Question

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

b. If f is not one-to-one on the interval [a, b], then the area of the surface generated when the graph of f on [a, b] is revolved about the x-axis is not defined.

Accepted Answer

Recall the definition of a one-to-one function: a function $$ f  is $$one-to-one on an interval $$ [a, b]  if $$for every $$ x_1, x_2 \in [a, b] $$, $$ f(x_1) = f(x_2) $$ implies $$ x_1 = x_2 . $$This means the function never takes the same value twice on that interval. Understand the formula for the surface area generated by revolving the graph of $$ f $$ about the x-axis on $$ [a, b] . It is $$given by:

$$
S = \int_a^b 2\pi |f(x)| \sqrt{1 + (f'(x))^2} \, dx
$$

This formula requires $$ f  to be $$continuous and differentiable on $$ [a, b] $$, but it does not require $$ f  to be $$one-to-one. Analyze the statement: "If $$ f  is $$not one-to-one on $$ [a, b] $$, then the surface area is not defined." Since the surface area formula depends on $$ f(x) $$ and $$ f'(x) $$, as long as these are well-defined and integrable, the surface area can be computed regardless of whether $$ f  is $$one-to-one. Consider a counterexample: a function like $$ f(x) = \sin x  on$$ $$ [0, 2\pi]  is $$not one-to-one, but it is continuous and differentiable. The surface area generated by revolving $$ \sin x $$ about the x-axis on this interval is well-defined and can be calculated using the formula. Conclude that the statement is false because the one-to-one property is not necessary for the surface area to be defined. The key requirements are continuity and differentiability of $$ f  on$$ $$ [a, b] .$$

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

b. If f is not one-to-one on the interval [a, b], then the area of the surface generated when the graph of f on [a, b] is revolved about the x-axis is not defined.

Verified video answer for a similar problem:

Key Concepts

One-to-One Functions

Surface Area of Revolution

Conditions for Surface Area Integral to be Defined

Determine whether the following statements are true and give an explanation or counterexample. b. If f is not one-to-one on the interval [a, b], then the area of the surface generated when the graph of f on [a, b] is revolved about the x-axis is not defined.

Verified video answer for a similar problem:

Key Concepts

One-to-One Functions

Surface Area of Revolution

Conditions for Surface Area Integral to be Defined

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

b. If f is not one-to-one on the interval [a, b], then the area of the surface generated when the graph of f on [a, b] is revolved about the x-axis is not defined.