Question

Let \(f\) be differentiable and positive on the interval \([a, b] .\) Let \(S\) be the surface generated when the graph of \(f\) on \([a, b]\) is revolved about the \(x\) -axis. Use Theorem 17.14 to show that the area of \(S\) (as given in Section 6.6 ) is $$ \int_{a}^{b} 2 \pi f(x) \sqrt{1+f^{\prime}(x)^{2}} d x $$

Short answer

Based on the step by step solution explained above, the surface area, \(S\), generated by revolving a differentiable and positive function \(f\) around the x-axis on the interval \([a, b]\) can be represented by the integral: $$ S = \int_{a}^{b} 2 \pi f(x) \sqrt{1+f'(x)^{2}} dx. $$ This result was obtained using Theorem 17.14 and the formula for the arc length of a curve.

Step-by-step solution

Recall Theorem 17.14 Surface Area Formula

Theorem 17.14 provides a formula to compute the surface area generated by revolving a differentiable function around the x-axis. For a function \(f(x)\) being revolved around the x-axis, the surface area is given by: $$ S = \int_{a}^{b} 2 \pi f(x) ds, $$ where \(ds\) represents the differential arc length. Now our goal is to find the \(ds\) using the given function and its derivative.

Find the differential arc length \(ds\)

To find \(ds\), we will use the formula for the arc length of a curve, which is given by: $$ ds = \sqrt{1+\left(\frac{dy}{dx}\right)^2} dx, $$ where \(\frac{dy}{dx}\) represents \(f'(x)\). Now we have: $$ ds = \sqrt{1+f'(x)^{2}} dx. $$

Substitute the arc length \(ds\) in the surface area formula

With \(ds =\sqrt{1+f'(x)^{2}} dx\), we can substitute this into the surface area formula given by Theorem 17.14: $$ S = \int_{a}^{b} 2 \pi f(x) \sqrt{1+f'(x)^{2}} dx, $$ which is the integral that we were asked to prove. So, the area of surface \(S\) is given by the integral: $$ S = \int_{a}^{b} 2 \pi f(x) \sqrt{1+f'(x)^{2}} dx. $$

Key concepts

Differentiable Function

A differentiable function is a function that has a derivative at each point in its domain. This means the function is smooth enough so that there's a tangent line at any given point on its graph. The key characteristic of differentiable functions is that they do not have sharp corners or discontinuities within the considered interval.
Differentiable functions have continuous derivatives, which means they do not "jump" suddenly. This property is crucial when working with calculus, especially when revolving functions about the x-axis.
In our context, having a differentiable function ensures that we can find the derivative, denoted as \(f'(x)\). This derivative helps us understand how the function changes and is foundational for calculating both the arc length and the surface area when the function is revolved about the x-axis.

Arc Length

The arc length of a curve represents the distance along the curve between two points. To find the arc length of a curve described by a function \(f(x)\), calculus allows us to use an integral. This is crucial when we revolve a function around an axis, as it helps us take into account the curve's shape fully.

• The formula for the arc length \(ds\) is: \(ds = \sqrt{1+igg(\frac{dy}{dx}\bigg)^2}\, dx\)
• In our case, substituting \(\frac{dy}{dx} = f'(x)\) gives us: \(ds = \sqrt{1+f'(x)^2}\, dx\)

This expression accounts for the curve's tendency to move away from the axis, which is essential for precisely calculating not just the simple length of the curve but also complexities introduced during its revolution around an axis.

Theorem 17.14

Theorem 17.14 is a valuable tool in calculus that provides a method to determine the surface area of a three-dimensional shape created when a function is revolved around an axis. This theorem constructs a bridge between the two-dimensional function and the three-dimensional object derived from it.

• The theorem's formula for surface area \(S\) is: \(S = \int_{a}^{b} 2 \pi f(x)\, ds\)
• Since \(ds\) corresponds to the arc length differential, the theorem relies heavily on having a valid expression of \(ds\)

By integrating over \([a, b]\), Theorem 17.14 ensures that every conceivable infinitesimal slice of the revolving shape is accounted for in the surface area calculation. This theorem works seamlessly with arc length and differentiable functions to deliver robust evaluations of surfaces.

Integral Calculus

Integral calculus is the branch of calculus that deals with integrals, fundamentally regarding accumulation and area. It allows us to find the total sum when dealing with continuous rates of change, such as area under a curve, volumes, and in our exercise, the surface area of revolution.
Here, we apply integral calculus to evaluate the surface area of a solid revolution:

• We compute the integral: \(\int_{a}^{b} 2 \pi f(x) \sqrt{1+f'(x)^2}\, dx\)
• This integral denotes the accumulation of infinitesimal surface elements \(2 \pi f(x) ds\).

Integral calculus not only gives tools for handling these variables but seamlessly incorporates them into calculating significant geometric properties, such as surface areas, by breaking complex shapes into manageable, small pieces summed through integration.