Chapter 15: Problem 40
\(\mathbf{F}(x, y)=\left(e^{-x}+3 y\right) \mathbf{i}+x \mathbf{j} ; C\) is the boundary of the re- gion \(R\) inside the circle \(x^{2}+y^{2}=16\) and outside the circle \(x^{2}-2 x+y^{2}=3\)
Short Answer
Expert verified
The line integral is \(-24\pi\).
Step by step solution
01
Understand the Problem
We need to evaluate the line integral of the vector field \( \mathbf{F} \) along the closed curve \( C \), which is the boundary of region \( R \). The region \( R \) is bounded by the circle \( x^2 + y^2 = 16 \) (a circle with radius 4 centered at origin) and lies outside the circle \( x^2 - 2x + y^2 = 3 \), which can be rewritten as \((x-1)^2 + y^2 = 4\) (a circle with radius 2 centered at (1, 0)).
02
Apply Green's Theorem
Since \( C \) is a closed curve and we need to calculate the line integral \( \oint_C \mathbf{F} \cdot d\mathbf{r} \), we can use Green's Theorem. Green's Theorem relates a line integral around a simple closed curve \( C \) to a double integral over the region \( R \) bounded by \( C \). It is given by: \[ \oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_R \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) \, dA \] where \( \mathbf{F}(x, y) = M\mathbf{i} + N\mathbf{j} \). In our case, \( M = e^{-x} + 3y \) and \( N = x \).
03
Compute Partial Derivatives
Calculate the partial derivatives needed for Green's Theorem: \( \frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(x) = 1 \), \( \frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(e^{-x} + 3y) = 3 \).
04
Set Up Double Integral
Substitute the partial derivatives into Green's Theorem: \[ \iint_R \left( 1 - 3 \right) \, dA = \iint_R (-2) \, dA \] This simplifies to: \[ -2 \iint_R \, dA \] which means we need to find the area of the region \( R \) and multiply it by \(-2\).
05
Calculate the Area of Region \( R \)
The area of region \( R \) is the difference between the area of the larger circle (radius 4) and the smaller circle (radius 2). The area of the larger circle is \( \pi \times 4^2 = 16\pi \) and the area of the smaller circle is \( \pi \times 2^2 = 4\pi \). Thus, the area of \( R \) is \( 16\pi - 4\pi = 12\pi \).
06
Evaluate the Double Integral
Finally, multiply the area of \( R \) by \(-2\) to find the value of the integral: \[ -2 \times 12\pi = -24\pi \] Thus, the line integral \( \oint_C \mathbf{F} \cdot d\mathbf{r} = -24\pi \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Green's Theorem
Green's Theorem is a fundamental result in vector calculus that establishes a relationship between a line integral around a simple curve and a double integral over the plane region bounded by the curve. This theorem is very valuable when calculating work done by a force field or other line integrals over closed curves.
In essence, Green's Theorem states that for a continuously differentiable vector field \( \mathbf{F} = M\mathbf{i} + N\mathbf{j} \), the line integral around a closed curve \( C \) can be converted to a double integral over the region \( R \) that \( C \) encloses. The theorem is represented as:
\[ \oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_R \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) \, dA \] Here, \( \oint_C \mathbf{F} \cdot d\mathbf{r} \) represents the work around the curve \( C \), while the double integral calculates the rotation effect of the vector field over the area \( R \).
In essence, Green's Theorem states that for a continuously differentiable vector field \( \mathbf{F} = M\mathbf{i} + N\mathbf{j} \), the line integral around a closed curve \( C \) can be converted to a double integral over the region \( R \) that \( C \) encloses. The theorem is represented as:
\[ \oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_R \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) \, dA \] Here, \( \oint_C \mathbf{F} \cdot d\mathbf{r} \) represents the work around the curve \( C \), while the double integral calculates the rotation effect of the vector field over the area \( R \).
- It applies to regions that are simple, meaning their boundaries are non-intersecting closed curves.
- The domains must be continuously differentiable, and the curve must be positively oriented (counter-clockwise).
Line Integral
A line integral extends the concept of an integral to functions along a curve. It is used in various applications, including physics and engineering, to determine quantities like work, mass, and charge along a path.
In vector calculus, the line integral of a vector field \( \mathbf{F} \) over a path \( C \) is given by \( \int_C \mathbf{F} \cdot d\mathbf{r} \). This represents the cumulative effect of the vector field along the curve.
In vector calculus, the line integral of a vector field \( \mathbf{F} \) over a path \( C \) is given by \( \int_C \mathbf{F} \cdot d\mathbf{r} \). This represents the cumulative effect of the vector field along the curve.
Components of Line Integrals
- Path: The curve \( C \) along which the integral is evaluated. It can be open or closed.
- Vector Field: The function \( \mathbf{F} = M\mathbf{i} + N\mathbf{j} \) that assigns a vector to each point on the path.
- Dot Product: The integral uses the dot product of the vector field and the differential vector \( d\mathbf{r} \).
Double Integral
Double integrals allow us to compute the accumulation of quantities over a two-dimensional area. They extend single integrals from lines or paths to surfaces or regions. This is crucial in calculating areas, volumes, and other physical properties in multiple dimensions.
Typically, a double integral has the form: \[ \\iint_R f(x, y) \, dA \] Here, \( f(x, y) \) is the function being integrated over the region \( R \). The \( dA \) is the differential area element.
Typically, a double integral has the form: \[ \\iint_R f(x, y) \, dA \] Here, \( f(x, y) \) is the function being integrated over the region \( R \). The \( dA \) is the differential area element.
Applications and Process
- Region \( R \): It can be any shape in the 2D plane, often described using inequalities or boundary conditions.
- Evaluation: Often performed using iterated integrals in Cartesian coordinates or using polar coordinates for circular regions.
- Transforming with Green's Theorem: Converts line integrals into these integrals for easy calculation.
Vector Field
A vector field on a plane assigns a vector to every point in that plane. It is a fundamental concept in physics and engineering, representing various quantities like wind speed, magnetic field, or fluid flow.
Vector fields are usually noted as \( \mathbf{F}(x, y) = M\mathbf{i} + N\mathbf{j} \), where \( M \) and \( N \) are functions that describe the field's components along the \( x \)- and \( y \)-axes, respectively.
Vector fields are usually noted as \( \mathbf{F}(x, y) = M\mathbf{i} + N\mathbf{j} \), where \( M \) and \( N \) are functions that describe the field's components along the \( x \)- and \( y \)-axes, respectively.
- Flow: Demonstrates how particles would move through the plane.
- Divergence and Curl: Measure how much the field spreads out or rotates, important for applying Green's Theorem.
- Visualization: Often represented with arrows indicating direction and magnitude.