Problem 31 - Calculus Early Transcendentals

Question

Evaluate \(\oint_{C} y d x-x d y,\) where \(C\) is the cardioid $$ r=a(1+\cos \theta) \quad(0 \leq \theta \leq 2 \pi) $$

Step-by-Step Solution

Answer

The integral evaluates to \(-3\pi a^2\).

Step 1: Understanding the problem

We need to evaluate the line integral \( \oint_{C} y \, dx - x \, dy \) over the curve \( C \), which is given as a cardioid in polar coordinates: \( r = a(1 + \cos \theta) \).

Step 2: Convert the Curve to Cartesian Coordinates

In polar coordinates, we have \( x = r \cos \theta \) and \( y = r \sin \theta \). Substituting \( r = a(1 + \cos \theta) \), the parametric equations for the cardioid are \( x = a(1 + \cos \theta) \cos \theta \) and \( y = a(1 + \cos \theta) \sin \theta \).

Step 3: Calculate dx and dy

Differentiate the parametric equations with respect to \( \theta \) to find \( dx \) and \( dy \). - \( dx = \left( \frac{d}{d\theta} [a(1 + \cos \theta) \cos \theta] \right) \, d\theta \) - \( dy = \left( \frac{d}{d\theta} [a(1 + \cos \theta) \sin \theta] \right) \, d\theta \)

Step 4: Set Up and Simplify the Integral

Plug \( dx \) and \( dy \) into the integral \( \oint_{C} y \, dx - x \, dy \). Simplify the expression using trigonometric identities and substitutions to set up a manageable integral. Notice that the resulting integral can be difficult in its raw form, so check for symmetry and other properties to simplify it.

Step 5: Use Green's Theorem

Instead of evaluating directly via parameterized integration over \( \theta \), use Green’s Theorem: \[ \oint_{C} y \, dx - x \, dy = \iint_{R} \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dA \]Where \( P = y \) and \( Q = -x \), calculate: \[ \frac{\partial (-x)}{\partial x} - \frac{\partial (y)}{\partial y} = -1 - 1 = -2 \]Thus the integral over the region \( R \), enclosed by \( C \), is \( -2 \cdot \text{area of the cardioid} \).

Step 6: Calculate the Area of the Cardioid

The area \( A \) of a cardioid given by \( r = a(1 + \cos \theta) \) is known to be \( 3\pi a^2 / 2 \).

Step 7: Final Calculation and Solution

Using the area of the cardioid from Step 6 in the Green's Theorem result from Step 5:\[ \oint_{C} y \, dx - x \, dy = -2 \times \frac{3\pi a^2}{2} = -3\pi a^2 \].

Key Concepts

Line Integral

A line integral is a means of integrating functions along a curve. It is particularly useful in fields like physics and engineering, where you may want to find the work done by a force field along a path. In calculus, the line integral of a vector field over a curve \( C \), for example, can be expressed as \( \oint_{C} \mathbf{F} \cdot d\mathbf{r} \). Here, \( \mathbf{F} \) represents a vector function and \( d\mathbf{r} \) is a differential element along the curve.

In this exercise, the line integral \( \oint_{C} y \, dx - x \, dy \) is taken over a cardioid-shaped curve \( C \). This type of integral measures the circulation of the field around the path. By transforming from the curve's polar coordinates into Cartesian coordinates, one can convert the integral into a form that is easier to evaluate. This often involves calculating the derivatives of the parameterized equations for \( x \) and \( y \) in terms of \( \theta \), the angle in polar coordinates.

• Useful for field applications.
• Converts geometrical shapes into analytically solvable problems.
• Important for understanding physical behavior over paths.

Green's Theorem

Green's Theorem provides a powerful relation between a line integral around a simple, closed curve \( C \) and a double integral over the plane region \( R \) that is bounded by \( C \). Essentially, it translates the complexity of evaluating a line integral into a potentially simpler problem of evaluating a double integral.

The theorem is expressed as:
\[ \oint_{C} P \, dx + Q \, dy = \iint_{R} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA \]
In this formula, \( P \) and \( Q \) are functions of \( x \) and \( y \). In the given exercise, \( P = y \) and \( Q = -x \) are used. Applying Green's Theorem helps break down the original problem into more manageable calculations—calculating the partial derivatives and the area of the region \( R \).

The result heavily relies on the simplification provided by the theorem, which leverages the symmetry or specific properties of the region \( C \). Especially in polar coordinates, these relations can drastically simplify the computational effort.

• Links line integrals with double integrals.
• Simplifies complex calculations with region properties.
• Especially useful in evaluating circulations and fluxes.

Polar Coordinates

In mathematics, polar coordinates provide an alternative to the Cartesian x-y coordinate system and are often useful for problems involving circular or spiral shapes. In this system, a point in the plane is determined by a distance from a reference point (called the origin) and an angle from a reference direction.

The cardioid in the exercise is defined in polar coordinates as specified by \( r = a(1 + \cos \theta) \), where \( r \) is the radius or the distance from the origin, and \( \theta \) is the angle. Such equations are advantageous when dealing with naturally circular or heart-shaped paths, like cardioids, because they align with the natural curvature of the shape.

To convert polar coordinates to Cartesian, you use the transformations:
\[ x = r \cos \theta \]
\[ y = r \sin \theta \]
These equations allow you to express the position and shape of curves or figures in a more algebraically amenable form. Analyzing these curves using polar coordinates can significantly streamline calculations by leveraging the geometric properties of the curve.

• Facilitates handling circular and spiral designs.
• Converts complex curves into simpler expressions.
• Enhances solving integrals by matching shape symmetry.