Question

For the following exercises, find the area of the surface obtained by rotating the given curve about the \(x\) -axis\(x=a \cos ^{3} \theta, \quad y=a \sin ^{3} \theta, \quad 0 \leq \theta \leq \frac{\pi}{2}\)

Short answer

The area of the surface is \( \frac{6\pi a^2}{5} \).

Step-by-step solution

Express the curve in parametric form

The given curve is already in a parametric form with parameters: \[ x = a \cos^3 \theta, \quad y = a \sin^3 \theta \]where \(0 \leq \theta \leq \frac{\pi}{2}\). This will be useful for finding derivatives later.

Find the derivatives

We need the derivatives of \(x\) and \(y\) with respect to \(\theta\):\( \frac{dx}{d\theta} = -3a\cos^2\theta\sin\theta \)\( \frac{dy}{d\theta} = 3a\sin^2\theta\cos\theta \)

Set up and simplify the surface area formula

The formula for the surface area generated by rotating a parametric curve \((x(t), y(t))\) about the \(x\)-axis is:\[ A = \int 2\pi y \sqrt{(\frac{dx}{d\theta})^2 + (\frac{dy}{d\theta})^2} \, d\theta \]Substitute the given equations:\[ A = \int_0^{\frac{\pi}{2}} 2\pi (a\sin^3\theta) \sqrt{(-3a\cos^2\theta\sin\theta)^2 + (3a\sin^2\theta\cos\theta)^2} \, d\theta \]

Simplify the integrand

Simplify the expression under the square root:\[ (-3a\cos^2\theta\sin\theta)^2 + (3a\sin^2\theta\cos\theta)^2 = 9a^2\cos^4\theta\sin^2\theta + 9a^2\sin^4\theta\cos^2\theta \]Factor out the common term:\[ 9a^2\cos^2\theta\sin^2\theta(\cos^2\theta + \sin^2\theta) = 9a^2\cos^2\theta\sin^2\theta \]Because \(\cos^2\theta + \sin^2\theta = 1\), the expression simplifies to:\[ 3a^2|\sin2\theta| \]

Complete the surface area integral

Substitute the simplified expression back into the integral:\[ A = \int_0^{\frac{\pi}{2}} 2\pi a\sin^3\theta \cdot 3a|\sin2\theta|\cos\theta \sin\theta \, d\theta \]This simplifies to:\[ A = \int_0^{\frac{\pi}{2}} 6\pi a^2 \sin^4\theta \cos\theta \, d\theta \]

Solve the integral

Use substitution to solve the integral. Let \(u = \sin\theta\), then \(du = \cos\theta \, d\theta\).When \(\theta = 0\), \(u = 0\), and when \(\theta = \frac{\pi}{2}\), \(u = 1\).The integral becomes:\[ 6\pi a^2 \int_0^1 u^4 \, du = 6\pi a^2 \left[ \frac{u^5}{5} \right]_0^1 = 6\pi a^2 \left( \frac{1}{5} \right) = \frac{6\pi a^2}{5} \]

Conclusion

The area of the surface obtained by rotating the curve around the \(x\)-axis is: \[ \frac{6\pi a^2}{5} \]

Key concepts

Surface Area

Calculating the surface area of a 3D object is essential in geometry and calculus, especially for objects formed by rotating a curve. In this exercise, we explore the concept of surface area generated by rotating a parameterized curve about the x-axis. The formula for finding the surface area of a parametric curve

• rotated about the x-axis is: \[ A = \int 2\pi y \sqrt{(\frac{dx}{d\theta})^2 + (\frac{dy}{d\theta})^2} \, d\theta\].

By understanding this formula, students learn how to apply it to real-world problems involving rotations and three-dimensional shapes.
The expression under the square root is derived from the Pythagorean identity involving derivatives of parametric equations, ensuring the curve's accurate measurement.
Parametric curves, when rotated, create surfaces whose area is solved by integrating the function over a given interval. In this case, the interval for \(\theta\) is from 0 to \(\frac{\pi}{2}\), ensuring we cover one complete rotation section of the original curve.

Parametric Equations

Parametric equations are a way to express a curve by defining both x and y according to a third variable, often denoted as \(t\) or \(\theta\). This approach is particularly useful in calculus for describing complex curves not easily handled by standard Cartesian equations.
In this exercise, the curve is expressed parametrically as:

• \(x = a \cos^3 \theta\)
• \(y = a \sin^3 \theta\)

By parameterizing the curve, we gain flexibility in calculations and can easily find derivatives with respect to the parameter \(\theta\).
This simplification is crucial for calculating quantities such as the curve's surface area when it is rotated about an axis.
Furthermore, parametric equations simplify the process of differentiating and offer a compact form to calculate other attributes like tangents or arc length, encompassing a wide range of mathematical concepts within accessible formulas.

Curve Rotation

Curve rotation transforms a two-dimensional parametric curve into a three-dimensional surface by rotating it around an axis, in this case, the x-axis. This principle is foundational in creating models of rotational symmetry and is often used in physics and engineering to analyze rotational structures.
The rotation of the curve \(x = a\cos^3\theta\) and \(y = a\sin^3\theta\) generates a three-dimensional surface that is uniquely characterized by its stability and uniformity around the x-axis.
Understanding curve rotation involves understanding the dynamics of how each differential segment of the curve contributes to the entire surface area when revolved.
By solving and simplifying the integral for surface area, we effectively measure the extent of this rotation to get quantities like {\(\frac{6\pi a^2}{5}\), showcasing the inherent properties of the curve and its geometry in a full circular sweep.