Question

Determine all values of the integral $$ \int_{(1,0)}^{(2,2)} \frac{-y d x+x d y}{x^{2}+y^{2}} $$ B on a path not passing through the origin.

Short answer

Question: Verify if the given integral is path-independent and compute its value along a straight line path between (1, 0) and (2, 2). Answer: The integral is path-independent because the curl of the given vector field is zero. The integral along a straight line path between (1, 0) and (2, 2) is approximately equal to the value of the following numerical integral: $$ \int_{0}^{1} \frac{1}{1 + 20t^4} dt $$ The exact value of this integral can be found using numerical integration methods, such as Simpson's rule or the trapezoidal rule.

Step-by-step solution

Verify Path Independence

To verify if the given integral is path-independent, we will check if the given scalar field is conservative. In other words, we will calculate the curl of the vector field: $$ \textbf{F}(x, y) = \frac{-y}{x^2 + y^2} \textbf{i} + \frac{x}{x^2 + y^2} \textbf{j} $$ Let's compute the partial derivatives: $$ \frac{\partial}{\partial y} \frac{-y}{x^2 + y^2} = \frac{x^2 - y^2}{(x^2 + y^2)^2} \\ \frac{\partial}{\partial x} \frac{x}{x^2 + y^2} = \frac{x^2 - y^2}{(x^2 + y^2)^2} $$ Since these partial derivatives are equal, the curl of the vector field is zero, and thus, the vector field is conservative.

Choosing a Path

Since the vector field is conservative, the value of the integral will only depend on the initial and final points, not on the path itself. So, we can choose any path not passing through the origin. Let's pick a straight line path from \((1, 0)\) to \((2, 2)\). The parametric equations for this path are: $$ x(t) = 1 + t \\ y(t) = 2t $$ where \(0 \leq t \leq 1\). Now, let's find \(dx\) and \(dy\) in terms of \(dt\): $$ dx = \frac{d(x(t))}{dt} dt = dt \\ dy = \frac{d(y(t))}{dt} dt = 2 dt $$

Compute the Integral

Substituting for \(x(t)\), \(y(t)\), \(dx\), and \(dy\), we will compute the integral: $$ \begin{aligned} \int_{(1,0)}^{(2,2)} \frac{-y dx + x dy}{x^{2}+y^{2}} & = \int_{0}^{1} \frac{-(2t)(1) + (1+t)(2)}{((1+t)^2+(2t)^2)} dt \\ &= \int_{0}^{1} \frac{2t + 2}{(1 + 2t^2)^2 + (4t^2)^2} dt \end{aligned} $$ Now, we will integrate with respect to \(t\) using a standard antiderivative technique: $$ \begin{aligned} \int_{0}^{1} \frac{2t + 2}{(1 + 2t^2)^2 + (4t^2)^2} dt & = \int_{0}^{1} \frac{1}{1 + 20t^4} dt \\ \end{aligned} $$ Unfortunately, this integral does not have an elementary antiderivative. However, we can now use a numerical integration method, such as Simpson's rule or the trapezoidal rule, to approximate the value of the integral.

Key concepts

Path Independence

In multivariable calculus, path independence is a standout property that can drastically simplify the task of evaluating line integrals. The concept implies that the value of a line integral over a scalar or vector field is the same along any path between two points, as long as the field is conservative and the domain is simply-connected. This concept is particularly useful because, if a field is path-independent, you only need to know the endpoints of the path to calculate an integral, not the specific route taken.

When evaluating the line integral \[ \text{\int_{(1,0)}^{(2,2)} \frac{-y d x+x d y}{x^{2}+y^{2}} \] \[ B \] on a path not passing through the origin, path independence guarantees that the integral's value would be the same regardless of the path chosen. This is under the precondition that the vector field is conservative and does not include singularities, like the origin in this case, where the field becomes undefined.

Conservative Vector Fields

The term conservative vector fields pertains to a class of vector fields where the line integral from one point to another is independent of the path taken. Deducing whether a vector field is conservative can be done in several ways. One common method is to calculate the curl of the vector field; if it's zero everywhere in a simply connected domain, the field is considered conservative. However, it's important to note domains with holes, like punctures where the field is not defined, may complicate matters.

When examining the vector field \[ \textbf{F}(x, y) = \frac{-y}{x^2 + y^2} \textbf{i} + \frac{x}{x^2 + y^2} \textbf{j} \] we determined that it's conservative because the partial derivatives—which make up the components of the curl—are equal, indicating that the curl is zero. Therefore, for our example, any line integral of \textbf{F}\ along a path that doesn't cross a singularity will depend only on the endpoints and not the path itself.

Parameterization

To compute a line integral over a vector field, we often need to parameterize the path of integration. Parameterization involves expressing the x and y components as functions of a third variable, typically denoted as t. This variable t will often traverse an interval from 0 to 1 and serves as a means to describe the path in terms of one variable.

In the case of the integral \[ \text{\int_{(1,0)}^{(2,2)} \frac{-y d x+x d y}{x^{2}+y^{2}} \] we elected a straight-line trajectory from (1, 0) to (2, 2), depicted by parameterized functions \(x(t) = 1 + t\) and \(y(t) = 2t\), with \(t\) stretching from 0 to 1. Through this process, the path turns into a set of parametric equations, allowing us to transform the differential elements \(dx\) and \(dy\) into expressions in terms of \(dt\), facilitating the integration with respect to a single variable.

Numerical Integration Methods

When faced with an integral that does not succumb to elementary antiderivative techniques, we can turn to numerical integration methods to approximate its value. These methods include but are not limited to, the Trapezoidal Rule, Simpson's Rule, and the Midpoint Rule. They work by estimating the area under the curve by summing up areas of simple geometric shapes, typically rectangles or trapezoids, that approximate the function on a finite interval.

For our complex integral \[ \text{\int_{0}^{1} \frac{2t + 2}{(1 + 2t^2)^2 + (4t^2)^2} dt \] that lacks an easy antiderivative, Simpson's Rule or the Trapezoidal Rule offer the tools to estimate its value. These numerical approaches are immensely helpful especially when the function's behavior is well-understood, and a high level of precision is unnecessary. They provide a practical means to solve real-world problems where exact solutions may be elusive or impractical to compute.