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Chapter 16: Problem 21

Find the counterclockwise circulation and outward flux of the field \(\mathbf{F}=x y \mathbf{i}+y^{2} \mathbf{j}\) around and over the boundary of the region enclosed by the curves \(y=x^{2}\) and \(y=x\) in the first quadrant.

Short Answer

Expert verified
Circulation: \(-\frac{1}{12}\), Flux: \(\frac{1}{5}\).

Step by step solution

01

Sketch and Identify the Region

First, sketch the region enclosed by the curves. In the first quadrant, the parabola \( y = x^2 \) opens upwards and intersects the line \( y = x \) at the origin \((0,0)\) and at \((1,1)\). The region is a triangular-like area bounded by these curves. Ensure that visualizing this area makes it easier to apply Green's Theorem.
02

Green's Theorem for Circulation

Green's Theorem relates the counterclockwise circulation of a field \( \mathbf{F} = P\mathbf{i} + Q\mathbf{j} \) around a closed curve to a double integral over the region enclosed. It states \( \oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_R \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA \). For \( \mathbf{F} = xy\mathbf{i} + y^2\mathbf{j} \), \( P = xy \) and \( Q = y^2 \). Compute \( \frac{\partial Q}{\partial x} = 0 \) and \( \frac{\partial P}{\partial y} = x \), so the integrand becomes \( -x \).
03

Setup the Double Integral for Circulation

The double integral to compute is \( \iint_R -x \, dA \). In the first quadrant, the region \( R \) is bounded by \( y = x^2 \) and \( y = x \). Convert this into limits as \( x \) varies from 0 to 1, and for each \( x \), \( y \) varies from \( x^2 \) to \( x \). This gives the integral \( \int_0^1 \int_{x^2}^{x} -x \, dy \, dx \).
04

Evaluate the Circulation Integral

Calculate the inner integral: \( \int_{x^2}^{x} -x \, dy = -x(y) \Big|_{x^2}^{x} = -x(x - x^2) = -x^2 + x^3 \). Substitute this into the outer integral: \( \int_0^1 (-x^2 + x^3) \, dx \). Break it into two integrals: \( \int_0^1 -x^2 \, dx + \int_0^1 x^3 \, dx \), which evaluates to \( -\frac{1}{3} + \frac{1}{4} = -\frac{1}{12} \).
05

Green's Theorem for Flux

For outward flux, Green's Theorem states \( \oint_C \mathbf{F} \cdot \mathbf{n} \, ds = \iint_R \left( \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} \right) \, dA \). Using \( P = xy \) and \( Q = y^2 \), compute \( \frac{\partial P}{\partial x} = y \) and \( \frac{\partial Q}{\partial y} = 2y \). Thus, the integrand becomes \( y + 2y = 3y \).
06

Setup the Double Integral for Flux

The double integral is \( \iint_R 3y \, dA \). Using the same bounds as before, the integral is \( \int_0^1 \int_{x^2}^{x} 3y \, dy \, dx \).
07

Evaluate the Flux Integral

Calculate the inner integral: \( \int_{x^2}^{x} 3y \, dy = \frac{3}{2}y^2 \Big|_{x^2}^{x} = \frac{3}{2}(x^2 - x^4) \). Substitute to the outer integral: \( \int_0^1 \frac{3}{2}(x^2 - x^4) \, dx \), broken into two parts gives \( \frac{3}{2}(\int_0^1 x^2 \, dx - \int_0^1 x^4 \, dx) = \frac{3}{2}(\frac{1}{3} - \frac{1}{5}) = \frac{1}{5} \).

