Problem 22 - Thomas Calculus

Question

\(\begin{array}{l}{\text { Find the line integral of } f(x, y)=x-y+3 \text { along the curve }} \\ {\mathbf{r}(t)=(\cos t) \mathbf{i}+(\sin t) \mathbf{j}, 0 \leq t \leq 2 \pi}\end{array}\)

Step-by-Step Solution

Answer

The line integral is \( \pi \).

Step 1: Define the Problem

We need to find the line integral of the function \( f(x, y) = x - y + 3 \) along the curve defined by \( \mathbf{r}(t) = (\cos t) \mathbf{i} + (\sin t) \mathbf{j} \), where \( 0 \leq t \leq 2\pi \). This curve describes a full circle of radius 1.

Step 2: Parametrize the Curve

The given curve can be parametrized with \( x = \cos t \) and \( y = \sin t \). We will use these parametric equations to express the function \( f(x, y) \) in terms of \( t \).

Step 3: Substitute Parametric Equations into the Function

Substitute the parametric equations \( x = \cos t \) and \( y = \sin t \) into the function:\[f(x, y) = x - y + 3 = \cos t - \sin t + 3\]

Step 4: Find the Differential Element

The differential element \( \mathbf{dr} \) of the curve is given by:\[\mathbf{dr} = \left(- \sin t \cdot \mathbf{i} + \cos t \cdot \mathbf{j}\right) dt\]

Step 5: Evaluate the Dot Product

Compute the dot product of the vector function \( \cos t - \sin t + 3 \) with the differential path element:\[(\cos t - \sin t + 3) \cdot (- \sin t \mathbf{i} + \cos t \mathbf{j}) = - \cos t \sin t + 3\cos t + \sin^2 t\]

Step 6: Integrate Over the Interval

Integrate the expression from Step 5 over the interval \( t \) from 0 to \( 2\pi \):\[\int_{0}^{2\pi} (- \cos t \sin t + 3\cos t + \sin^2 t) \, dt\] This can be split into simpler integrals:\[I_1 = -\int_{0}^{2\pi} \cos t \sin t \, dt, \quad I_2 = 3 \int_{0}^{2\pi} \cos t \, dt, \quad I_3 = \int_{0}^{2\pi} \sin^2 t \, dt\]

Step 7: Compute Each Integral

Calculate each integral separately:1. \( I_1 = -\frac{1}{2}\int_{0}^{2\pi} \sin(2t) \, dt = 0 \).2. \( I_2 = 3 \left[ \sin t \right]_{0}^{2\pi} = 0 \).3. \( I_3 = \frac{1}{2}\int_{0}^{2\pi} (1 - \cos(2t)) \, dt = \pi \).

Step 8: Combine Calculations

Sum up the evaluated integrals to find the total line integral:\[I = I_1 + I_2 + I_3 = 0 + 0 + \pi = \pi\]

Step 9: Final Step: State the Result

The value of the line integral of \( f(x, y) = x - y + 3 \) along the given curve \( \mathbf{r}(t) \) is \( \pi \).

Key Concepts

Parametrization of Curves

When dealing with curves in mathematics, especially in the context of line integrals, understanding how to parametrize a curve is key. Parametrization involves expressing the coordinates of points on a curve as functions of a variable, often denoted as \( t \). This allows us to transform the problem of analyzing or integrating over a curve into a more straightforward one-dimensional problem.

For instance, in our example where we have a curve depicting a circle of radius 1, this can be parametrized using trigonometric functions:

• \( x = \cos t \)
• \( y = \sin t \)

The parameters \( x \) and \( y \) now explicitly depend on \( t \), meaning as \( t \) progresses from 0 to \( 2\pi \), the point traverses the circle. This transformation is beneficial because integrating over a simpler domain helps simplify complex problems.
Parametrizing curves in vector form, like \( \mathbf{r}(t) = (\cos t) \mathbf{i} + (\sin t) \mathbf{j} \), provides a clear path to integrate vector-valued functions and evaluate line integrals.

Vector Calculus

Vector calculus plays a significant role in analyzing vector fields and understanding the behavior of scalar and vector fields. In our given exercise, vector calculus is utilized to analyze and integrate a function over a path or curve. This function, \( f(x, y) = x - y + 3 \), is evaluated along the parametrized curve.We convert the function into parametric form via given parameter equations. The vector field is obtained as:

• \( f(x, y) = \cos t - \sin t + 3 \)

Now comes the differential element of the curve. It expresses small changes in direction along the curve using vectors. We denote these small changes as:
\[ \mathbf{dr} = (-\sin t \cdot \mathbf{i} + \cos t \cdot \mathbf{j}) \, dt \]
These are necessary to perform the dot product operation with the vector field function. The evaluation proceeds with calculating the dot product of the vector field with this differential element. This step forms an integral over which we perform the line integral around our curve. It showcases the power of vector calculus in managing directions and areas within scalar field functions.

Path Integrals

Path integrals are a central concept when calculating the line integral of a function across a specific path or curve. They allow one to accumulate quantities such as work done by a force field or accumulated values of a function along a path. In this context, the line integral over the path is decomposed into smaller segments, each contributing to the total value.

To evaluate an integral along a path, we:

• Define a path or curve parametric to integrate along, such as a circle in our case.
• Substitute the curve’s parametrization into the function to get an expression in terms of the parameter.
• Perform the dot product with \( \mathbf{dr} \), the differential element vector.

Finally, integrating the expression, while considering the bounds (0 to \( 2\pi \)), accumulates the total effect of the function over the path. The step-by-step integration results in:

• \( I_1 = 0 \) from a sine-identity simplification
• \( I_2 = 0 \), since the integral of \( \cos t \) over one full cycle yields zero
• \( I_3 = \pi \), capturing the effect of \( \sin^2 t \) over the interval

This total gives \( \pi \), illustrating how individual path segments contribute a piece that combines to form the complete solution of the line integral.