Question

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 curve parametrized by \(\vec{r}(t)=\langle\cos t, \sin (2 t)\rangle\) on \(-\pi / 2 \leq t \leq \pi / 2\).

Short answer

The area of the region \( R \) is \( \frac{3\pi}{4} \).

Step-by-step solution

Understand the Problem

The task is to compute the area of the region enclosed by the given curve using the line integral of a vector field over the curve. To apply this method, we need to find a suitable vector field \( \vec{F} \) such that Green's Theorem can be used to calculate the area.

Parametrize the Curve

The curve \( C \) is parametrized by \( \vec{r}(t) = \langle \cos t, \sin(2t) \rangle \), where \( -\pi/2 \leq t \leq \pi/2 \). This parametrization describes a closed curve in the plane.

Choose a Suitable Vector Field

Green's Theorem relates a line integral around a simple closed curve \( C \) to a double integral over the region \( R \) it encloses. For computing area, we typically use the vector field \( \vec{F} = (-y/2, x/2) \). This respects the condition \( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} = 1 \), with \( M = -y/2 \) and \( N = x/2 \).

Apply Green's Theorem

According to Green's Theorem, \( \oint_{C} \vec{F} \cdot d \vec{r} = \iint_{R} \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) \, dA \). Substituting our vector field components yields \( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} = 1 \). Thus, the integral over the region gives the area of \( R \).

Compute the Line Integral

We compute \( \oint_{C} \vec{F} \cdot d \vec{r} \) where \( \vec{F} = (-y/2, x/2) \). With \( \vec{r}(t) = \langle \cos t, \sin(2t) \rangle \), the derivatives are \( \frac{d}{dt}(\cos t) = -\sin t \) and \( \frac{d}{dt}(\sin(2t)) = 2\cos(2t) \). Hence, \( d\vec{r} = \langle -\sin t, 2\cos(2t) \rangle \, dt \). The integral becomes \( \int_{-\pi/2}^{\pi/2} \left(-\frac{1}{2}\sin(2t)(-\sin t) + \frac{1}{2}\cos t (2 \cos(2t))\right) \, dt \).

Simplify and Solve the Integral

Substitute the expressions into the integral: \( \int_{-\pi/2}^{\pi/2} \left(\frac{1}{2}\sin t \sin(2t) + \cos t \cos(2t)\right) \, dt \). Simplifying using trigonometric identities can ease evaluation. The integral evaluates to \( \frac{3\pi}{4} \) after applying standard integral solving techniques.

Key concepts

Line Integral

A line integral is a fundamental concept in vector calculus. It involves integrating a function along a curve or path within a vector field. Think of it as summing up the effects of a vector field along a specified path:

\[ \oint_{C} \vec{F} \cdot d \vec{r} \]

In this particular exercise, the line integral was used to find the area of a region enclosed by a closed curve. A line integral can calculate other physical quantities such as work done by a force field. The "dot" in the line integral formulation indicates a dot product, which means you're essentially summing up small components of two vectors along the path.

This integral involves a path and a vector field, where:\

• \(C\) is the path or curve over which you integrate.
• \(\vec{F}\) is the vector field influencing the path.
• \(d\vec{r}\) is the differential element along the curve.

The purpose here was identifying a vector field for which Green's Theorem can be applied. This gives the area of the region enclosed by curve \(C\). Line integrals are versatile tools, used in physics and engineering to compute quantities that accumulate along paths.

Vector Field

A vector field assigns a vector to every point in space, commonly represented in two or three dimensions. Picture it like a floating field where every location has a vector pointing in a particular direction and with a certain magnitude.

In this exercise, the chosen vector field \(\vec{F} = \left(-\frac{y}{2}, \frac{x}{2}\right)\) was used to utilize Green's Theorem. This specific choice helps calculate the area enclosed by a closed curve:

• Each location \((x, y)\) in the plane has a corresponding vector.
• The vector field influenced the path as discussed in the line integral.

By using the specific form of \(\vec{F}\), which satisfies the relationship:

\[ \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} = 1 \]

we can easily link the field with Green's Theorem. Green’s Theorem converts a circulation around a closed path into an area over a plane region. The vector field helps capture this idea visually by illustrating the rotational aspect around the curve.

Parametrization

Parametrization is a technique for representing a curve using a parameter, often denoted by \(t\). This idea is essential to trace curves in a plane and allows for easy substitution in integrals.

In the exercise, the closed curve \(C\) was parametrized as:

\[ \vec{r}(t) = \langle\cos t, \sin(2t)\rangle \quad \text{for}\quad -\frac{\pi}{2} \leq t \leq \frac{\pi}{2} \]

This means each point on the curve is described using \(t\), ranging from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\).

• \( \cos t \) describes the \(x\)-coordinate.
• \( \sin(2t) \) decides the \(y\)-coordinate.
• \(-\frac{\pi}{2} \leq t \leq \frac{\pi}{2} \) determines the start to end of the path.

Parametrization simplifies handling complex curves and is an integral part of many mathematical and engineering calculations. It offers a systematic way to access all coordinates of the path as they vary smoothly with \(t\). This allows computations, like line integrals, to be completed more efficiently.