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Magazine Chapter 16: Problem 11
Use Green's Theorem to evaluate \(\int_{C} \mathbf{F} \cdot d \mathbf{r} .\) (Check the orientation of the curve before applying the theorem.) \(\mathbf{F}(x, y)=\left\langle\sqrt{x}+y^{3}, x^{2}+\sqrt{y}\right\rangle\) \(C\) consists of the arc of the curve \(y=\sin x\) from \((0,0)\) to \((\pi, 0)\) and the line segment from \((\pi, 0)\) to \((0,0)\)
Short Answer
Expert verified
The line integral evaluates to 0 using Green's Theorem.
Step by step solution
01
Verify Curve Orientation
The curve is closed and consists of the arc of the curve \(y=\sin x\) going from \((0,0)\) to \((\pi,0)\) and then returning via the line segment from \((\pi, 0)\) to \((0,0)\). Ensure that the orientation is counterclockwise. The described path is indeed oriented counterclockwise, satisfying the condition for Green's Theorem.
02
State Green's Theorem
Green's Theorem states that for a positively oriented, simple closed curve \(C\) and a vector field \(\mathbf{F}(x, y) = (P(x, y), Q(x, y))\), the line integral around the curve can be converted to a double integral over the region \(R\) bounded by \(C\): \[ \oint_{C} \mathbf{F} \cdot d\mathbf{r} = \iint_{R} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA \]
03
Determine \(\frac{\partial Q}{\partial x}\) and \(\frac{\partial P}{\partial y}\)
Given \(\mathbf{F} = \langle \sqrt{x}+y^{3}, x^{2}+\sqrt{y} \rangle\), let \(P(x,y) = \sqrt{x} + y^3\) and \(Q(x,y) = x^2 + \sqrt{y}\). Calculate:\[ \frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(x^2 + \sqrt{y}) = 2x \]\[ \frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(\sqrt{x} + y^3) = 3y^2 \]
04
Set Up the Double Integral
The region \(R\) is the area under the sine wave \(y=\sin x\) between \(x=0\) and \(x=\pi\). We need to evaluate:\[ \iint_{R} \left( 2x - 3y^2 \right) dA \]
05
Define Limits and Evaluate
Express the double integral with limits:\[ \int_{0}^{\pi} \int_{0}^{\sin x} (2x - 3y^2) \, dy \, dx \]Integrate with respect to \(y\):\[ \int_{0}^{\sin x} (2x - 3y^2) \, dy = [2xy - y^3]_{0}^{\sin x} = 2x\sin x - (\sin x)^3 \]This simplifies to:\[ \int_{0}^{\pi} [2x\sin x - (\sin x)^3] \, dx \]
06
Solve the Integral
Evaluate:\[ \int_{0}^{\pi} 2x \sin x \, dx - \int_{0}^{\pi} (\sin x)^3 \, dx \]The integral \(\int 2x\sin x\, dx\) can be solved using integration by parts. Let:- \(u = 2x\) and \(dv = \sin x \, dx\)- \(du = 2 \, dx\) and \(v = -\cos x\)Then by parts:\[ -2x\cos x \bigg|_0^\pi + 2 \int \cos x \, dx \bigg|_0^\pi \]Which evaluates to \(0\), due to sine and cosine periodic limits.The integral \(\int_0^\pi (\sin x)^3 \, dx\) also results in \(0\), after using the power-reducing formula for sine cube.
07
Conclusion
Combine results, both integrals evaluate to zero, hence:\[ \oint_{C} \mathbf{F} \cdot d\mathbf{r} = 0 \]
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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 generalization of definite integrals to functions of multiple variables. They are used to calculate the accumulation of a field along a curve. In essence, line integrals extend the notion of integration to paths in a vector field. This concept is central to physics and engineering because it captures how fields interact along paths in space.
- Concept: The line integral of a vector field \( \mathbf{F}(x, y) \) along a curve \( C \) is written as \( \int_{C} \mathbf{F} \cdot d \mathbf{r} \). It essentially measures the work done by the field along the curve.
- Parametrization: To compute it, the path, or the curve \( C \), is usually parametrized by a vector function \( \mathbf{r}(t) = (x(t), y(t)) \). The limits of integration come from the start and end points of the curve.
- Vector Field Interaction: The dot product in \( \mathbf{F} \cdot d\mathbf{r} \) combines the vector field and the tangent to the curve \( C \), highlighting how aligned the field is with the direction of the curve.
Vector Fields
Vector fields assign a vector to every point in a space. These fields are essential in describing various physical phenomena such as gravitational, electric, and magnetic fields. Each vector in a vector field represents the force direction and magnitude that a particle would experience at any given point.
- Definition: A vector field \( \mathbf{F}(x, y) = (P(x, y), Q(x, y)) \) is a function that assigns a vector to each position \( (x, y) \) in the plane.
- Components: The components \( P(x, y) \) and \( Q(x, y) \) describe the vector's horizontal and vertical influences at any point.
- Visual Representation: Vector fields can be visualized as arrows plotted over a region, where the direction and length of these arrows indicate the field's action at those points.
- Example: In the provided exercise, the vector field \( \mathbf{F}(x, y) = \langle \sqrt{x} + y^{3}, x^{2} + \sqrt{y} \rangle \) forms the basis for evaluating the line integral using Green's Theorem.
Double Integrals
Double integrals extend the concept of single integrals to functions of two variables. They are crucial for finding the volume under a surface defined over a two-dimensional region. When applied in Green's Theorem, double integrals are used to convert line integrals into area integrals for easier computation.
- Purpose: Double integrals compute the accumulation of quantities like volume or mass across a two-dimensional region.
- Application: In the context of Green's Theorem, a line integral around a simple, closed curve can be rewritten as a double integral over the region it encloses, simplifying many computations.
- Example: For the exercise, after verifying the curve's orientation and using Green's theorem, the line integral \( \oint_{C} \mathbf{F} \cdot d\mathbf{r} \) transforms into the double integral \( \iint_{R} \left(2x - 3y^2\right) dA \).
- Computation: Define the region \( R \)'s limits, i.e., the area under \( y = \sin x \) from \( x = 0 \) to \( x = \pi \), and integrate over this region.
