Question

Find the area enclosed by the ellipse \(x=a \cos \theta, y=b \sin \theta, 0 \leq \theta<2 \pi\).

Short answer

The area enclosed by the ellipse is \( \pi ab \).

Step-by-step solution

Recognize the Ellipse Equation

The given parametric equations represent an ellipse in the form of \( x = a \cos \theta \) and \( y = b \sin \theta \). The ellipse is centered at the origin with semi-major axis \( a \) parallel to the x-axis and semi-minor axis \( b \) parallel to the y-axis.

Area Formula Derivation

The area \( A \) enclosed by the ellipse can be derived using calculus. The integral for the area in terms of \( \theta \) is given by: \[A = \int_{0}^{2\pi} x \frac{dy}{d\theta}\, d\theta\]Substitute \( x = a\cos\theta \) and \( y = b\sin\theta \), then \( \frac{dy}{d\theta} = b\cos\theta \).

Setup the Integral

Substituting the parametric expressions into the integral, we have:\[A = \int_{0}^{2\pi} a\cos\theta \cdot b\cos\theta \ \, d\theta\]This simplifies to:\[A = ab \int_{0}^{2\pi} \cos^2\theta \, d\theta\]

Use Trigonometric Identity

Utilize the trigonometric identity \( \cos^2 \theta = \frac{1 + \cos(2\theta)}{2} \) to simplify the integral:\[A = ab \int_{0}^{2\pi} \frac{1 + \cos(2\theta)}{2} \, d\theta\]

Evaluate the Integral

The integral simplifies to:\[A = \frac{ab}{2} \left( \int_{0}^{2\pi} 1 \, d\theta + \int_{0}^{2\pi} \cos(2\theta) \, d\theta \right)\]The integral of 1 over \( 0 \) to \( 2\pi \) is \( 2\pi \), and the integral of \( \cos(2\theta) \) over \( 0 \) to \( 2\pi \) is zero. Thus, \[ A = \frac{ab}{2} (2\pi) = \pi ab \]

Conclusion

The area enclosed by the ellipse is \( \pi ab \), which is the standard area formula for an ellipse with semi-major axis \( a \) and semi-minor axis \( b \).

Key concepts

Ellipse

An ellipse is a smooth, closed curve resembling an elongated circle. It can be visualized as the shape formed when a plane intersects a cone, resulting in a symmetric, oval-like structure. The key axes of an ellipse are:

• Semi-major axis (a): The longest diameter of the ellipse, stretching from one end of the ellipse to the other, passing through the center.
• Semi-minor axis (b): The shortest diameter, perpendicular to the semi-major axis, extending across the ellipse.

The characteristics of an ellipse can be captured using parametric equations, which come in handy when calculating properties like area and arc length. In the given exercise, the ellipse is represented by the parametric equations \( x = a \cos \theta \) and \( y = b \sin \theta \). Here, \( a \) and \( b \) represent the lengths of the semi-major and semi-minor axes, respectively. Understanding the geometry of the ellipse is crucial for grasping concepts related to its area, perimeter, and other properties.

Parametric Equations

Parametric equations express the coordinates of the points making up a geometric object, using one or more parameters. Instead of defining \( y \) directly as a function of \( x \), parametric equations specify both \( x \) and \( y \) in terms of another variable, often denoted as \( t \) or \( \theta \). This approach is beneficial in various scenarios:

• Describing curves: Parametric equations often describe more complex curves, such as ellipses or circles, that cannot be expressed as a simple function \( y = f(x) \).
• Animating objects: By manipulating the parameter value, it's possible to animate or trace the path of an object over time.

In the exercise, the parameter \( \theta \) is employed to describe the ellipse's points using the equations \( x = a \cos \theta \) and \( y = b \sin \theta \). Such expressions allow precise calculation of the ellipse's features, like its enclosed area, by setting the parameter range \( 0 \leq \theta < 2\pi \). Navigating through a curve efficiently requires understanding parametric representations and their transformations.

Integral Calculus

Integral calculus is a branch of calculus concerned with the accumulation of quantities and the areas under and between curves. In the context of the ellipse, integral calculus helps compute the area enclosed by the shape, utilizing the parametric representation. The key steps involved in the exercise include:

• Setting up the Integral: Using the parametric equations for \( x \) and \( y \), the integral to find the area becomes \( A = \int_{0}^{2\pi} x \frac{dy}{d\theta} \ d\theta \).
• Solving the Integral: By substituting the derivative \( \frac{dy}{d\theta} = b\cos\theta \), the area simplifies to \( A = ab \int_{0}^{2\pi} \cos^2\theta \ d\theta \).

Integral calculus not only assists in finding areas but also in calculating volumes, work done by a force, and most importantly, solving differential equations. It is a powerful tool for numerous applications across mathematics and physics. By recognizing patterns within integrals and applying appropriate techniques, such as the substitution of parametric equations, surprisingly complex problems can become manageable.

Trigonometric Identities

Trigonometric identities are equalities involving trigonometric functions that hold true for all values in their domains. These identities simplify potentially complicated trigonometric expressions, proving indispensable in calculus, particularly for integration or differentiation of trigonometric integrals.

• Pythagorean Identity: \( \sin^2\theta + \cos^2\theta = 1 \)
• Double Angle Identity: \( \cos^2\theta = \frac{1 + \cos(2\theta)}{2} \)

These identities are leveraged in solving the given problem and simplifying integrals. In the exercise, the double angle identity transforms \( \cos^2\theta \) into an integrable form, breaking the integral into straightforward parts that are easier to evaluate:

• \( \int_{0}^{2\pi} 1 \ d\theta = 2\pi \)
• \( \int_{0}^{2\pi} \cos(2\theta) \ d\theta = 0 \)

The resulting expression leads to the elegant solution \( A = \pi ab \). Mastery of trigonometric identities is essential, as they often unlock more efficient pathways to solutions, especially when dealing with periodic functions and curves described by these functions.