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Key Concepts

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

Circulation
Circulation in vector calculus refers to the line integral of a vector field around a closed curve. Imagine circulation as the amount of the field that 'flows' along a path. This concept is crucial when using Green's Theorem, which connects line integrals around a curve to double integrals within the region it encloses.
  • Green's Theorem states that the circulation around a closed curve can be expressed as a double integral over the region enclosed by the curve.
  • The formula used is: \(\oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_R \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA\).
  • This means that instead of evaluating the curve on the boundary directly, we calculate the change in the vector field inside the region.
In our problem, with \(\mathbf{F} = xy\mathbf{i} + y^2\mathbf{j}\), the circulation is evaluated as a double integral over the triangular region in the first quadrant, where the line \(y = x\) and parabola \(y = x^2\) intersect. This involves setting up and calculating the corresponding integrals.
Flux
Flux measures how much of a field passes through a surface. Think of flux like how much water flows through or across a boundary, and it is computed using Green's Theorem. Flux across a closed curve relates to changes in the field within the region.
  • To find outward flux using Green's Theorem, the equation is: \(\oint_C \mathbf{F} \cdot \mathbf{n} \, ds = \iint_R \left( \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} \right) \, dA\).
  • Here, \( \mathbf{n} \) represents the outward normal, which simplifies to computing partial derivatives in our double integral.
  • In our example, for \( \mathbf{F} = xy\mathbf{i} + y^2\mathbf{j} \), we compute \( y + 2y \), and integrate over the region given by the parabola and line intersection.
Using Green's Theorem helps turn complex line integrals into simpler double integrals over bounded regions, offering a powerful tool for calculating flux efficiently.
Double Integral
Double integrals expand the calculus of single integrals, allowing us to compute volume under a surface over a region. They're crucial in calculus and are widely used to solve problems involving areas bounded by curves, like when applying Green's Theorem.
  • A double integral is expressed as \( \iint_R f(x, y) \, dA \), where \( R \) is the region of integration.
  • This integral evaluates a function over a two-dimensional area, thus calculating 'accumulated' values like area or volume.
  • Setting up a double integral involves defining the right limits for integration, essentially tracing how \( x \) and \( y \) vary over the region.
In our scenario, we examined \( \iint_R -x \, dA \) for circulation and \( \iint_R 3y \, dA \) for flux over the triangular region. Proper limits are found by analyzing the region bounded by the curves \( y = x^2 \) and \( y = x \) from 0 to 1.
Vector Field
A vector field assigns a vector to each point in a region of space. Picture a vector field as an array of arrows representing direction and magnitude of influence at different points. It's integral to analyzing realistic models in physics and engineering.
  • A vector field is typically expressed as \( \mathbf{F} = P\mathbf{i} + Q\mathbf{j} \), where \( P \) and \( Q \) are functions of \( x \) and \( y \).
  • Understanding vector fields involves looking at how vectors change across a region, which can indicate circulation or flux.
  • In our task, the given vector field \( \mathbf{F} = xy\mathbf{i} + y^2\mathbf{j} \) helps determine the behavior of \( x \) and \( y \) components at each point and is foundational to calculating integral values over regions.
Vector fields are fundamental to visualizing and solving dynamic systems, and they allow us to apply theorems like Green's to find circulation and flux efficiently.

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Most popular questions from this chapter

Curl component of conservative fields Can anything be said about the curl component of a conservative two-dimensional vector field? Give reasons for your answer.

A revealing experiment By experiment, you find that a force field \(\mathbf{F}\) performs only half as much work in moving an object along path \(C_{1}\) from \(A\) to \(B\) as it does in moving the object along path \(C_{2}\) from \(A\) to \(B\) . What can you conclude about \(\mathbf{F}\) ? Give reasons for your answer.

Find the work done by \(\mathbf{F}\) in moving a particle once counterclockwise around the given curve. \(\mathbf{F}=(4 x-2 y) \mathbf{i}+(2 x-4 y) \mathbf{j}\) \(C :\) The circle \((x-2)^{2}+(y-2)^{2}=4\)

Let \(\mathbf{F}_{1}\) and \(\mathbf{F}_{2}\) be differentiable vector fields and let \(a\) and \(b\) be arbitrary real constants. Verify the following identities. $$\begin{array}{l}{\text { a. } \nabla \cdot\left(a \mathbf{F}_{1}+b \mathbf{F}_{2}\right)=a \nabla \cdot \mathbf{F}_{1}+b \nabla \cdot \mathbf{F}_{2}} \\ {\text { b. } \nabla \times\left(a \mathbf{F}_{1}+b \mathbf{F}_{2}\right)=a \nabla \times \mathbf{F}_{1}+b \nabla \times \mathbf{F}_{2}} \\ {\text { c. } \nabla \cdot\left(\mathbf{F}_{1} \times \mathbf{F}_{2}\right)=\mathbf{F}_{2} \cdot \nabla \times \mathbf{F}_{1}-\mathbf{F}_{1} \cdot \nabla \times \mathbf{F}_{2}}\end{array}$$

Find the area of the band cut from the paraboloid \(x^{2}+y^{2}-z=\) 0 by the planes \(z=2\) and \(z=6 .\)
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