Problem 27
Question
Suppose that \(f(x)\) and \(g(x)\) are continuous functions with \(g(x) \leq f(x) .\) Let \(R\) denote the region bounded by the graph of \(f,\) the graph of \(g,\) and the vertical lines \(x=a\) and \(x=b .\) Let \(C\) denote the boundary of \(R\) oriented counterclockwise. What familiar formula results from applying Green's Theorem to \(\int_{C}(-y) d x ?\)
Step-by-Step Solution
Verified Answer
The line integral \( \int_{C} (-y) \, dx \) equals the area of the region \( R \).
1Step 1: Understanding the Problem
We are given two continuous functions, \( f(x) \) and \( g(x) \), with \( g(x) \leq f(x) \). We need to find a familiar formula using Green's Theorem applied to a line integral \( \int_{C} (-y) \, dx \) over the region \( R \) bounded by \( f(x) \), \( g(x) \), and vertical lines at \( x = a \) and \( x = b \).
2Step 2: Setting Up Green's Theorem
Green's Theorem relates a line integral around a simple, positively oriented, closed curve \( C \) to a double integral over the plane region \( R \) it encloses. If \( C \) is the boundary of \( R \), and \( P(x, y) \, dx + Q(x, y) \, dy \) is a vector field, the theorem states: \[ \oint_{C} P \, dx + Q \, dy = \iint_{R} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA. \]
3Step 3: Identifying Components for the Theorem
For the given integral \( \int_{C} (-y) \, dx \), identify \( P(x, y) = -y \) and \( Q(x, y) = 0 \). Compute the partial derivatives: \[ \frac{\partial Q}{\partial x} = \frac{\partial 0}{\partial x} = 0 \] and \[ \frac{\partial P}{\partial y} = \frac{\partial (-y)}{\partial y} = -1. \]
4Step 4: Applying Green's Theorem
Substitute the partial derivatives into Green's Theorem: \[ \oint_{C} (-y) \, dx = \iint_{R} \left( 0 - (-1) \right) \, dA = \iint_{R} 1 \, dA. \] This simplifies to the expression for the area of region \( R \), i.e., \( \iint_{R} 1 \, dA \) is simply the area of \( R \).
5Step 5: Conclusion
By applying Green's Theorem, we find that the line integral \( \oint_{C} (-y) \, dx \) equates to the double integral representing the area of the region \( R \). Thus, the familiar formula resulting from the application is the formula for calculating the area of a region \( R \).
Key Concepts
Line IntegralContinuous FunctionsArea of a Region
Line Integral
A line integral, sometimes referred to as a path integral, is an extension of the concept of an integral to cover functions defined along a curve. Imagine you are walking along a path and measuring something along the way, like elevation or temperature.
A line integral helps accumulate these values along the path. In mathematics, this is expressed as \[\int_{C} P \, dx + Q \, dy,\]where \(C\) is the path or curve, and \(P\) and \(Q\) are functions that define the vector field.
Line integrals are useful in various fields, including physics and engineering, and help solve numerous problems related to forces and fields.In Green's Theorem, line integrals are used to relate the values on a boundary, like the enclosing line of a region, to properties within the region itself.
This allows the conversion of a detailed pathway measurement into an area integral over a surface.
A line integral helps accumulate these values along the path. In mathematics, this is expressed as \[\int_{C} P \, dx + Q \, dy,\]where \(C\) is the path or curve, and \(P\) and \(Q\) are functions that define the vector field.
Line integrals are useful in various fields, including physics and engineering, and help solve numerous problems related to forces and fields.In Green's Theorem, line integrals are used to relate the values on a boundary, like the enclosing line of a region, to properties within the region itself.
This allows the conversion of a detailed pathway measurement into an area integral over a surface.
Continuous Functions
Continuous functions are a fundamental concept in calculus. They are functions where small changes in input lead to small changes in output. Imagine a smooth road without any bumps or breaks.
That's what a continuous function is like—it has no sudden jumps or gaps.In formal terms, a function \(f(x)\) is continuous over an interval \([a, b]\) if, for every point \(x\) in that interval, the function's limit as it approaches \(x\) is equal to its actual value at \(x\).
This means that you can draw the graph of the function without lifting your pen from the paper.
Continuous functions have several important properties, such as being integrable and having well-defined limits.
That's what a continuous function is like—it has no sudden jumps or gaps.In formal terms, a function \(f(x)\) is continuous over an interval \([a, b]\) if, for every point \(x\) in that interval, the function's limit as it approaches \(x\) is equal to its actual value at \(x\).
This means that you can draw the graph of the function without lifting your pen from the paper.
Continuous functions have several important properties, such as being integrable and having well-defined limits.
- They ensure that integral and limit operations behave predictably.
- Green's Theorem crucially relies on continuity to guarantee the existence of a defined enclosed area and its line integral.
Area of a Region
The area of a region refers to the measure of the size of a two-dimensional space enclosed within a boundary. Calculating this area can be straightforward for regular shapes, but becomes more complex for irregular regions.
Mathematicians use calculus, specifically integration, to find these areas.In the context of Green's Theorem, the area is represented by the double integral \[\iint_{R} 1 \, dA,\]where this assesses every tiny piece \(dA\) within the region \(R\).
When adding up these tiny pieces over the entire region, it produces the total area.Green's Theorem allows translating a line integral around a region's boundary into an area calculation.
This is a powerful application because it turns a potentially complex path integral into a more straightforward task of finding an area.
Mathematicians use calculus, specifically integration, to find these areas.In the context of Green's Theorem, the area is represented by the double integral \[\iint_{R} 1 \, dA,\]where this assesses every tiny piece \(dA\) within the region \(R\).
When adding up these tiny pieces over the entire region, it produces the total area.Green's Theorem allows translating a line integral around a region's boundary into an area calculation.
This is a powerful application because it turns a potentially complex path integral into a more straightforward task of finding an area.
- This is particularly useful in physics and engineering where areas can represent quantities like work or circulation.
- Understanding this principle helps in solving practical problems involving areas of irregular shapes or domains.
Other exercises in this chapter
Problem 26
Find \(\nabla \cdot(\nabla \times \mathbf{F})\) $$ \mathbf{F}(x, y, z)=e^{x z} \mathbf{i}+3 x e^{y} \mathbf{j}-e^{y z} \mathbf{k} $$
View solution Problem 26
Evaluate the line integral along the curve C. $$ \begin{array}{l}{\int_{C} x^{2} d x+x y d y+z^{2} d z} \\ {C: x=\sin t, y=\cos t, z=t^{2} \quad(0 \leq t \leq \
View solution Problem 27
Find \(\nabla \times(\nabla \times \mathbf{F})\) $$ \mathbf{F}(x, y, z)=x y \mathbf{j}+x y z \mathbf{k} $$
View solution Problem 27
Use a CAS to evaluate the line integrals along the given curves. $$ \begin{array}{l}{\text { (a) } \int_{C}\left(x^{3}+y^{3}\right) d s} \\\ {\quad C: \mathbf{r
View solution