Problem 23 - Thomas Calculus

Question

Find the area enclosed by the ellipse $$ x=a \cos t, \quad y=b \sin t, \quad 0 \leq t \leq 2 \pi $$

Step-by-Step Solution

Answer

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

Step 1: Understand the Problem

Understand that the problem requires us to find the area enclosed by an ellipse defined parametrically by the equations for x and y. The parametric equations are: \( x = a \cos t \) and \( y = b \sin t \), and \( t \) ranges from 0 to \( 2\pi \), which means a full rotation around the ellipse.

Step 2: Formula for the Area of an Ellipse

Recall the formula for the area enclosed by an ellipse in the standard form \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). The area \( A \) of this ellipse is given by \( A = \pi ab \). In the problem, \( a \) and \( b \) are the semi-major and semi-minor axes of the ellipse defined by the parametric equations.

Step 3: Compute the Area Using the Formula

Since the given parametric equations represent an ellipse with semi-major axis \( a \) and semi-minor axis \( b \), substitute \( a \) and \( b \) into the area formula. Therefore, the area \( A \) is \( A = \pi ab \).

Step 4: Verification by Parametric Integration

To verify the result, consider using parametric integration. The integral for the area is \( \int_{0}^{2\pi} \frac{1}{2} (x \frac{dy}{dt} - y \frac{dx}{dt}) \, dt \). Calculate \( \frac{dx}{dt} = -a \sin t \) and \( \frac{dy}{dt} = b \cos t \). Substitute to get \( \frac{1}{2} \int_{0}^{2\pi} (a b \sin^2 t + ab \cos^2 t) \, dt = \pi ab \), which matches the formula-derived area.

Key Concepts

Parametric Equations

Parametric equations are an elegant way to define mathematical curves using a parameter, typically denoted as \( t \). Instead of expressing one variable in terms of another, parametric equations express each variable as a function of \( t \). For an ellipse, the parametric equations are given by:

• \( x = a \cos t \)
• \( y = b \sin t \)

This means that as \( t \) varies, \( (x, y) \) traces out the shape of an ellipse. Here \( a \) and \( b \) represent the lengths of the semi-major and semi-minor axes of the ellipse, respectively. This approach lets you easily handle more complex curves and transformations that are difficult to express in traditional Cartesian coordinates.
These equations play a crucial role in analyzing the motion around an ellipse, as the parameter \( t \) typically ranges from 0 to \( 2\pi \) for a full traversal of the shape.

Area of an Ellipse

The area enclosed by an ellipse can be calculated easily once we understand its mathematical representation. An ellipse can be described in the standard form equation: \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\]Mythologically, this shape is known as a squished circle and its area can be derived from integrating over its boundaries. However, a convenient formula, without computing a complex integral, for finding the area is: \[ A = \pi ab\]Here \( a \) represents the semi-major axis and \( b \) represents the semi-minor axis. This formula is derived considering a full traversal around the ellipse, making it an efficient method for calculating the area directly from the parameters of the ellipse. Understanding this formula allows us to skip complex calculations when dealing directly with standard ellipses.

Integration

Integration is a powerful mathematical tool used to accumulate quantities over a given range. In terms of geometry, it helps in finding areas under curves and enclosed shapes. When considering parametric curves, like an ellipse, we use a specific kind of integration known as parametric integration.
In our problem, to find the area enclosed by the ellipse using integration, we use parametric integration with a parametric equation setup. The formula used is:

• \( \int_{0}^{2\pi} \frac{1}{2} (x \frac{dy}{dt} - y \frac{dx}{dt}) \, dt \)

For the given parametric equations of the ellipse, \( \frac{dx}{dt} = -a \sin t \) and \( \frac{dy}{dt} = b \cos t \). When these are substituted into the integral, the calculation simplifies to:

• \( \int_{0}^{2\pi} \frac{1}{2} (a b \sin^2 t + ab \cos^2 t) \, dt \)

This expression then resolves to \( \pi ab \), verifying the formula-derived area. Integration thus not only confirms existing formulas but also provides a systematic way of finding areas, particularly when formulas might be complex or unknown.