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Chapter 16: Problem 14

Use Green's Theorem to find the counterclockwise circulation and outward flux for the field \(\mathbf{F}\) and curve \(C .\) \(\mathbf{F}=(x+y) \mathbf{i}-\left(x^{2}+y^{2}\right) \mathbf{j}\) \(C :\) The triangle bounded by \(y=0, x=1,\) and \(y=x\)

Short Answer

Expert verified
The counterclockwise circulation is \(-\frac{7}{6}\) and the outward flux is \(\frac{1}{6}\).

Step by step solution

01

Understanding Green's Theorem

Green's Theorem relates the line integral over a closed curve to a double integral over the region it encloses. There are two forms: circulation form, \( \oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_R \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA \) and flux form, \( \oint_C \mathbf{F} \cdot \mathbf{n} \, ds = \iint_R \left( \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} \right) dA \). For a vector field \( \mathbf{F} = P\mathbf{i} + Q\mathbf{j} \).
02

Identify P and Q

For the given vector field \( \mathbf{F} = (x+y) \mathbf{i} - (x^2 + y^2) \mathbf{j} \), identify \( P = x+y \) and \( Q = -(x^2 + y^2) \).
03

Geometrically Determine Region R

The region \( R \) is a triangle with vertices at (0,0), (1,0), and (1,1), bounded by the line \( y=0 \), \( x=1 \), and \( y=x \).
04

Set up Circulation Double Integral

For circulation, compute \( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \). Start with derivatives: \( \frac{\partial Q}{\partial x} = -2x \) and \( \frac{\partial P}{\partial y} = 1 \). The circulation double integral becomes \( \iint_R (-2x - 1) \, dA \).
05

Compute Circulation Double Integral

Evaluate \( \iint_R (-2x - 1) \, dA \) over the region. The limits for x are 0 to 1, and for a fixed x, y varies from 0 to x. Thus, \[ \int_{0}^{1} \int_{0}^{x} (-2x - 1) \, dy \, dx \]. Calculate: \( \int_{0}^{x} (-2x - 1) \, dy = (-2xy - y) \bigg|_0^x = -2x^2 - x \). Integrate with respect to x: \( \int_{0}^{1} (-2x^2 - x) \, dx = \left[ -\frac{2}{3}x^3 - \frac{1}{2}x^2 \right]_0^1 = -\frac{2}{3} - \frac{1}{2} = -\frac{7}{6} \).
06

Set up Flux Double Integral

For flux, compute \( \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} \). Start with derivatives: \( \frac{\partial P}{\partial x} = 1 \) and \( \frac{\partial Q}{\partial y} = -2y \). The flux double integral becomes \( \iint_R (1 - 2y) \, dA \).
07

Compute Flux Double Integral

Evaluate \( \iint_R (1 - 2y) \, dA \) over the region. As determined before, integrate over \( \int_{0}^{1} \int_{0}^{x} (1 - 2y) \, dy \, dx \). First, \( \int_{0}^{x} (1 - 2y) \, dy = (y - y^2) |_0^x = x - x^2 \). Now, integrate with respect to x: \( \int_{0}^{1} (x - x^2) \, dx = \left[ \frac{1}{2}x^2 - \frac{1}{3}x^3 \right]_0^1 = \frac{1}{2} - \frac{1}{3} = \frac{1}{6} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vector Field
A vector field is a mathematical structure where a vector is assigned to every point in a region. It's like a map showing the wind speed and direction at different locations on the Earth's surface.
In two dimensions, a vector field \( \mathbf{F} = P\mathbf{i} + Q\mathbf{j} \) consists of two functions, often denoted as \( P(x, y) \) and \( Q(x, y) \). They tell us how the vector's components change across the plane.
For the given exercise, the vector field is \(\mathbf{F} = (x+y) \mathbf{i} - (x^2+y^2) \mathbf{j}\), where \( P = x+y \) and \( Q = -(x^2 + y^2) \). This representation helps in analyzing physical phenomena such as fluid flow or electromagnetic fields.
Line Integral
A line integral is an extension of the concept of an integral to integrate along curves or paths in a field. It is used to find the work done by a field along a path or the cumulative effect along a curve.
For a vector field \( \mathbf{F} = P\mathbf{i} + Q\mathbf{j} \) around a closed curve \( C \), the line integral is denoted by \( \oint_C \mathbf{F} \cdot d\mathbf{r} \).
This involves integrating the vector field's components along the given path, taking into account the direction and magnitude of each component relative to the path.
Double Integral
A double integral allows us to compute the volume under a surface or to accumulate values over a two-dimensional area. It's like finding the amount of paint needed to cover a region.
The double integral \[ \iint_R f(x, y) \, dA \] over a region \( R \) sums a function over that area. In the context of Green's Theorem, we use the double integral for both circulation and flux calculations.
First identify the region's boundaries, such as the triangle in the exercise, then set up appropriate limits for integration to compute the accumulation of quantities.
Circulation and Flux
Circulation and flux are two fundamental concepts describing the behavior of vector fields over a region.
  • Circulation: This measures the tendency of a field to loop around a region. For a given vector field \( \mathbf{F} = P\mathbf{i} + Q\mathbf{j} \), the circulation form of Green's Theorem is \[ \oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_R \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA \], where the result reflects how strongly and in what direction the field loops around the region.
  • Flux: This measures the tendency of a field to push through a surface. It relates to outward flow across the boundary. For the same vector field, the flux form of Green's Theorem is \[ \oint_C \mathbf{F} \cdot \mathbf{n} \, ds = \iint_R \left( \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} \right) dA \], describing the net flow out of the region.
Partial Derivatives
Partial derivatives focus on the rate of change of a function with respect to one variable, holding the other variables constant.
This is crucial in fields like physics and engineering, where changes in one direction don't necessarily affect other directions in the same way.
In the context of Green鈥檚 Theorem, partial derivatives are used to calculate components necessary for circulation and flux.
  • For the circulation form, we find \( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \), which measures how much the field turns.
  • For the flux form, calculate \( \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} \) to determine how much field flows through the region's boundary.
Understanding these derivatives helps in analyzing how vector fields behave in space.

