Problem 25 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the area of the surface generated by revolving the given curve about the \(x\) -axis \(y=6 x, 0 \leq x \leq 1\)

Step-by-Step Solution

Answer

The surface area is \( 6\pi \sqrt{37} \).

Step 1: Define the Surface Area Formula

The formula to find the surface area when a curve is rotated about the x-axis is given by \[ S = \int_a^b 2\pi y \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \]where \( y \) is the function in terms of \( x \), and the limits \( a \) to \( b \) are the range of \( x \). In this case, \( y = 6x \), \( a = 0 \), and \( b = 1 \).

Step 2: Differentiate the Function

Differentiate \( y = 6x \) with respect to \( x \). The derivative is: \[ \frac{dy}{dx} = 6 \]

Step 3: Plug into the Surface Area Integral

Substitute \( y \) and \( \frac{dy}{dx} \) into the surface area formula:\[ S = \int_0^1 2\pi (6x) \sqrt{1 + 6^2} \, dx \]which simplifies to:\[ S = \int_0^1 12\pi x \sqrt{37} \, dx \]

Step 4: Evaluate the Integral

Factor out constants and compute the integral:\[ S = 12\pi \sqrt{37} \int_0^1 x \, dx \]The integral of \( x \) from 0 to 1 is:\[ \int_0^1 x \, dx = \left[ \frac{x^2}{2} \right]_0^1 = \frac{1^2}{2} - \frac{0^2}{2} = \frac{1}{2} \]

Step 5: Calculate the Final Answer

Substitute back into the expression:\[ S = 12\pi \sqrt{37} \cdot \frac{1}{2} \]Multiply to get the final result:\[ S = 6\pi \sqrt{37} \]

Key Concepts

Definite Integrals

A definite integral is a crucial concept in calculus that helps to find the area under a curve between two points. In our exercise, we utilized the definite integral to calculate the surface area of a curve revolved around an axis. When you hear "definite integral," think of capturing the entire sum of tiny areas between a specific interval:

• The limits of the integral, usually denoted as \(a\) and \(b\), determine the range.
• The function being integrated, in our case, is "correlated" with the geometric problem's formula.
• The process provides a concrete numerical value.

In the exercise, we specifically integrated from \(x = 0\) to \(x = 1\) to capture the surface area while the curve \(y = 6x\) was revolved around the \(x\)-axis. This utilization of a definite integral transforms calculus into a powerful tool.

Differentiation

Differentiation is the process of finding the derivative of a function, which measures how a function changes as its input changes. It's like finding the function's speed or slope at any given point. For our problem:

• The function \(y = 6x\) was differentiated with respect to \(x\).
• The derivative, \(\frac{dy}{dx} = 6\), represents a constant slope.
• Knowing the derivative is essential when using the surface area formula because it helps adjust the arc length element \(\sqrt{1 + \left(\frac{dy}{dx}\right)^2}\).

By finding this derivative, we understood how steep the curve was, which is vital in assessing the surface area when the curve is rotated.

Calculus Applications

The magic of calculus lies in its varied and practical applications, allowing us to solve real-world problems like calculating volumes, areas, and surface areas. Here's how it applies in our exercise:

• Revolving a curve around an axis to create a 3D surface is a typical problem in calculus applications.
• The formula \(S = \int_a^b 2\pi y \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx\) combines both integration and differentiation to deal with such revolutions.
• It also highlights concepts of geometry and algebra by using them together to solve problems beyond flat surfaces and lines.

This exercise exemplifies the beauty and utility of calculus, making it indispensable for engineering, physics, and more, effectively carving out the intricacies of shape and space.