Ch. 6 - Applications of Integration

Briggs - Calculus: Early Transcendentals 3rd Edition

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Briggs 3rd Edition

Ch. 6 - Applications of Integration

Problem 6.6.23c

Chapter 6, Problem 6.6.23c

Determine whether the following statements are true and give an explanation or counterexample.

c. Let f(x)=12x^2. The area of the surface generated when the graph of f on [−4, 4] is revolved about the x-axis is twice the area of the surface generated when the graph of f on [0, 4] is revolved about the x-axis.

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Recall the formula for the surface area generated by revolving a curve $y = f(x)$ about the x-axis over the interval $[a, b]$: \[S = \int_a^b 2\pi f(x) \sqrt{1 + (f'(x))^2} \, dx\]

Identify the function and its derivative: Given $f(x) = 12x^2$, then \[f'(x) = 24x\]

Set up the surface area integrals for both intervals: For $[-4, 4]$: \[S_1 = \int_{-4}^4 2\pi (12x^2) \sqrt{1 + (24x)^2} \, dx\] For $[0, 4]$: \[S_2 = \int_0^4 2\pi (12x^2) \sqrt{1 + (24x)^2} \, dx\]

Analyze the integrand's symmetry: Since $f(x) = 12x^2$ is an even function and $f'(x) = 24x$ is an odd function, the term inside the square root, $1 + (f'(x))^2 = 1 + (24x)^2$, is even. The product $f(x) \sqrt{1 + (f'(x))^2}$ is therefore even because it is the product of an even function and an even function.

Use the property of even functions in integrals: For an even function $g(x)$, \[\int_{-a}^a g(x) \, dx = 2 \int_0^a g(x) \, dx\] Therefore, \[S_1 = 2 S_2\] This shows that the surface area generated on $[-4,4]$ is twice the surface area generated on $[0,4]$, confirming the statement.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Surface Area of Revolution

The surface area generated by revolving a curve y = f(x) about the x-axis over an interval [a, b] is found using the formula S = ∫ from a to b 2π f(x) √(1 + (f'(x))^2) dx. This integral accounts for the circumference of circular slices and the curve's slope, providing the total surface area.

Symmetry of Functions and Intervals

When a function is even (f(-x) = f(x)) and the interval is symmetric about zero, properties of symmetry can simplify calculations. For surface areas, symmetry does not always imply the surface area over [-a, a] is twice that over [0, a], because the integrand may not be an even function.

Derivative and Its Role in Surface Area

The derivative f'(x) measures the slope of the function and affects the surface area integral through the term √(1 + (f'(x))^2). This term adjusts the length element to account for the curve's steepness, influencing the total surface area generated by revolution.

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