Problem 32 - Thomas Calculus

Question

Integral along different paths Evaluate the line integral \(\int_{C} 2 x \cos y d x-x^{2} \sin y d y\) along the following paths \(C\) in the \(x y\) -plane. \begin{equation}\begin{array}{l}{\text { a. The parabola } y=(x-1)^{2} \text { from }(1,0) \text { to }(0,1)} \\ {\text { b. The line segment from }(-1, \pi) \text { to }(1,0)} \\ {\text { c. The } x \text { -axis from }(-1,0) \text { to }(1,0)} \\ {\text { d. The astroid } \mathbf{r}(t)=\left(\cos ^{3} t\right) \mathbf{i}+\left(\sin ^{3} t\right) \mathbf{j}, 0 \leq t \leq 2 \pi, \text { counterclockwise }} \\ {\text { from }(1,0) \text { back to }(1,0)}\end{array}\end{equation}

Step-by-Step Solution

Answer

Solution requires evaluating line integrals along four different paths.

Step 1

Recognize that the given vector field for this line integral is \(\mathbf{F}(x,y) = (2x\cos y, -x^2 \sin y)\). We are asked to compute the line integral \(\int_{C} \mathbf{F} \cdot d\mathbf{r}\).

Step 2 (Path a - Parabola)

For the parabola \(y = (x-1)^2\) from \((1,0)\) to \((0,1)\), parameterize the curve as \(\mathbf{r}(t) = (1-t, (1-t)-1)^2)\), where \(t\) goes from 0 to 1. Compute \(d\mathbf{r} = (-1, 2(1-t)(-1)) dt = (-1, -2(1-t)) dt\). Plug the parametric equations into \(\mathbf{F}(x, y)\) and integrate.

Step 3 (Path a - Calculate Integral)

Substitute \(x=1-t\) and \(y=t^2\) into \(\mathbf{F}(x,y)\) and calculate the line integral \(\int_{0}^{1} (2(1-t)\cos(t^2))( -1) + (-(1-t)^2 \sin(t^2))(-2t) dt\). Evaluate the integral.

Step 4 (Path b - Line Segment)

Parameterize the line segment from \((-1, \pi)\) to \((1, 0)\) as \(\mathbf{r}(t) = (-1+2t, \pi-\pi t)\), where \(t\) goes from 0 to 1. Find \(d\mathbf{r} = (2, -\pi) dt\).

Step 5 (Path b - Calculate Integral)

Substitute \(x = -1+2t\) and \(y = \pi-\pi t\) into \(\mathbf{F}(x,y)\) and compute \(\int_{0}^{1} (2(-1+2t)\cos(\pi - \pi t))(2) + (-(1+2t)^2 \sin(\pi - \pi t))(-\pi) dt\). Evaluate.

Step 6 (Path c - x-axis)

For the x-axis from \((-1,0)\) to \((1,0)\), parameterize \(\mathbf{r}(t) = (t, 0)\) where \(t\) goes from -1 to 1. Calculate \(d\mathbf{r} = (1, 0) dt\).

Step 7 (Path c - Calculate Integral)

Substitute \(y=0\) into \(\mathbf{F}(x,y) = (2x, 0)\). Evaluate the integral \(\int_{-1}^{1} 2x \, dx\).

Step 8 (Path d - Astroid)

Parameterize the astroid using the given parametric equations. Find \(d\mathbf{r} = (d(\cos^3 t)/dt, d(\sin^3 t)/dt) dt\).

Step 9 (Path d - Calculate Integral)

Substitute \(x=\cos^3 t\) and \(y=\sin^3 t\) into \(\mathbf{F}(x,y)\) and integrate \(\int_{0}^{2\pi} (2(\cos^3 t)\cos(\sin^3 t))(d(\cos^3 t)/dt) + (-(\cos^6 t)\sin(\sin^3 t))(d(\sin^3 t)/dt) dt\). Then evaluate.

Key Concepts

Vector Fields

In the context of line integrals, vector fields are essential as they provide a field of vectors upon which you perform the integration. A vector field is typically denoted by \( \mathbf{F}(x, y) \), representing vectors at each point in a plane or space. For our exercise, the vector field is given by \( \mathbf{F}(x, y) = (2x\cos y, -x^2 \sin y) \). This means that at any point \((x, y)\) in the plane, the vector

• has a component \(2x\cos y\) in the \(x\) direction
• and a component \(-x^2 \sin y\) in the \(y\) direction.

This information helps us understand how the vectors vary across the field, which is crucial for determining how they will interact with different paths or curves over which we wish to perform the line integral.

Parameterization

Parameterization is the process of expressing some mathematical elements using a parameter, typically denoted by \( t \). It simplifies the integration process by describing curves in terms of a single variable, rather than two separate variables \( (x, y) \).
For instance, while examining the parabola \( y = (x-1)^2 \) from point \((1, 0)\) to \((0, 1)\), we use the parameter \( t \) to express these points as functions of \( t \). The parameterization of the curve is \( \mathbf{r}(t) = (1-t, (1-t)^2) \), with \( t \) going from 0 to 1.

• This means at \( t = 0 \), \( \mathbf{r}(0) = (1, 0) \).
• At \( t = 1 \), \( \mathbf{r}(1) = (0, 1) \).

This way, parameterization converts a curve in the plane into a straight line segment in the 'parameter space', making calculations straightforward during integration.

Curve Integration

Curve integration, or line integration, refers to the integration of a vector field along a curve. The purpose is to sum the vector field's influence along the path described by the curve.
When performing curve integration, the notation \( \int_{C} \mathbf{F} \cdot d\mathbf{r} \) represents integrating the vector field \( \mathbf{F} \) along the curve \( C \). The dot \( \cdot \) indicates that we're computing the dot product between the vector field and the differential element of the curve, leading to the field's projection along the curve.

• A differential element \( d\mathbf{r} \) is derived via parameterization as \( (dx, dy) \).
• It reflects an infinitesimal change following the curve's path.

This process helps determine how much of the vector field is directed along the curve, which directly impacts the integral's result.

Integral Evaluation

Once a curve is parameterized, and you understand the vector field, you can perform the integral evaluation. This involves calculating the actual line integral by substituting the parameterized expressions into the integral.
Take the parameterized form and substitute it into the expressions for \( \mathbf{F}(x, y) \), then evaluate the integral over the specified interval of \( t \). For example, with our parabola parameterization, we substitute the variables as \( x = 1-t \) and \( y = (1-t)^2 \) into the vector field.

• The line integral becomes an ordinary integral in terms of \( t \).
• Carry out integration by applying integral calculus techniques like substitution and evaluating it over the limits of \( t \).

Through precise evaluation, you derive the total effect the vector field has along the specified path, yielding the value of the integral.