Problem 22 - Calculus: Early Transcendental Functions

Question

Use a line integral to compute the area of the given region. The ellipse \(4 x^{2}+y^{2}=4\)

Step-by-Step Solution

Answer

The area of the given region described by the ellipse is \(\pi\).

Step 1: Convert into Polar Coordinates

To simplify the computations, it's best to convert the given ellipse equation into polar form. The following substitutions are used to achieve this: \(x = r \cos(\theta)\) and \(y = r \sin(\theta)\). Substituting these into the ellipse equation \(4 x^{2}+y^{2}=4\), we get that \(4 (r \cos(\theta))^{2}+(r \sin(\theta))^{2}=4\), which simplifies to \(4r^{2}=4\) or \(r=1\). This represents a circle of radius 1 in polar coordinates as the area we need to compute.

Step 2: Parametrizing the Circle for the Line Integral

A parameterization is formed for the circle is given by \(r(t)=\left(\cos(t), \sin(t)\right)\) with t from 0 to \(2 \pi\). The derivative \(r'(t)\) is then computed to be \(-\sin(t), \cos(t)\).

Step 3: Compute the Line Integral

The line integral to find the area is \(A=\frac{1}{2} \int_{C}(x d y-y d x)\). This is translated to polar coordinates as \(A=\frac{1}{2} \int_{0}^{2 \pi} r(t) \cdot r^\prime(t) d t\). By using the calculated values and evaluating the integral, the result is \(\pi\).

Step 4: Apply Green's Theorem

Green's Theorem allows for translating line integrals into double integrals. It gives the relationship \(\int_{C} P d x+Q d y=\iint_{D}\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right) d A\). It can be used to confirm the results obtained in the line integral computation.

Key Concepts

Polar Coordinates

To comprehend the idea behind polar coordinates, imagine a point in a plane. However, unlike the standard Cartesian coordinate system, where this point is defined by its horizontal and vertical distances from a reference point (the origin), in polar coordinates, the location of the point is given by two values: the angle \theta from the positive x-axis and the distance r from the origin.

In terms of a mathematical equation, any point (x, y) in Cartesian coordinates can be translated to polar coordinates \textrm{{(r, }}\theta\textrm{{)}} by using the replacements: \[ x = r \times \textrm{{cos}}(\theta), \] and \[ y = r \times \textrm{{sin}}(\theta). \] This system proves particularly useful when dealing with curves and shapes that are naturally circular or radial, like the ellipse from our exercise. By using polar coordinates, we simplified the problem and revealed the true nature of the ellipse as a circle, making the subsequent steps easier.

Parametrization

Parametrization involves describing a curve or surface in terms of a variable, usually denoted as t, which ranges over an interval. It's like providing a set of instructions that define how to trace out the shape in space as time evolves. For our original problem involving an ellipse, we define a parametrization for the resulting circle in polar coordinates by \[ r(t) = (\textrm{{cos}}(t), \textrm{{sin}}(t)), \] where t ranges from 0 to \( 2\theta \). The beauty of parametrization is that it simplifies complex shapes to a form that's easier to work with mathematically, and in this case, it sets the stage for computing a line integral in a more straightforward way.

Green's Theorem

Linking Line and Double Integrals

Green's Theorem is a powerful tool in calculus that serves as a bridge between line integrals over a curve C and double integrals over a region D enclosed by C. Stated simply, Green's Theorem asserts that the circulation around C can be related to the curl inside D:

\[ \textbf{Green's Theorem:}\ \ \textbf{Circulation-Curl Form}\ \textbf{Equation: }\ \textbf{\( \int_{C} P dx + Q dy = \textbf{\iint_{D} \textbf{\left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) }dA \)}} \] Using this theorem, we can either verify the area calculations made through line integrals or, at times, find a more accessible path to solve the problem by converting it into a double integral over the region. For the ellipse example, we could use Green's Theorem as an alternative means to confirm the area that we originally computed using a line integral.

Double Integral

Understanding Double Integrals

A double integral extends the concept of an integral to functions of two variables. The primary goal is to compute the volume under the surface defined by the function, over a certain region in the xy-plane. In essence, a double integral adds up 'infinitesimally' small regions to find the total volume.

When it pertains to area calculations, as in the case of the ellipse exercise, the double integral formula can be simplified to \( \textbf{\iint_{D} dA \)} where D denotes the region over which we’re integrating. The practicality of double integrals lies in their ability to handle complex regions that might be difficult to navigate using single integrals. In our exercise, if we were to apply Green's Theorem, we could swap the original line integral for a double integral to compute the same area. By adjusting our approach based on the problem at hand, we either use parametrization with a line integral or apply Green's Theorem to shift to a double integral, allowing us to solve and verify our region's area calculation.