Question

In Exercises \(17-20,\) a closed curve \(C\) enclosing a region \(R\) is given. Find the area of \(R\) by computing \(\oint_{C} \vec{F} \cdot d \vec{r}\) for an appropriate choice of vector field \(\vec{F}\). \(C\) is the ellipse parametrized by \(\vec{r}(t)=\langle 4 \cos t, 3 \sin t\rangle\) on \(0 \leq t \leq 2 \pi\).

Short answer

The area of the region enclosed by the ellipse is \( 6\pi \).

Step-by-step solution

Identify Suitable Vector Field

To find the area enclosed by the curve using Green's Theorem, we need a vector field \( \vec{F} \) where \( abla \times \vec{F} = 1 \). A simple choice is \( \vec{F} = \left(-\frac{y}{2}, 0\right) \).

Parameterize and Differentiate

The curve \( C \) is given parametrically by \( \vec{r}(t) = \langle 4 \cos t, 3 \sin t \rangle \). The derivative \( \vec{r}\,'(t) \) is \( \langle -4 \sin t, 3 \cos t \rangle \).

Compute the Vector Field Dot Product

Substitute \( \vec{r}(t) \) into the vector field \( \vec{F} \) to get \( \vec{F}(\vec{r}(t)) = \left(-\frac{3 \sin t}{2}, 0\right) \). Compute the dot product with \( \vec{r}\,'(t) \), which is \(-\frac{3}{2} \sin t (-4 \sin t) = 6 \sin^2 t \).

Set Up and Evaluate the Integral

The line integral \( \oint_{C} \vec{F} \cdot d \vec{r} \) becomes \( \int_{0}^{2\pi} 6 \sin^2 t\, dt \). Use the trigonometric identity \( \sin^2 t = \frac{1 - \cos 2t}{2} \) to simplify the integral to \( 3 \int_{0}^{2\pi} (1 - \cos 2t) \, dt \).

Solve the Simplified Integral

The integral becomes \( 3 \left[ t - \frac{1}{2} \sin 2t \right]_{0}^{2\pi} \). Evaluating this gives \( 3 \left[ 2\pi \right] = 6\pi \). Thus, the area of \( R \) is \( 6\pi \).

Key concepts

Vector Fields

In mathematics, a vector field assigns a vector to every point in a space. Think of it like a map showing wind speed and direction across a geographic area. Each point on the map indicates the wind blowing in a particular direction and at a certain speed. In calculus, a vector field is often represented as \( \vec{F}(x, y) = (P(x, y), Q(x, y)) \), where each component function \( P \) and \( Q \) can vary with \( x \) and \( y \).
Green's Theorem utilizes vector fields by integrating along a closed curve in a plane, translating the complex line integrals into simpler double integrals over the region enclosed by the curve. For our exercise, the choice of the vector field \( \vec{F} = \left(-\frac{y}{2}, 0\right) \) is strategic. It simplifies finding the area of the region enclosed by the ellipse specified in the problem.

Line Integrals

Line integrals are used to calculate values along a curve, involving scalar or vector fields. When dealing with vector fields, like in our exercise, the line integral gives us information about how the field accumulates as you move along a path.
The line integral of a vector field \( \vec{F} \) over a curve \( C \) is given by \( \oint_{C} \vec{F} \cdot d \vec{r} \). This dot product combines the vector field's direction and magnitude with the direction of the curve. For our specific task, this integral calculates the area enclosed by the curve, an application of the famous Green's Theorem, bridging line integrals with double integrals over regions.

Parametric Equations

Parametric equations allow curves to be described using parameters, usually making it simpler to describe complex shapes like ellipses or circles. In our problem, the curve \( C \) is an ellipse described with parametric equations \( \vec{r}(t) = \langle 4 \cos t, 3 \sin t \rangle \), where \( t \) ranges from 0 to \( 2\pi \).
This parameterization uses trigonometric functions to define the ellipse's x and y coordinates in terms of a single variable, \( t \). Differentiating these equations provides the tangent to the curve, which is crucial for setting up the line integral by detailing \( d \vec{r} \), or the differential displacement along the curve, as \( \vec{r}\,'(t) = \langle -4 \sin t, 3 \cos t \rangle \).

Area Calculation

Area calculation using Green's Theorem involves the transformation of a difficult line integral into an easier double integral over a region. In our exercise, once we've expressed our vector field and parameterized the curve, the task shifts to setting up and solving the integral \( \oint_{C} \vec{F} \cdot d \vec{r} \).
Substituting our parametric equations into the chosen vector field, we compute the dot product. This results in the integral \( \int_{0}^{2\pi} 6 \sin^2 t\, dt \). Using identities like \( \sin^2 t = \frac{1 - \cos 2t}{2} \), we rework the integral to a simpler form: \( 3 \int_{0}^{2\pi} (1 - \cos 2t) \, dt \).

• The first part, \( 3 \int_{0}^{2\pi} 1\, dt \), calculates to \( 6\pi \), representing the total area.
• The second part, \( 3 \int_{0}^{2\pi} -\cos 2t \, dt \), averages out to zero over a full period, simplifying the solution.
  Thus, the technique provides us with the coveted result: the area of region \( R \), equating to \( 6\pi \).