Problem 24 - Thomas Calculus

Question

Evaluate \(\int_{C}(x-y) d x+(x+y) d y\) counterclockwise around the triangle with vertices \((0,0),(1,0),\) and \((0,1) .\)

Step-by-Step Solution

Answer

The line integral evaluates to 1.

Step 1: Identify Path and Function

We are given the vector field \( \mathbf{F} = (x-y, x+y) \). We need to evaluate the line integral \( \int_C (x-y) \, dx + (x+y) \, dy \) counterclockwise around the triangular path with vertices \( (0,0), (1,0), (0,1) \).

Step 2: Check if the Field is Conservative

To determine if the vector field is conservative, we calculate the partial derivatives: \( \frac{\partial F_2}{\partial x} = \frac{\partial}{\partial x} (x+y) = 1 \) and \( \frac{\partial F_1}{\partial y} = \frac{\partial}{\partial y} (x-y) = -1 \). Since these partial derivatives are not equal, the field is not conservative.

Step 3: Apply Green's Theorem

Green's Theorem relates a line integral around a simple closed curve \(C\) to a double integral over the region \(R\) it encloses: \( \oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_R \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) \, dA \).

Step 4: Compute the Double Integral

First, calculate the expression inside the integral: \( \frac{\partial N}{\partial x} = 1 \) and \( \frac{\partial M}{\partial y} = -1 \). The integral becomes \( \iint_R (1 - (-1)) \, dA = \iint_R 2 \, dA \).

Step 5: Evaluate the Integral Over the Triangular Region

The triangular region has vertices \( (0,0), (1,0), (0,1) \). The area \( A \) of this right triangle is \( \frac{1}{2} \cdot base \cdot height = \frac{1}{2} \cdot 1 \cdot 1 = \frac{1}{2} \). Thus, \( \iint_R 2 \, dA = 2 \cdot \frac{1}{2} = 1 \).

Step 6: Conclusion

According to Green's Theorem, the line integral \( \oint_C (x-y) \, dx + (x+y) \, dy = 1 \).

Key Concepts

Green's Theorem

Green's Theorem is a fundamental principle in vector calculus. It provides a relationship between a line integral around a closed curve and a double integral over the region enclosed by the curve. This theorem is particularly useful when dealing with vector fields and can simplify complicated line integrals.

Mathematically, Green's Theorem is expressed as:\[ \oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_R \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) \, dA \]In this equation:

• \( C \) represents a positively oriented simple closed curve in the plane.
• \( R \) is the region bounded by \( C \).
• \( \mathbf{F} = (M, N) \) is a continuously differentiable vector field.

Green’s Theorem essentially tells us that the circulation of the vector field around the curve \( C \) is equal to the sum of the sources and sinks within the region \( R \). This is incredibly useful for simplifying calculations, especially when the area can be easily defined as in the case of polygons like triangles.

Vector Field

A vector field is an assignment of a vector to each point in a subset of space. Vector fields are typically used to model physical quantities that have both a magnitude and direction, such as velocity fields, magnetic fields, or force fields.

For our exercise, we are given the vector field:\[\mathbf{F} = (x-y, x+y)\]In this case, the vector field assigns a vector to each point in the plane, where the components of each vector depend on the position \( (x,y) \).

Studying vector fields helps us understand the behavior of these quantities in space, their potential interactions, and how their properties might affect specific curves and regions. They are fundamental to solving various real-world physics and engineering problems.

Conservative Vector Field

A conservative vector field is a special type of vector field where the line integral between any two points is path-independent. In simpler terms, the field can be described entirely by a scalar potential function, meaning that the work done along a path in the field depends only on the end points, not the particular path taken.

For a vector field \( \mathbf{F} = (M, N) \) to be conservative, the partial derivatives, \( \frac{\partial N}{\partial x} \) and \( \frac{\partial M}{\partial y} \) must be equal:\[\frac{\partial N}{\partial x} = \frac{\partial M}{\partial y}\]In the given exercise, we calculated:

• \( \frac{\partial N}{\partial x} = 1 \)
• \( \frac{\partial M}{\partial y} = -1 \)

Since these are not equal, the field \( \mathbf{F} \) is not conservative.

Being able to identify whether a vector field is conservative is important as it determines whether certain simplifications can be applied to calculating integrals among other uses.

Double Integral

A double integral is a way to integrate over a two-dimensional area. Just like single integrals can find the area under a curve, double integrals can be used to find the volume under a surface and, in certain contexts, can calculate the total sum or mass of a field over a region.

In the application of Green's Theorem, we compute a double integral over the region \( R \) enclosed by the curve \( C \). Our exercise transforms the line integral into a double integral:\[\iint_R (1 - (-1)) \, dA = \iint_R 2 \, dA\]To evaluate this integral, one must determine the limits of integration, which are described by the geometry of the region. In our triangular area defined by vertices \( (0,0) \), \( (1,0) \), and \( (0,1) \), we calculate the area as:\[\text{Area } A = \frac{1}{2} \cdot 1 \cdot 1 = \frac{1}{2}\]
Using this area calculation, the double integral simplifies to:\[2 \cdot \frac{1}{2} = 1\]Double integrals offer a robust toolset for evaluating the cumulative impacts of a field over a two-dimensional area, making them particularly valuable in fields like physics and engineering.