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Most popular questions from this chapter

The tangent plane at a point \(P_{0}\left(f\left(u_{0}, v_{0}\right), g\left(u_{0}, v_{0}\right), h\left(u_{0}, v_{0}\right)\right)\) on a parametrized surface \(\mathbf{r}(u, v)=f(u, v) \mathbf{i}+g(u, v) \mathbf{j}+h(u, v) \mathbf{k}\) is the plane through \(P_{0}\) normal to the vector \(\mathbf{r}_{u}\left(u_{0}, v_{0}\right) \times \mathbf{r}_{v}\left(u_{0}, v_{0}\right),\) the cross product of the tangent vectors \(\mathbf{r}_{u}\left(u_{0}, v_{0}\right)\) and \(\mathbf{r}_{v}\left(u_{0}, v_{0}\right)\) at \(P_{0} .\) In Exercises \(27-30,\) find an equation for the plane tangent to the surface at \(P_{0} .\) Then find a Cartesian equation for the surface and sketch the surface and tangent plane together. $$ \begin{array}{l}{\text { Cone The cone } \mathbf{r}(r, \theta)=(r \cos \theta) \mathbf{i}+(r \sin \theta) \mathbf{j}+r \mathbf{k}, r \geq 0} \\ {0 \leq \theta \leq 2 \pi \text { at the point } P_{0}(\sqrt{2}, \sqrt{2}, 2) \text { corresponding to }} \\ {(r, \theta)=(2, \pi / 4)}\end{array} $$

In Exercises 29 and 30 , find the surface integral of the field \(\mathbf{F}\) over the portion of the given surface in the specified direction. $$\mathbf{F}(x, y, z)=-\mathbf{i}+2 \mathbf{j}+3 \mathbf{k}$$ S: rectangular surface \(z=0, \quad 0 \leq x \leq 2, \quad 0 \leq y \leq 3\) \(\quad\) direction \(\mathbf{k}\)

Let \(S\) be the portion of the cylinder \(y=e^{x}\) in the first octant that projects parallel to the \(x\) -axis onto the rectangle \(R_{y z} : 1 \leq y \leq 2\) \(0 \leq z \leq 1\) in the \(y z\) -plane (see the accompanying figure). Let \(\mathbf{n}\) be the unit vector normal to \(S\) that points away from the \(y z\) -plane. Find the flux of the field \(\mathbf{F}(x, y, z)=-2 \mathbf{i}+2 y \mathbf{j}+z \mathbf{k}\) across \(S\) in the direction of \(\mathbf{n} .\)

Green's Theorem Area Formula Area of \(R=\frac{1}{2} \oint_{C} x d y-y d x\) Use the Green's Theorem area formula given above to find the areas of the regions enclosed by the curves. The ellipse \(\mathbf{r}(t)=(a \cos t) \mathbf{i}+(b \sin t) \mathbf{j}, \quad 0 \leq t \leq 2 \pi\)

Find the area of the surface cut from the paraboloid \(x^{2}+y^{2}-z=\) 0 by the plane \(z=2\) .
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