Chapter 6: Problem 174
For the following exercises, use Green's theorem to calculate the work done by force \(\mathbf{F}\) on a particle that is moving counterclockwise around closed path \(C .\) $$\mathbf{F}(x, y)=x y \mathbf{i}+(x+y) \mathbf{j}, \quad C : x^{2}+y^{2}=4$$
Short Answer
Expert verified
The work done by the force is \(4\pi\).
Step by step solution
01
Identify the Components of Vector Field F
Given the vector field \(\mathbf{F}(x, y) = xy \mathbf{i} + (x+y) \mathbf{j}\), identify the components \(P(x, y) = xy\) and \(Q(x, y) = x + y\). These are essential for applying Green's Theorem.
02
Express Green's Theorem
Green's Theorem relates a line integral around a closed curve to a double integral over the plane region it encloses. The theorem states:\[ \oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_R \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA \]where \(C\) is the positively oriented, or counterclockwise, curve, and \(R\) is the region enclosed by \(C\).
03
Calculate Partial Derivatives
Calculate the partial derivatives needed for Green's Theorem:- \(\frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(x + y) = 1\)- \(\frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(xy) = x\)
04
Set Up the Double Integral
Substitute the partial derivatives into Green's Theorem:\[ \oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_R \left( 1 - x \right) \, dA \]The region \(R\) is a circle where \(x^2 + y^2 \leq 4\).
05
Convert to Polar Coordinates
To evaluate the double integral over the circular region, convert to polar coordinates:- \(x = r\cos\theta\)- \(y = r\sin\theta\)- \(dA = r\,dr\,d\theta\)Thus, the integral becomes:\[ \iint_R (1 - r\cos\theta) \, r \, dr \, d\theta \]
06
Evaluate the Integral Limits
The limits for \(r\) are from 0 to 2 (radius of circle) and for \(\theta\) from 0 to \(2\pi\):\[ \int_0^{2\pi} \int_0^2 (1 - r\cos\theta)\, r\, dr\, d\theta \]
07
Solve the Inner Integral
First, solve the inner integral:\[ \int_0^2 (1 - r\cos\theta) r \, dr = \int_0^2 (r - r^2\cos\theta) \, dr \]Calculate:- \(\int_0^2 r\, dr = \left[ \frac{1}{2} r^2 \right]_0^2 = 2\)- \(\int_0^2 r^2\cos\theta \ dr = \cos\theta \left[ \frac{1}{3}r^3 \right]_0^2 = \frac{8}{3} \cos\theta\)Thus:\[ \int_0^2 (r - r^2\cos\theta)\ dr = 2 - \frac{8}{3}\cos\theta \]
08
Solve the Outer Integral
Next, solve the outer integral:\[ \int_0^{2\pi} \left( 2 - \frac{8}{3}\cos\theta \right) \, d\theta = 2\int_0^{2\pi} 1 \, d\theta - \frac{8}{3} \int_0^{2\pi} \cos\theta \, d\theta \]As \(\int_0^{2\pi} \cos\theta \, d\theta = 0\), we have:\[ = 2 \cdot (2\pi) = 4\pi \]
09
State Final Answer
Thus, the work done by the force \(\mathbf{F}\) on the particle moving counterclockwise around the path \(C\) is \(4\pi\).
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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 way to integrate over a curve or path in a vector field. They help us find quantities like work done by a force field on a particle moving along a given path. For a vector field \( \mathbf{F}(x, y) = P(x, y) \mathbf{i} + Q(x, y) \mathbf{j} \), the line integral around a closed curve \( C \) is given by: \[ \oint_C \mathbf{F} \cdot d\mathbf{r} = \oint_C P(x, y) \, dx + Q(x, y) \, dy \] This involves integrating the components of the vector field along the path. In the context of Green's Theorem, the line integral can be transformed into a double integral over the region enclosed by the path, which can simplify calculations when the path and vector field have certain symmetries.
Vector Fields
Vector fields are functions that assign a vector to every point in space. They can represent many physical phenomena, such as the flow of liquids or forces acting at various points in a region. For example, our vector field \( \mathbf{F}(x, y) = xy \mathbf{i} + (x+y) \mathbf{j} \) assigns a force vector at every point \((x, y)\). Key concepts about vector fields:
- Components: Each vector field has components \( P(x, y) \) and \( Q(x, y) \), which can be functions of \( x \) and \( y \).
- Visual Representation: Vector fields are often visualized as arrows on a grid, where the direction and length represent the direction and magnitude of the vector.
- Applications: They are used in physics, engineering, and computer graphics to model complex systems.
Polar Coordinates
Polar coordinates are a different way to describe points in a plane, particularly useful for circular or rotational symmetries. In polar coordinates, each point is determined by a distance \( r \) from the origin and an angle \( \theta \) from the positive x-axis. Transformation from Cartesian coordinates \((x, y)\) to polar coordinates is achieved using:
- \( x = r\cos\theta \)
- \( y = r\sin\theta \)
- The area element \( dA \) becomes \( r \, dr \, d\theta \).
Partial Derivatives
Partial derivatives represent how a multivariable function changes as one of its variables changes, keeping others constant. They are essential in vector calculus for analyzing functions defined over a plane or space. When applying Green's Theorem, we calculate the partial derivatives of the vector field components. For our vector field \( \mathbf{F}(x, y) = xy \mathbf{i} + (x+y) \mathbf{j} \):
- \( \frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(x+y) = 1 \)
- \( \frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(xy) = x \)