91Ó°ÊÓ

When we talk about curve parametrization, we're referring to the process of describing a curve using a parameter. This parameter is usually denoted by a variable like \( t \), and it makes it easier to perform calculations involving the curve. In our exercise, the curve \( C \) is parametrized by \( \vec{r}(t) = \langle 2 \cos t + \frac{1}{10} \cos (10 t), 2 \sin t + \frac{1}{10} \sin (10 t) \rangle \). This equation describes how points on the curve move as \( t \) changes from 0 to \( 2\pi \). The first component \( 2 \cos t + \frac{1}{10} \cos(10t) \) gives the \( x \)-coordinates, and the second component \( 2 \sin t + \frac{1}{10} \sin(10t) \) gives the \( y \)-coordinates. Since \( t \) varies smoothly, it helps in computing other properties like line integrals easily. So, curve parametrization transforms complicated curves into manageable mathematical forms.

A line integral is a way of adding up values along a curve. In our context, we use it to calculate properties like the area enclosed by a curve. Imagine walking along the curve and collecting a value at each point; this integral adds up all those values. In our exercise, the line integral is expressed as \( \oint_{C} \vec{F} \cdot d \vec{r} \). This indicates we are finding the line integral of the vector field \( \vec{F} \) along the curve \( C \). To find the area using Green's Theorem, we choose a special form of \( \vec{F} = \langle 0, x \rangle \), so the line integral simplifies to \( \oint_{C} x \, dy \). By parametrizing the curve and substituting \( x \) and \( dy \) from the parametrization, we can evaluate the integral to find the desired area.

A vector field \( \vec{F} \) assigns a vector to every point in a plane, and it is a crucial concept in this problem. By selecting the appropriate vector field, we can utilize Green's Theorem to find the area enclosed by a curve. In this problem, we used \( \vec{F} = \langle 0, x \rangle \) to ensure the theorem outputs the area. Choosing \( \vec{F} = \langle 0, x \rangle \) ensures that \( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 1 \). This results in the line integral around the curve yielding the total area of the region \( R \):

• \( \vec{F} = \langle P, Q \rangle = \langle 0, x \rangle \)
• \( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 1 \)
• Area = \( \oint_{C} x \, dy \)

The vector field plays a critical role as it helps translate the geometric problem into an algebraic one, which is then solved using calculus tools.

Calculating the area of a region enclosed by a curve can be achieved efficiently using Green's Theorem. The theorem links the line integral around a closed curve to a double integral over the region inside the curve. In the context of this exercise, we use the vector field \( \vec{F} = \langle 0, x \rangle \) and conclude that the line integral equals the area of region \( R \). First, we set up the integral:

• Integral: \( \oint_{C} x \, dy = \int_{0}^{2 \pi} \left(2 \cos t + \frac{1}{10} \cos(10t)\right) \cdot \left(2 \cos t + \cos(10t)\right) \, dt \)

By expanding and integrating, we focus on terms that are not products of odd functions, which cancel over the symmetric interval. The dominant term \( 4 \cos^2 t \) integrates to provide the area since others contribute zero due to their periodicity:

• Final Area = \( \frac{1}{2} \int_{0}^{2\pi} 4 \, dt \)
• Result: Area = \( 4\pi \)

This method reduces geometric complexity to systematic calculus, demonstrating the power of using mathematical theorems and parametrized curves.