Problem 24 - Thomas Calculus

Question

Write an integral for the area of the surface generated by revolving the curve \(y = \cos x , - \pi / 2 \leq x \leq \pi / 2 ,\) about the \(x\) -axis. In Section 8.4 we will see how to evaluate such integrals.

Step-by-Step Solution

Answer

\( A = \int_{-\pi/2}^{\pi/2} 2\pi \cos x \sqrt{1 + \sin^2 x} \, dx \).

Step 1: Understanding the Surface Area Formula

When a curve defined by a function \( y = f(x) \) is revolved around the \( x \)-axis, the surface area \( A \) generated can be found using the formula \( A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx \). Here, we need to identify \( f(x) \) and determine its derivative.

Step 2: Identifying Function and Calculating Derivative

Identify the function \( f(x) = \cos x \). Now, calculate the derivative \( \frac{dy}{dx} = -\sin x \). This derivative will be used to find \( \left( \frac{dy}{dx} \right)^2 \) as it appears in the surface area formula.

Step 3: Calculating the Derivative Square

Compute \( \left( \frac{dy}{dx} \right)^2 = (-\sin x)^2 = \sin^2 x \). This will be plugged into the surface area formula to simplify the integrand.

Step 4: Setting Up the Integral

Substitute into the surface area formula: \( A = \int_{-\pi/2}^{\pi/2} 2\pi \cos x \sqrt{1 + \sin^2 x} \, dx \). This integral represents the surface area of the curve \( y = \cos x \) revolved around the \( x \)-axis from \( x = -\pi/2 \) to \( x = \pi/2 \).

Key Concepts

Integral Calculus

Integral calculus is a branch of mathematics that is crucial for solving problems involving areas, volumes, and other concepts that require summation of small quantities. It involves finding the integral of functions, which can be understood as the reverse of differentiation. In the context of surface areas, integral calculus helps us sum up the infinitely small surface pieces generated when a curve revolves around an axis. By doing so, we can determine the total surface area of the shape created by this revolution.

• The integral of a function over an interval gives the area under the curve of that function within that specific interval.
• In our exercise, we are dealing with the trigonometric function cosine, and calculating the surface area generated when it is revolved about the x-axis.

Integrating over the range from i.e., the function from i.e., will give us the total surface area. It's a way to handle complex shapes and functions systematically and precisely.

Surface Area Formula

The surface area formula is essential when calculating the area of surfaces generated by revolving a curve around an axis. When a curve described by a function is spun around the x-axis, the surface area can be computed using the formula:
\[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx \]

• \( f(x) \) is the function describing the curve you're revolving.
• \( \frac{dy}{dx} \) is the derivative of the function, indicating how the function's value changes with respect to x.
• \( a \) and \( b \) are the limits of integration, defining the interval over which you're revolving the curve.

This formula takes into account not just the basic length of the curve, but also the way it twists and turns by incorporating the derivative.

Derivatives

A derivative represents the rate of change of a function with respect to a variable. In basic terms, it's like understanding how the slope of a curve changes at any given point. Derivatives are crucial when dealing with the surface area formula because they help calculate how the curve's gradient adds to the overall surface area when the curve spins around an axis.
For the function in our exercise, \( y = \cos x \), the derivative is \( \frac{dy}{dx} = -\sin x \).

• This tells us how rapidly \( y \) changes as \( x \) changes, impacting the surface area.
• Squaring the derivative, \((-\sin x)^2 = \sin^2 x\), simplifies the integrand in the formula, making it easier to solve the integral.

Understanding derivatives in this context helps us unpack how they contribute to shaping the surface formed by revolution.

Trigonometric Functions

Trigonometric functions, like sine and cosine, are fundamental in calculus, especially when dealing with curves related to oscillatory scenarios or circular motion. These functions describe relationships in triangles, commonly appearing when dealing with repetitive or circular patterns.
In this exercise, we are working with the cosine function, \( y = \cos x \), which is one of the primary trigonometric functions.

• When revolutionizing around the x-axis, its unique periodicity influences the shape and size of the surface generated.
• By understanding the behavior of \( \cos x \), we can predict the shape of the surface area computed over a specific interval \(-\pi/2 \) to \( \pi/2 \).

Trigonometric identities and relationships, such as \( \sin^2 x + \cos^2 x = 1 \), play a pivotal role in simplifying calculations during integration processes. These functions help frame our understanding of periodic functions and their applications in calculus.