curve as a path traced out by a set of equations that define the coordinates as functions of some variable, usually \(t\). This is immensely useful when computing line integrals. It allows us to transform the complicated problem of integrating along a curve into a simpler one involving integration with respect to the parameter.
For example, in our exercise, Sandra's path - the straight line directly from \((0,0)\) to \((1,1)\) - can be parameterized with \(x = t\) and \(y = t\) where \(t\) ranges from 0 to 1. Similarly, David's path can be broken down into two parameterized segments - one moving horizontally and the other vertically.
This parameterization is crucial for evaluating their respective line integrals. It helps in simplifying and converting the paths into equations we can easily handle through standard integration techniques.

Calculus

Calculus, particularly integral calculus, is at the heart of understanding and solving line integral problems. It provides us with tools like integration which allow us to find the cumulative effect of a vector field along a path. Line integrals extend the concept of traditional integrals to integrate functions over paths or curves.
In our exercise, calculus helps David and Sandra evaluate the effect of the vector field \(\mathbf{F}(x, y)\) along different paths. By using integration, they find the line integral, which gives insight into how the vector field influences or interacts with different paths.
Line integrals not only account for the functions being integrated but also include the direction and path taken across the plane. This makes integral calculus a powerful method in fields like physics and engineering, where such path-dependent interactions frequently arise.