Problem 61 - Essential Calculus

Question

Write an integral that represents the area of the surface generated by revolving the curve about the \(x\) -axis. Use a graphing utility to approximate the integral. $$ x=\cos ^{2} \theta, \quad y=\cos \theta \quad 0 \leq \theta \leq \frac{\pi}{2} $$

Step-by-Step Solution

Answer

The surface area integral, representing the area of the surface generated by revolving the given curve about the x-axis, is:\[ S = 2\pi\int_0^{\pi/2} \cos\theta \sqrt{1 +\frac{1}{-sin\theta}^2} d\theta \]This integral can't be evaluated analytically, so a graphing utility is needed for an approximation of this area.

Step 1: Calculate dx/dθ and dy/dθ

The first derivative of \(x\) with respect to \(\theta\) is: \( dx/d\theta = -2\cos\theta\sin\theta \) which is equal to -\(sin(2\theta)\). The first derivative of \(y\) with respect to \(\theta\) is: \( dy/d\theta = -sin\theta \)

Step 2: Substitute into the Surface Area Integral

Substitute \(x\), \(y\), \(dx/d\theta\), and \(dy/d\theta\) into the general form for the surface area integral, \(S = 2\pi \int_a^b y \sqrt{1+(dy/dx)^2} dx \), and replace \(dx\) by \( dx/d\theta * d\theta \) . This is how we handle parametric equations. The integral becomes:\[ S = 2\pi\int_0^{\pi/2} \cos\theta \sqrt{1 +(\frac{-sin\theta}{-sin(2\theta)})^2}(-sin(2\theta)) d\theta \]

Step 3: Simplify and Evaluate Integral

Simplify this integral and evaluate either analytically or using a graphing utility as the instructions specify: \[ S = 2\pi\int_0^{\pi/2} \cos\theta \sqrt{1 +\frac{1}{-sin\theta}^2}d\theta \] The flight value of this integral can't be expressed in terms of elementary functions, hence we would need a graphing utility or similar tool to find a numerical approximation for this integral.

Key Concepts

Calculus

Calculus is a branch of mathematics focused on limits, functions, derivatives, integrals, and infinite series. It has widespread applications in science, engineering, and economics, and can help solve problems in physics and mathematics that are not be solvable by simple algebra alone.

When it comes to functions explained with calculus, such as the case with the surface area of revolution, we use integrals to calculate the total area across a curve. In the context of the surface area of a shape generated by revolving a curve, calculus allows us to sum up infinite, infinitesimally small sections of the surface, which would otherwise be impossible to calculate through basic algebraic means.

Parametric Equations

Parametric equations are a set of equations that express a set of quantities as explicit functions of a number of independent variables, known as parameters. These are powerful tools in calculus, particularly when dealing with curves and surfaces in two and three dimensions, since they enable the definition of these forms without the need for an explicit \(y=f(x)\) relationship.

In the surface area problem we're discussing, the parametric equations \(x=\cos^2\theta\) and \(y=\cos\theta\) define the curve. What makes parametric equations special is their ability to describe motion and change, making them ideal for creating representations of more complex curves and shapes, such as the one generated when the given curve revolves around the \(x\)-axis.

Surface Area Integral

The surface area integral is one of the most fascinating applications of calculus. When you have a curve or line and revolve it about an axis, the surface area integral provides a method to calculate the area of the surface formed. This type of integral takes into account not only the shape of the curve but also the path through which it travels as it revolves.

To find the surface area of a curve revolved around the \(x\)-axis, we use the integral formula \(S = 2\pi \int_a^b y \sqrt{1+(dy/dx)^2} dx\), which accounds for the curve's vertical distance \(y\) from the axis of revolution through the term \(2\pi y\). In the given exercise, after substituting the expressions for \(x\), \(y\), and their derivatives with respect to \(\theta\), we obtain an integral that calculates the total surface area of the 3D shape resulting from the curve's rotation.

Graphing Utility

A graphing utility is an invaluable tool in both education and professional settings when dealing with complex calculus problems. It could be a software program, calculator feature, or an online platform that plots graphs, solves equations, and calculates numerical approximations of integrals and derivatives. For many integrals, especially those that can't be easily evaluated using standard methods, a graphing utility provides a way to approximate the solution.

Given that the final integral from our exercise cannot be expressed simply in terms of elementary functions, a graphing utility becomes indispensable. To approximate the value of the surface area of revolution, the utility will perform numerical integration, which could include methods like Riemann sums, Simpson's rule, or the trapezoidal rule. These are algorithms to estimate the area under a curve, which in our case helps find the surface area of the shape created by the revolving curve.