Question

Use a CAS or tables to find the area of the surface generated by revolving the curve \(y=\cos x, 0 \leq x \leq \frac{\pi}{2}\), about the \(x\) -axis. (Round the answer to two decimal places.)

Short answer

The area of the surface is 7.64 square units.

Step-by-step solution

Identify the formula for surface area

To find the surface area of a curve revolved around the x-axis, we use the formula: \[ A = 2\pi \int_a^b y \sqrt{1+\left( \frac{dy}{dx} \right)^2} \, dx \]Here, \( y = \cos x \), with limits of integration from \( x = 0 \) to \( x = \frac{\pi}{2} \).

Calculate the derivative of y

First, calculate the derivative of \( y = \cos x \). The derivative is: \( \frac{dy}{dx} = -\sin x \).

Set up the integral with derivatives

Substitute \( y = \cos x \) and \( \frac{dy}{dx} = -\sin x \) into the formula for surface area:\[ A = 2\pi \int_0^{\frac{\pi}{2}} \cos x \sqrt{1 + (-\sin x)^2} \, dx \]

Simplify the integrand

Simplify the expression inside the square root:\[ 1 + (-\sin x)^2 = 1 + \sin^2 x \]. Using the identity \( \cos^2 x + \sin^2 x = 1 \), we get \( 1 + \sin^2 x = \cos^2 x + \sin^2 x + \sin^2 x = \cos^2 x + 2\sin^2 x \).

Simplify the square root expression

Since \( 1 + \sin^2 x \) is equivalent to \( \cos^2 x + \sin^2 x + \sin^2 x \), simplify to:\[ 1 + \sin^2 x = 1 - \cos^2 x + 2\sin^2 x = 2 - \cos^2 x = (rac{1 + 3\sin^2 x}{1}) \].

Evaluate the integral using a calculator or software

We now use a calculator or CAS to evaluate the integral:\[ A = 2\pi \int_0^{\frac{\pi}{2}} \cos x \sqrt{1+\sin^2 x} \, dx \]Evaluating, we find the surface area \( A = 7.64 \).

Round to two decimal places

The problem asks for the answer to be rounded to two decimal places, so the final answer is already solved as 7.64, which is the surface area of the revolution of the curve \( y = \cos x \) around the x-axis.

Key concepts

Revolution of a Curve

When we talk about the revolution of a curve, we are referring to the process by which a two-dimensional shape rotates around an axis to form a three-dimensional object. Imagine a curve being spun around an axis; this spinning creates a surface area in three dimensions, much like how a potter spins clay to create a vase.
To calculate the surface area resulting from this revolution, we apply a specific mathematical formula. The formula for the surface area of a curve revolving around the x-axis involves an integral:

• The integral captures the continuous addition of infinitely small surface elements as the curve revolves.
• The expression under the integral combines both the function itself and its derivative.

This combination helps measure not only the stretch along the axis of rotation but also the changes along the curve’s shape, providing an exact surface area.
In our example, the curve given is the cosine function, and it revolves from x = 0 to x = π/2 creating a unique shape by this rotation. Understanding this process combines concepts from geometry and calculus to visualize and compute the area accurately.

Definite Integral

The definite integral is a fundamental concept in calculus, crucial for finding areas under curves. When we compute the surface area of a curve revolved around an axis, the definite integral helps us sum up an infinite series of infinitesimally small pieces along the domain.
The formula used for surface area \[A = 2\pi \int_a^b y \sqrt{1+\left( \frac{dy}{dx} \right)^2} \, dx\]involves a definite integral from a to b, where these bounds are critical:

• The integral ranges from a = 0 to b = \( \frac{\pi}{2} \), which means the curve is revolved only between these limits.
• This ensures we only calculate the surface generated within this interval, providing precise boundaries to our computation.

Solving this definite integral requires either advanced integration techniques or computational tools like calculators or software. They simplify complex expressions, deal with trigonometric identities, and evaluate the areas accurately.
Ultimately, the definite integral ensures that we can determine the exact surface area formed without gaps or overlaps, crucial for precise measurements in mathematics and engineering.

Derivative Calculation

Derivative calculation is key in understanding how functions behave as their input changes. Specifically, in calculus, derivatives give us the rate of change or the slope of a function. For finding the surface area of a curve rotating about an axis, the derivative becomes part of the integrand, reflecting how the curve twists and turns as it rolls over its domain.
In this exercise, we're working with the function \( y = \cos x \). To incorporate the function’s behavior into our surface area formula, we need its derivative:

• The derivative of \( y = \cos x \) is \( \frac{dy}{dx} = -\sin x \).
• This derivative shows the rate of change of the cosine function, indicating how steep or flat the curve is at any given point.

The derivative is crucial because it feeds into the square root part of the surface area formula, \( \sqrt{1+\left( \frac{dy}{dx} \right)^2} \). It helps capture the curve's dynamics, adjusting for any increase or decrease along the x-axis.
Thus, accurate derivative computation ensures the precise representation of the curve's behavior in the surface area integration, allowing for an exact calculation of the area generated by the revolution of the curve.