Chapter 6: Problem 443
Use Green’s theorem to evaluate the following integrals. \(\oint_{C} 3 y d x+\left(x+e^{y}\right) d y,\) where \(C\) is a circle centered at the origin with radius 3
Short Answer
Expert verified
The integral evaluates to \(-18\pi\).
Step by step solution
01
Understanding the Problem
We need to evaluate the line integral \( \oint_{C} \left( 3y \, dx + (x + e^y) \, dy \right) \) where \( C \) is a circle centered at the origin with radius 3.
02
Green's Theorem Setup
Green's Theorem relates a line integral around a closed curve \( C \) to a double integral over the plane region \( D \) it encloses. According to Green's Theorem: \[ \oint_{C} P \, dx + Q \, dy = \int \int_{D} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA \] where \( P = 3y \) and \( Q = x + e^y \).
03
Compute Partial Derivatives
Find the partial derivatives needed for Green's Theorem: \( \frac{\partial Q}{\partial x} = 1 \) and \( \frac{\partial P}{\partial y} = 3 \).
04
Set Up the Double Integral
Plug the partial derivatives into Green's Theorem formula: \[ \int \int_{D} (1 - 3) \, dA = \int \int_{D} -2 \, dA \]
05
Convert to Polar Coordinates
Since \( D \) is a circle of radius 3, we convert to polar coordinates. The area element \( dA \) becomes \( r \, dr \, d\theta \), with \( r \) ranging from 0 to 3 and \( \theta \) from 0 to \( 2\pi \).
06
Evaluate the Double Integral
Calculate the double integral: \[ \int_{0}^{2\pi} \int_{0}^{3} -2r \, dr \, d\theta \] Integrate with respect to \( r \) first: \[ \int_{0}^{3} -2r \, dr = -r^2 \bigg|_0^3 = -(9) + (0) = -9 \] Then with respect to \( \theta \): \[ \int_{0}^{2\pi} -9 \, d\theta = -9\theta \bigg|_0^{2\pi} = -18\pi \]
07
Conclusion
Thus, the value of the line integral \( \oint_{C} 3y \, dx + (x + e^y) \, dy \) is \( -18\pi \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Line Integrals
Line integrals are a type of integral where we integrate a function along a curve. This can be thought of as a way to accumulate values over a path. We generally use line integrals when dealing with vector fields and path-dependent energies.
In mathematical terms, a line integral can be expressed as \( \oint_{C} P \, dx + Q \, dy \), where \( C \) is the curve over which the integral is taken. In our example, \( C \) is a circle, making it a closed curve.
Line integrals have wide applications, particularly in physics to evaluate work done by a force field along a path. By using Green's theorem, line integrals over closed curves can be converted into much easier-to-evaluate double integrals over the region enclosed by the curve.
In mathematical terms, a line integral can be expressed as \( \oint_{C} P \, dx + Q \, dy \), where \( C \) is the curve over which the integral is taken. In our example, \( C \) is a circle, making it a closed curve.
Line integrals have wide applications, particularly in physics to evaluate work done by a force field along a path. By using Green's theorem, line integrals over closed curves can be converted into much easier-to-evaluate double integrals over the region enclosed by the curve.
Double Integrals
Double integrals are used for calculating the volume under a surface within a certain region in the plane. They extend the concept of integration to functions of two variables. These integrals help us compute areas and accumulate values across two-dimensional spaces.
For Green's theorem, double integrals play a key role in relating the circulation around a region—expressed as a line integral—to the values within the region expressed by a double integral. The theorem helps us convert \( \oint_{C} P \, dx + Q \, dy \) into \( \int \int_{D} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA \).
This change is crucial because, in many scenarios, evaluating a double integral is easier than a line integral, especially when a region has symmetry or simple boundaries. Our calculation of the circle centered at the origin with radius 3 effectively leverages this conversion for simplicity.
For Green's theorem, double integrals play a key role in relating the circulation around a region—expressed as a line integral—to the values within the region expressed by a double integral. The theorem helps us convert \( \oint_{C} P \, dx + Q \, dy \) into \( \int \int_{D} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA \).
This change is crucial because, in many scenarios, evaluating a double integral is easier than a line integral, especially when a region has symmetry or simple boundaries. Our calculation of the circle centered at the origin with radius 3 effectively leverages this conversion for simplicity.
Polar Coordinates
Polar coordinates offer a way to describe locations on a plane using the distance from a reference point and an angle from a reference direction. This system is useful in dealing with circular regions or curves centered at a known point, like \( D \) in Green's theorem applications where the enclosed region is a circle.
Coordinates \((r, \theta)\) describe each point within the circle, where \( r \) is the radial distance from the origin, and \( \theta \) is the angle from the positive x-axis.
In our exercise, converting the Cartesian coordinates of double integrals to polar coordinates (i.e., \( dA \) becomes \( r \, dr \, d\theta \)) simplifies the calculation. It is especially effective for circular regions, allowing to neatly integrate over the circular area \( D \). This conversion simplifies the integration process, especially over circular regions, leading to straightforward computation of \( \int_{0}^{2\pi} \int_{0}^{3} -2r \, dr \, d\theta \).
Coordinates \((r, \theta)\) describe each point within the circle, where \( r \) is the radial distance from the origin, and \( \theta \) is the angle from the positive x-axis.
In our exercise, converting the Cartesian coordinates of double integrals to polar coordinates (i.e., \( dA \) becomes \( r \, dr \, d\theta \)) simplifies the calculation. It is especially effective for circular regions, allowing to neatly integrate over the circular area \( D \). This conversion simplifies the integration process, especially over circular regions, leading to straightforward computation of \( \int_{0}^{2\pi} \int_{0}^{3} -2r \, dr \, d\theta \).
Partial Derivatives
Partial derivatives help us understand how a multi-variable function changes with respect to one variable while keeping others constant. They are crucial for applying Green's theorem, where we need derivatives of the component functions \( P \) and \( Q \) with respect to their variables.
In applying Green's theorem to line integrals, we found that \( \frac{\partial Q}{\partial x} = 1 \) and \( \frac{\partial P}{\partial y} = 3 \) for \( Q = x + e^y \) and \( P = 3y \), respectively.
These partial derivatives are part of the expression \( \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \) that ultimately integrates over the area \( D \). By calculating these derivatives, we determine the value of \( -2 \) for the integrand in our converted double integral, simplifying the overall computation extensively.
In applying Green's theorem to line integrals, we found that \( \frac{\partial Q}{\partial x} = 1 \) and \( \frac{\partial P}{\partial y} = 3 \) for \( Q = x + e^y \) and \( P = 3y \), respectively.
These partial derivatives are part of the expression \( \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \) that ultimately integrates over the area \( D \). By calculating these derivatives, we determine the value of \( -2 \) for the integrand in our converted double integral, simplifying the overall computation extensively.