Problem 40 - Calculus Early Transcendentals

Question

Evaluate the integral \(\int_{C} \mathbf{F} \cdot d \mathbf{r},\) where \(C\) is the boundary of the region \(R\) and \(C\) is oriented so that the region is on the left when the boundary is traversed in the direction of its orientation. \(\mathbf{F}(x, y)=\left(e^{-x}+3 y\right) \mathbf{i}+x \mathbf{j} ; C\) is the boundary of the region \(R\) inside the circle \(x^{2}+y^{2}=16\) and outside the circle \(x^{2}-2 x+y^{2}=3\)

Step-by-Step Solution

Answer

The integral evaluates to \(-24\pi\).

Step 1: Understand the Problem

We need to evaluate the line integral of the vector field \( \mathbf{F}(x, y) = (e^{-x} + 3y) \mathbf{i} + x \mathbf{j} \) along the curve \( C \), which is the boundary of the region \( R \). \( R \) is the area inside the circle \( x^2 + y^2 = 16 \) and outside the circle \( x^2 - 2x + y^2 = 3 \). The curve \( C \) is oriented counterclockwise, having the region \( R \) on the left.

Step 2: Use Green's Theorem

Green's Theorem relates the line integral of a vector field around a closed curve to a double integral over the region enclosed by the curve. Given:\[ \int_C \mathbf{F} \cdot d\mathbf{r} = \int\int_R \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA \]where \( P(x, y) = e^{-x} + 3y \) and \( Q(x, y) = x \). Therefore, calculate the partial derivatives: \( \frac{\partial Q}{\partial x} = 1 \) and \( \frac{\partial P}{\partial y} = 3 \). Substitute into Green's representation:\[ \int_C \mathbf{F} \cdot d\mathbf{r} = \int\int_R (1 - 3) \, dA \]Thus, the formula simplifies to:\[ \int_C \mathbf{F} \cdot d\mathbf{r} = \int\int_R (-2) \, dA \]

Step 3: Compute Region's Area

For the region \( R \), calculate the area formed by the difference between the two circles. The larger circle is \( x^2 + y^2 = 16 \), with radius 4, having an area \( A_1 = \pi \times 4^2 = 16\pi \). The smaller circle, \( (x-1)^2 + y^2 = 4 \), has radius 2, resulting in area \( A_2 = \pi \times 2^2 = 4\pi \). The area of \( R \) is thus \( A_R = A_1 - A_2 = 16\pi - 4\pi = 12\pi \).

Step 4: Integrate Over the Region

Using the area obtained in Step 3, calculate the double integral:\[ \int_C \mathbf{F} \cdot d\mathbf{r} = \int\int_R (-2) \, dA = -2 \times (12\pi) = -24\pi \]

Step 5: Conclusion

Thus, the line integral \( \int_C \mathbf{F} \cdot d\mathbf{r} \) evaluates to \( -24\pi \).

Key Concepts

Line Integral

A line integral is a powerful mathematical tool used to calculate the integral of a function along a curve. In vector calculus, it often involves integrating a vector field over a path or curve. This is particularly useful for work done by a force field or the flow of a fluid along a path.

To compute a line integral of a vector field \( \mathbf{F} \) along a curve \( C \), we express it as \( \int_C \mathbf{F} \cdot d\mathbf{r} \). Here, \( d\mathbf{r} \) represents a small segment along the path \( C \), and \( \mathbf{F} \cdot d\mathbf{r} \) is the dot product between the vector field and the differential path element. This measures how much the vector field is aligned with the path itself.

In practical terms:

• Determine the vector field \( \mathbf{F}(x,y) \).
• Identify the path or curve \( C \).
• Calculate \( \mathbf{F} \cdot d\mathbf{r} \).
• Integrate the results over the curve \( C \).

Line integrals extend the idea of ordinary integrals to calculating values along a line or curve.

Vector Field

A vector field is a function that assigns a vector to every point in a region of space. Picture a weather map where each point has a wind velocity represented by arrows. This is a simple example of a vector field, where each arrow (or vector) indicates direction and magnitude.

In mathematics, vector fields are crucial in studying differential equations, dynamics, and calculus. A vector field \( \mathbf{F}(x, y) \) can be expressed as \( P(x, y)\mathbf{i} + Q(x, y)\mathbf{j} \), where \( P(x, y) \) and \( Q(x, y) \) are functions that determine the components of vectors in the field.

Key features to note:

• Direction: Given by the direction of vectors at different points.
• Magnitude: Represented by the length of vectors, showing the strength or intensity.
• Components: \( P(x, y) \) and \( Q(x, y) \) as the field’s horizontal and vertical components, respectively.

Understanding vector fields helps you to visualize and analyze the behaviors of vectors in physical systems, such as fluid flow or electromagnetic fields.

Double Integral

A double integral extends the concept of a single integral to functions of two variables, allowing for integration over an area in the plane. This type of integral is central to finding areas, volumes, and other quantities that accumulate over two-dimensional regions.

For a region \( R \) in the two-dimensional plane, the double integral is written as \( \iint_R f(x, y) \, dA \), where \( f(x, y) \) is the function being integrated, and \( dA \) is a small area element.

Benefits and applications:

• Calculate the total value of a function over a region \( R \), such as mass, charge, or probability.
• Used in physics for solving problems involving density and concentration over areas.
• Essential in Green's Theorem for converting line integrals into double integrals.

In practice, double integrals often require evaluating iterated integrals, where integration is performed sequentially with respect to each variable over specified intervals.

Area of a Region

In calculus, the area of a region can be calculated using integration methods, particularly when the region has complex boundaries or shapes. For circular or path-defined regions, integration becomes a vital tool.

The area of a region can be determined by setting up a double integral. For example, if your region \( R \) is bounded by simple shapes like circles or rectilinear edges, use the relevant equations to describe these boundaries.

Steps to find the area of a region:

• Identify the boundaries and set up the limits of integration for the region \( R \).
• Set up a double integral to cover the entire area within these boundaries.
• Solve the integral to find the total area.

This method is used effectively in Green's Theorem. It connects line integrals and the area of regions enclosed by curves, enhancing our understanding and capability in multivariable calculus.