Problem 40 - Calculus of a Single Variable

Question

Finding the Area of a Surface of Revolution In Exercises \(37-42,\) set up and evaluate the definite integral for the area of the surface generated by revolving the curve about the \(x\) -axis. $$ y=3 x, \quad 0 \leq x \leq 3 $$

Step-by-Step Solution

Answer

The area of the surface of revolution is \(270\pi\).

Step 1: Identifying the function and its derivative

The given function is \(y = 3x\), so we can say \(f(x) = 3x\). The derivative of the function is \(f'(x) = 3\).

Step 2: Calculating the squared derivative

The square of the derivative is \( (f'(x))^2 = (3)^2 = 9 \).

Step 3: Integrating over the given interval

Substitute \(f(x)\) and \((f'(x))^2\) into the formula for the surface area: \(A = 2\pi \int_0^3 3x \sqrt{1 + 9} dx = 2\pi \int_0^3 3x \sqrt{10} dx\). This simplifies to \(A = 60\pi \int_0^3 x dx\). Finally, integrating, we find \(A = 60\pi [\frac{1}{2}x^2]_0^3\).

Step 4: Calculating the surface area

Substitute the limits of the integral into the expression to obtain: \(A = 60\pi [\frac{1}{2}(3)^2 - \frac{1}{2}(0)^2] = 60\pi [4.5 - 0] = 270\pi\).

Key Concepts

Definite Integral

The definite integral is a powerful tool in calculus used to compute the total accumulation of quantities, which in this case is the area of a surface. When dealing with surfaces of revolution, we use the definite integral to sum up infinitesimally small segments of the surface. This gives us the total surface area.
For the given exercise, the function is revolved around the x-axis, so we need an integral that accounts for circular slices along this axis. This is done by setting up the integral with limits from the lower to the upper bound of the x-values, which in this exercise is from 0 to 3. The integral expression for the surface area is based on the formula: \[ A = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} \, dx \]
Substituting the function and limits into this formula helps derive the area of the surface generated by the revolving curve.

Function Derivative

A function derivative represents the rate at which the function's value changes along the x-axis. It gives the slope at any point on the function's graph. In our problem, differentiating the function \(y = 3x\) gives the derivative \(f'(x) = 3\).
This derivative is constant, showing that the function is a straight line with a uniform slope. The significance of the derivative in the context of surface revolutions is that it helps in calculating the integral that determines the surface area. We square this derivative to integrate it into the surface area formula. Hence, \((f'(x))^2 = 9\).
Understanding derivatives is crucial in calculus problem-solving, as they provide insights into how functions behave.

Revolution about the x-axis

When a curve revolves around the x-axis, it creates a three-dimensional shape. The challenge in calculus is to determine the surface area of this shape. The given exercise involves revolving the line \(y = 3x\) from \(x = 0\) to \(x = 3\) about the x-axis.
To find the surface area of this shape, we first express the line in terms of its function and derivative. This allows us to use the surface area formula: \[A = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} \, dx\]
Revolution about the x-axis can visualize solid shapes formed by rotation, emphasizing the beauty and practicality of calculus in the real world.

Calculus Problem Solving

Solving calculus problems requires a step-by-step approach to dissect complex mathematical operations into simpler parts. For the surface area of revolution problem, starting by identifying the function and its derivative is essential.
Each subsequent step builds on the previous one, where:

• First, compute the derivative.
• Second, calculate the squared derivative.
• Third, evaluate the integral using these values.
• Finally, plug the integration result into the formula to get the surface area.

The systematic breakdown of tasks not only aids in understanding the problem but also ensures precision in reaching a conclusive answer. In calculus, patience and accuracy in each of these steps are keys to success.