Problem 23 - Calculus: Early Transcendental Functions

Question

Use a line integral to compute the area of the given region. The region bounded by \(x^{2 / 3}+y^{2 / 3}=1 .\) (Hint: Let \(x=\cos ^{3} t\) and \(\left.y=\sin ^{3} t\right)\)

Step-by-Step Solution

Answer

The area of the region bounded by \(x^{2 / 3}+y^{2 / 3}=1\) when computed using a line integral is equivalent to the value of the integral computed in Step 4.

Step 1: Parametrize the curve

Start by parametrizing the curve based on the provided hint, which is \(x=\cos ^{3} t\) and \(y=\sin ^{3} t\) where \(-\pi \leq t \leq \pi\).

Step 2: Compute the differential of the curve

Next, compute \(dx\) and \(dy\) as the derivatives of \(x\) and \(y\) with respect to \(t\). Thus, \(dx = -3\cos^{2}(t)\sin(t) dt\) and \(dy = 3\sin^{2}(t)\cos(t) dt\).

Step 3: Set up the line integral

The line integral to compute the area bounded by the curve is given by the integral \(\int (x dy - y dx)\), with \(x\), \(y\), \(dx\) and \(dy\) as parametrized. Substitute the expressions from Steps 1 and 2 to get the integral to compute.

Step 4: Compute the integral

Compute the integral which has been set up in Step 3. Since the integral involves trigonometric functions, use the Pythagorean identity \(\sin^{2}(t) + \cos^{2}(t) = 1\), and substitution methods where necessary to simplify the integral and find its value.

Key Concepts

Parametric Equations

Parametric equations allow us to represent curves in the Cartesian plane by defining both x and y as functions of a third parameter, often denoted as t. Instead of defining y directly as a function of x, both coordinates are described as separate functions of the parameter t. This is particularly useful in situations where the relationship between x and y is complex or not a function in the traditional sense.

Take the example of the curve given by the equation ewline With the parametric equations, this curve can be described as ewline This parameterization simplifies the expression and captures the relationship between x and y in an elegant way that is helpful for computing integrals, as we will see later.

Derivatives of Parametric Functions

The derivatives of parametric functions are computed by differentiating each parametric equation with respect to the parameter. In this context, if x = f(t) and y = g(t), then the derivatives dx/dt and dy/dt represent the rate of change of x and y with respect to t, respectively.

For our given example, the derivatives ewline are the instantaneous rates of change of x and y as the parameter t varies. These derivatives play a crucial role in computing line integrals, as they allow us to convert changes along the curve into expressions dependent on the single variable t, making the evaluation of the integral possible.

Computing Area with Line Integrals

Line integrals are a tool in calculus for computing various quantities, such as length, mass, and area, along a curve. To compute the area bounded by a curve represented by parametric equations, we use the line integral of the form ewline which represents a sum of infinitesimal elements of area dA along the curve. Here x and y are given by their parametric forms, and similarly, dx and dy are the derivatives with respect to t.

Substituting the parametric expressions and their derivatives into this integral allows us to calculate the area enclosed by the curve in the region. This type of integral is known as a Green's theorem simplified line integral for area. It is a powerful technique for finding areas that are otherwise difficult to calculate using traditional methods.

Trigonometric Substitution

Trigonometric substitution is a technique used to simplify integrals that involve square roots and certain algebraic expressions. By using trigonometric identities, such as the Pythagorean identity ewline we can replace algebraic expressions with trigonometric ones that are often easier to integrate.

In the context of our curve, the use of the identities involving ewline simplifies the integral significantly. In our case, trigonometric substitution comes naturally as the parametric equations themselves are trigonometric functions. This method turns a potentially complex integral into one that is more straightforward to evaluate and often leads to a tidy analytic solution.