Problem 5 - Thomas Calculus

Question

Find the areas of the regions in Exercises \(1-8\) Inside one leaf of the four-leaved rose \(r=\cos 2 \theta\)

Step-by-Step Solution

Answer

The area of one leaf is \(\frac{\pi}{8}\).

Step 1: Understand the Problem

We need to find the area of one leaf of the polar curve given by the equation \(r = \cos 2\theta\). This curve is a four-leaved rose.

Step 2: Determine the Range for One Leaf

The function \(r = \cos 2\theta\) completes one leaf when \(\theta\) ranges from \(-\frac{\pi}{4}\) to \(\frac{\pi}{4}\) due to its periodic nature and symmetry.

Step 3: Set Up the Area Integral

The area \(A\) enclosed by a polar curve \(r(\theta)\) from \(\theta = a\) to \(\theta = b\) is given by the formula \(A = \frac{1}{2} \int_{a}^{b} (r(\theta))^2 \, d\theta\).

Step 4: Integrate to Find the Area

Substitute \(r = \cos 2\theta\) and the limits \(-\frac{\pi}{4}\) to \(\frac{\pi}{4}\) into the formula: \[ A = \frac{1}{2} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} (\cos 2\theta)^2 \, d\theta \]Use the identity \(\cos^2 x = \frac{1}{2}(1 + \cos 2x)\) to transform the integrand:\[ A = \frac{1}{2} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{1}{2} (1 + \cos 4\theta) \, d\theta \]This simplifies to:\[ A = \frac{1}{4} \left[\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} 1 \, d\theta + \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \cos 4\theta \, d\theta \right] \]

Step 5: Solve the Integrals

Compute the definite integrals separately:- For \(\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} 1 \, d\theta\), the result is \(\theta\bigg|_{-\frac{\pi}{4}}^{\frac{\pi}{4}} = \frac{\pi}{4} - (-\frac{\pi}{4}) = \frac{\pi}{2}\).- For \(\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \cos 4\theta \, d\theta\), recognize the antiderivative is \(\frac{1}{4}\sin 4\theta\) and evaluate it between the limits: \(\frac{1}{4}[\sin 4\theta]\bigg|_{-\frac{\pi}{4}}^{\frac{\pi}{4}} = \frac{1}{4}[0 - 0] = 0\) (since \(\sin 4(\pm \frac{\pi}{4}) = \sin (\pm \pi) = 0\))

Step 6: Calculate the Final Area

Combine the results from Step 5:\[ A = \frac{1}{4} \left( \frac{\pi}{2} + 0 \right) = \frac{\pi}{8} \]

Step 7: Conclusion

Thus, the area of one leaf of the rose \(r = \cos 2\theta\) is \(\frac{\pi}{8}\).

Key Concepts

Polar Coordinates

Polar coordinates provide a system for graphing points on a plane using a radius and angle, rather than the traditional Cartesian coordinates. This approach is very effective for problems involving circular and spiral patterns.
In this specific case, we're dealing with the polar curve described by the equation \( r = \cos 2\theta \), which forms a four-leaved rose. The radius \( r \) is dependent on the angle \( \theta \), indicating that as the angle changes, so does the distance from the origin to any point on the curve.

It's often helpful to visualize these curves by plotting various points as \( \theta \) increases, seeing how \( r \) changes. For the curve \( r = \cos 2\theta \), we notice it completes its pattern as \( \theta \) ranges from \( -\frac{\pi}{4} \) to \( \frac{\pi}{4} \) for one leaf. Each leaf is symmetrical, an important concept when finding areas enclosed by these curves.

Integral Calculus

Integral calculus is the mathematical technique used to calculate areas under curves and is essential for determining the area of regions within polar graphs.

To find the area enclosed by one leaf of the rose curve \( r = \cos 2\theta \), we use the formula for the area \( A \) in polar coordinates:

• \( A = \frac{1}{2} \int_{a}^{b} (r(\theta))^2 \, d\theta \)

By substituting \( r(\theta) = \cos 2\theta \) and choosing the appropriate limits of integration (\( -\frac{\pi}{4} \) to \( \frac{\pi}{4} \)), we can set up our integral to calculate the area of one leaf. Having the right limits ensures only one leaf is considered in the integration range. Understanding these setups is vital in solving polar area problems correctly.

Trigonometric Identities

Trigonometric identities are critical tools in simplifying integrals, especially those involving trigonometric functions.
To solve the integral \( \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} (\cos 2\theta)^2 \, d\theta \), it's necessary to use the identity:

• \( \cos^2 x = \frac{1}{2}(1 + \cos 2x) \)

This identity allows us to transform the integral of \( (\cos 2\theta)^2 \) into a more manageable form, splitting it into two simpler integrals: one constant and another involving \( \cos 4\theta \).

By evaluating these integrals separately, we simplify our calculations. This identity, and others like it, are pivotal in transforming complex trigonometric integrals into computable forms. Knowing when and how to use these identities can greatly simplify the process of finding areas using integral calculus.