Problem 61 - Calculus Early Transcendentals

Question

Find the exact area of the surface obtained by rotating the given curve about the \(x\) -axis. $$x=a \cos ^{3} \theta, \quad y=a \sin ^{3} \theta, \quad 0 \leqslant \theta \leqslant \pi / 2$$

Step-by-Step Solution

Answer

\(\frac{6 \pi a^2}{5}\)

Step 1: Understand the Problem

We need to find the surface area of the curve defined in parametric form when it is rotated around the \(x\)-axis. The given parametric equations are \(x = a\cos^3\theta\) and \(y = a\sin^3\theta\).

Step 2: Formula for Surface Area

The surface area \(S\) of a curve rotated around the \(x\)-axis from \(x=a\) to \(x=b\) is given by: \[ S = 2\pi \int_{a}^{b} y \sqrt{\left( \frac{dx}{d\theta} \right)^2 + \left( \frac{dy}{d\theta} \right)^2} \, d\theta \] where \(y\) is expressed as a function of \(\theta\).

Step 3: Parametric Derivatives

First, calculate \(\frac{dx}{d\theta}\) and \(\frac{dy}{d\theta}\). For \(x = a\cos^3\theta\), \(\frac{dx}{d\theta} = -3a\cos^2\theta\sin\theta\). For \(y = a\sin^3\theta\), \(\frac{dy}{d\theta} = 3a\sin^2\theta\cos\theta\).

Step 4: Plug into Surface Area Formula

Insert the derivatives and the function \(y = a\sin^3\theta\) into the surface area formula. Compute: \[ S = 2\pi \int_{0}^{\pi/2} a\sin^3\theta \sqrt{(-3a\cos^2\theta\sin\theta)^2 + (3a\sin^2\theta\cos\theta)^2} \, d\theta \].

Step 5: Simplify the Expression Under the Integral

Simplify the expression inside the square root: \((-3a\cos^2\theta\sin\theta)^2 + (3a\sin^2\theta\cos\theta)^2 = 9a^2\cos^4\theta\sin^2\theta + 9a^2\sin^4\theta\cos^2\theta = 9a^2\cos^2\theta\sin^2\theta(\cos^2\theta + \sin^2\theta)\).

Step 6: Use Trigonometric Identity

Use \(\cos^2\theta + \sin^2\theta = 1\) to simplify: \[ S = 2\pi \int_{0}^{\pi/2} a\sin^3\theta \cdot 3a\cos\theta\sin\theta \, d\theta = 6\pi a^2 \int_{0}^{\pi/2} \cos\theta\sin^4\theta \, d\theta \].

Step 7: Solve the Integral

Use the substitution \( u = \sin\theta, \, du = \cos\theta d\theta \). Then \(\int \cos\theta\sin^4\theta \, d\theta\) becomes \(\int u^4 \, du = \frac{u^5}{5}\). Substitute back \(u=\sin\theta\): \(\frac{\sin^5\theta}{5}\).

Step 8: Evaluate Definite Integral

Evaluate from \(0\) to \(\pi/2\): \[ 6 \pi a^2 \left[ \frac{\sin^5\theta}{5} \right]_{0}^{\pi/2} = 6 \pi a^2 \left( \frac{1}{5} - 0 \right) = \frac{6 \pi a^2}{5}. \]

Step 9: Final Result

Thus, the exact area of the surface obtained by rotating the curve around the \(x\)-axis is \(\frac{6 \pi a^2}{5}\).

Key Concepts

Parametric Equations

Parametric equations are a powerful tool in calculus to define a set of equations in terms of a parameter, usually denoted by \( \theta \) or \( t \). In this exercise, the equations \( x = a \cos^3 \theta \) and \( y = a \sin^3 \theta \) define a curve in the plane in terms of the parameter \( \theta \). Here, \( a \) is a constant that scales the curve, and \( \theta \) typically varies within a given interval, in this case from \( 0 \) to \( \pi/2 \).
This method allows us to explore curves that are difficult to describe with standard Cartesian equations. As \( \theta \) varies, it traces out a path on the coordinate system, creating an elegant representation of the curve's shape.
Parametric equations simplify the calculation of derivatives, which are essential for determining the surface area of the curve when rotated. By differentiating these parametric equations concerning \( \theta \), we get measures of change for both coordinates, \( \frac{dx}{d\theta} \) and \( \frac{dy}{d\theta} \). These derivatives help us understand the behavior of the curve as \( \theta \) changes and are vital in applying the surface area formula effectively.

Trigonometric Identities

Trigonometric identities can hugely simplify calculations involving parametric equations. One fundamental identity used in this exercise is \( \cos^2 \theta + \sin^2 \theta = 1 \). This identity describes the Pythagorean theorem in trigonometric terms, relating the sine and cosine of an angle.
By using this identity, it becomes easier to simplify the expression within the integral that emerges when calculating the surface area. Specifically, recognizing that \( \cos^2 \theta + \sin^2 \theta = 1 \) drastically reduces complexity, allowing the integral to focus purely on the parts that vary.
Such identities are crucial when finding concise and manageable ways to solve complex integrals within calculus. Familiarizing oneself with these identities enables quicker and more efficient problem solving.

Definite Integrals

Definite integrals play a critical role in calculating the surface area of a curve rotated about an axis. In this exercise, we use a definite integral to accumulate the infinitesimal surface portions along the interval \( \theta = 0 \) to \( \theta = \pi/2 \).
The integral calculates the total surface area by integrating the product of the radius function, \( y = a\sin^3\theta \), and the arc length differential under the square root, weighed by the circle's circumference, \( 2\pi \).
Using substitution, such as \( u = \sin\theta \), greatly simplifies the integral's computation process. This substitution changes the original integral involving trigonometric functions into a polynomial form that is far simpler to evaluate.

• The substitution \( du = \cos\theta \, d\theta \) provides a direct mapping of the \( \cos\theta \sin^4\theta \, d\theta \) integral into the form \( u^4 \, du \).
• Evaluating this integral lets us compute the total surface efficiently, ultimately simplifying the solution to \( \frac{6 \pi a^2}{5} \).

Definite integrals thus not only approximate quantities like areas and volumes but also serve as the foundational tools for exact calculations in calculus.