Finding the Area Enclosed by Parametric Curves Using Double Integrals: A Practical Example

Introduction

When dealing with complex geometrical shapes defined by parametric equations, determining the enclosed area often requires the application of advanced integration techniques. This article explores the process of finding the enclosed area of a parametric curve utilizing double integrals, specifically focusing on the parametric equations:

x 2cos^3 t,
y 2sin^3 t.

Understanding the Parametric Curve

The parametric curve is defined by the equations:

x 2cos^3 t,
y 2sin^3 t, for 0 ≤ t ≤ 2π.

To find the area enclosed by this curve, we will employ a double integral method and highlight the key steps, including parameter transformation, integration, and the use of symmetry.

Step-by-Step Solution

1. Identify the Bounds for t

The parameter t varies from 0 to 2π to describe the entire curve.

2. Convert to Cartesian Coordinates

Although the curve can be expressed in polar coordinates, it is often simpler to use the parametric equations directly. The given parametric equations can be rewritten to find the area:

x 2cos^3 t,
y 2sin^3 t.

The area A enclosed by a parametric curve can be calculated using the formula:

A ∫ab y (dx/dt) dt

3. Calculate (dx/dt)

The derivative of x with respect to t is:

(dx/dt) -6cos^2 t sin t.

4. Set Up the Integral

The area can thus be calculated as:

A ∫02π 2sin^3 t(-6cos^2 t sin t) dt.

This simplifies to:

A -12 ∫02π sin^4 t cos^2 t dt.

5. Use Symmetry

Since the integrand sin^4 t cos^2 t is periodic with a period of π, we can evaluate it from 0 to π and double it:

A -24 ∫0π sin^4 t cos^2 t dt.

6. Use a Substitution

Let u sin t, then du cos t dt. The limits change as follows:

- When t 0, u 0
- When t π, u 0

The integral becomes:

A -24 ∫00 u^4(1 - u^2) du.

This integral is zero, so we need to evaluate from 0 to 1 and multiply by 2:

A -24 middot; 2 ∫01 u^4(1 - u^2) du.

7. Calculate the Integral

The integral simplifies to:

A -48 ∫01 (u^4 - u^6) du.

Evaluating the integral:

A -48 left[ (u^5/5 - u^7/7) right]_{0}^{1} -48 left(1/5 - 1/7 right).

Finding a common denominator 35:

1/5 - 1/7 7/35 - 5/35 2/35.

Thus:

A -48 middot; 2/35 96/35.

Conclusion

The area enclosed by the curve defined by the parametric equations x 2cos^3 t and y 2sin^3 t is 96/35 square units.

This detailed example demonstrates the process of finding the enclosed area using double integrals and highlights the importance of leveraging symmetry and substitution techniques in solving complex parametric equations.