Problem 192 - Calculus Volume 3

Question

Determine a definite integral that represents the area. Region enclosed by one petal of \(r=\cos (3 \theta)\)

Step-by-Step Solution

Answer

The area of one petal is \(\frac{\pi}{12}\).

Step 1: Understand the Equation

The given equation describes a polar curve, specifically a rose curve, defined by \(r = \cos(3\theta)\). This curve has petals, and we need to find the area of one petal.

Step 2: Determine the Limits of Integration

Each petal of the rose curve \(r = \cos(3\theta)\) is completed in the interval \([0, \pi/3]\). This interval is derived because \(3\theta\) covers a full cycle from 0 to \(2\pi\) when \(\theta\) is 0 to \(\pi/2\), so each petal spans a sixth of that (i.e., \(\pi/6\)). However, due to the symmetry and origin crossing of the cosine function, we use \([0, \pi/3]\).

Step 3: Set Up the Integral for the Area

The formula to find the area of a region in polar coordinates is \(A = \int \frac{1}{2} r^2 \, d\theta\). For this curve, the area of one petal is \(\frac{1}{2} \int_{0}^{\pi/3} (\cos(3\theta))^2 \, d\theta\).

Step 4: Simplify and Solve the Integral

The integral becomes \(\frac{1}{2} \int_{0}^{\pi/3} \cos^2(3\theta) \, d\theta\). Use the trigonometric identity \(\cos^2(x) = \frac{1 + \cos(2x)}{2}\) to transform the integral to \(\frac{1}{2} \int_{0}^{\pi/3} \frac{1 + \cos(6\theta)}{2} \, d\theta\). Simplifying, this becomes \(\frac{1}{4} \int_{0}^{\pi/3} (1 + \cos(6\theta)) \, d\theta\).

Step 5: Evaluate the Integral

Evaluate \(\frac{1}{4} [\theta + \frac{1}{6} \sin(6\theta)]_{0}^{\pi/3}\). Substituting the limits gives \(\frac{1}{4} [\frac{\pi}{3} + 0 - (0 + 0)] = \frac{\pi}{12}\).

Step 6: Conclusion

The area of one petal of the rose curve \(r = \cos(3\theta)\) is \(\frac{\pi}{12}\).

Key Concepts

Polar Coordinates

Polar coordinates provide a different way to describe locations in a plane compared to traditional Cartesian coordinates. Instead of using a grid of x and y values to specify points, polar coordinates use a combination of a radius and an angle. This system is particularly useful for dealing with circular or rotational shapes.

• The radius, denoted as \(r\), tells us how far away the point is from the origin (the center of the coordinate system).
• The angle, typically represented by \(\theta\), tells us the direction of the point with respect to the origin, measured in radians from the positive x-axis.

When plotting curves like circles or spirals, polar coordinates can simplify equations and calculations. Understanding polar coordinates is crucial when dealing with equations like \(r = \cos(3\theta)\), where the geometry is inherently circular.

Rose Curve

A rose curve is a type of mathematical curve that resembles the petals of a rose. These curves are defined in polar coordinates by equations of the form \(r = \cos(n\theta)\) or \(r = \sin(n\theta)\), where \(n\) is an integer. The number of petals a rose curve has depends on whether \(n\) is even or odd:

• If \(n\) is odd, the curve has \(n\) petals.
• If \(n\) is even, the curve has \(2n\) petals.

For the equation \(r = \cos(3\theta)\), since 3 is odd, the rose curve will have 3 petals. Each petal is symmetrically placed around the origin and can be calculated efficiently using polar coordinates. When calculating specific regions, such as one petal, understanding this symmetry helps determine the limits of integration.

Trigonometric Identity

Trigonometric identities are equations involving trigonometric functions that hold true for all values of the involved variables. These identities are indispensable tools for simplifying expressions and solving equations in calculus and trigonometry.
A key identity used in solving the definite integral of the rose curve is \(\cos^2(x) = \frac{1 + \cos(2x)}{2}\). This identity helps transform the integrand into a more manageable form for integration. By applying this identity:

• We turn \(\cos^2(3\theta)\) into \(\frac{1 + \cos(6\theta)}{2}\).
• This simplification makes the integration process straightforward by breaking it into easier parts.

Mastering such identities is critical for conducting successful mathematical transformations and calculations.

Integral Evaluation

Integral evaluation refers to the process of finding the area under the curve represented by a given function. In calculus, this is achieved using definite integrals. The area enclosed by one petal of the rose curve described in polar coordinates is found using the definite integral \[ A = \frac{1}{2} \int_{0}^{\pi/3} (\cos(3\theta))^2 \, d\theta \] After applying the trigonometric identity, the integral simplifies to \[ \frac{1}{4} \int_{0}^{\pi/3} (1 + \cos(6\theta)) \, d\theta \] The final step involves evaluating this integral over the interval \([0, \pi/3]\), which considers the complete span of one petal.
Using basic integration techniques:

• The integral of \(1\) with respect to \(\theta\) is straightforward: \(\theta\).
• The integral of \(\cos(6\theta)\) is \(\frac{1}{6} \sin(6\theta)\).

After evaluating the definite integral and applying the limits, we find the area of one petal of the curve to be \(\frac{\pi}{12}\), representing the enclosed space effectively. Understanding the step-by-step breakdowns in integral evaluation is vital for solving complex calculus problems.