Question

In Exercises \(13-16,\) a vector field \(\vec{F}\) and a closed curve \(C,\) enclosing a region \(R,\) are given. Verify Green's Theorem by evaluating \(\oint_{C} \vec{F} \cdot d \vec{r}\) and \(\iint_{R}\) curl \(\vec{F} d A,\) showing they are equal. \(\vec{F}=\langle x-y, x+y\rangle ; C\) is the closed curve composed of the parabola \(y=x^{2}\) on \(0 \leq x \leq 2\) followed by the line segment from (2,4) to (0,0).

Short answer

Both the line integral and the double integral evaluate to 8, confirming Green's Theorem.

Step-by-step solution

Understand Green's Theorem

Green's Theorem relates the line integral around a simple closed curve \(C\) to a double integral over the plane region \(R\) bounded by \(C\). It states: \(\oint_{C} \vec{F} \cdot d\vec{r} = \iint_{R} (\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}) \, dA\), where \(\vec{F} = \langle M,N \rangle\). In this problem, \(\vec{F} = \langle x-y, x+y \rangle\).

Calculate the line integral \(\oint_{C} \vec{F} \cdot d\vec{r}\)

The curve \(C\) is composed of two parts: the parabola \(y = x^2\) from \(x=0\) to \(x=2\), and the line segment from (2,4) to (0,0). For the parabola, parameterize as \(\vec{r}_1(t) = \langle t, t^2 \rangle\) where \(t\) goes from 0 to 2. The line segment can be parameterized as \(\vec{r}_2(t) = \langle 2-2t, 4-4t \rangle\) with \(t\) from 0 to 1. Compute \(\vec{F}(\vec{r}_1(t)) \cdot \frac{d\vec{r}_1}{dt}\) and \(\vec{F}(\vec{r}_2(t)) \cdot \frac{d\vec{r}_2}{dt}\), and integrate over their respective intervals.

Compute curl \(\vec{F}\) and the double integral \(\iint_{R} (\text{curl } \vec{F}) \, dA\)

First, calculate the curl of \(\vec{F} = \langle M, N \rangle = \langle x-y, x+y \rangle\): \(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} = 1 - (-1) = 2\). The curl is constant over the region. Set up the double integral: \(\iint_{R} 2 \, dA\). The region \(R\) is bounded by \(y=x^2\) and the line \(y=2x\), from \(x=0\) to \(x=2\). Integrate \(2\) with respect to \(y\) first, then \(x\).

Compare the results of both integrals

The line integral \(\oint_{C} \vec{F} \cdot d\vec{r}\) should equal the double integral \(\iint_{R} (\text{curl } \vec{F}) \, dA = \int_0^2 \int_{x^2}^{2x} 2\, dy \, dx\). Check that both evaluations result in the same numerical value.

Key concepts

Calculating Line Integrals

To visualize a line integral, think about walking along a curve in a vector field where each vector represents a force, like wind.
The work done against this force is the line integral. This means you're measuring the sum of vector field strengths applied in the direction of the path.

• For the parabola segment, parameterize it: \( \vec{r}_1(t) = \langle t, t^2 \rangle \) with \( t \) ranging from 0 to 2.
• The line segment from (2,4) to (0,0) is parameterized as \( \vec{r}_2(t) = \langle 2-2t, 4-4t \rangle \) with \( t \) from 0 to 1.
• Calculate the derivative \( \frac{d\vec{r}}{dt} \) for both segments and substitute in the vector field to find \( \vec{F}(\vec{r}(t)) \cdot \frac{d\vec{r}}{dt} \).
• Integrate the dot product over the interval for each part of the curve.

By summing these integrals, you determine the total work done along the path \( C \).

Curl of a Vector Field

The curl of a vector field measures the rotation or swirling strength of the field at a point.
Imagine stirring a liquid—the water swirls differently at different points.
For the field \( \vec{F} = \langle M, N \rangle = \langle x-y, x+y \rangle \), compute:

• Partial derivative \( \frac{\partial N}{\partial x} = 1 \)
• Partial derivative \( \frac{\partial M}{\partial y} = -1 \)

Thus, the curl \( abla \times \vec{F} \) is calculated as:\[\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} = 1 - (-1) = 2\]Because it's constant, the field rotates with the same intensity everywhere in region \( R \). This uniform curl value simplifies computing double integrals, as you can apply it over the whole region directly.

Double Integrals

Double integrals let us compute the area, or more generally, properties of a region by stacking infinitesimally small tiles over a plane.
The region \( R \) specified is between the parabola \( y=x^2 \) and the line \( y=2x \).
The bounds of integration are from \( x=0 \) to \( x=2 \), computing over \( y \) first.

• Set up as \( \int_0^2 \int_{x^2}^{2x} 2 \, dy \, dx \).
• First integrate with respect to \( y \), evaluating from \( x^2 \) to \( 2x \).
• Then integrate because \( 2 \) is constant, so \( 2(y) \) from \( x^2\) to \( 2x \) evaluates easily.
• Finish by solving the single integral remaining in terms of \( x \).

This gives you the total swirl (rotation force) in \( R \), which should match the line integral around \( C \).

Parameterization of Curves

Parameterizing a curve means expressing it in terms of a parameter, usually \( t \), which makes the curve more manageable for integration.
Like describing a moving point along the curve over time.For the exercise:

• The parabola \( y = x^2 \) is represented as \( \vec{r}_1(t) = \langle t, t^2 \rangle \) for \( t \) between 0 and 2.
• The line from (2,4) to (0,0) translates to \( \vec{r}_2(t) = \langle 2-2t, 4-4t \rangle \) for \( t \) between 0 and 1.

Using these parameterizations:

• Substitute into vector field \( \vec{F} \).
• Substitute into \( d\vec{r} = \langle 1, 2t \rangle dt \) for parabola and \( \langle -2, -4 \rangle dt \) for line segment.
• Evaluate the integral to find total work along curve \( C \).

Parameterizing helps us break complex shapes into simpler, integrable segments.