Problem 39 - Thomas Calculus

Question

An alternative derivation of the surface area formula Assume \(f\) is smooth on \([a, b]\) and partition \([a, b]\) in the usual way. In the \(k\) th subinterval \(\left[x_{k-1}, x_{k}\right]\) construct the tangent line to the curve at the midpoint \(m_{k}=\left(x_{k-1}+x_{k}\right) / 2,\) as in the figure here. a. Show that \(r_{1}=f\left(m_{k}\right)-f^{\prime}\left(m_{k}\right) \frac{\Delta x_{k}}{2}\) and \(r_{2}=f\left(m_{k}\right)+\) \(\quad f^{\prime}\left(m_{k}\right) \frac{\Delta x_{k}}{2}\) b. Show that the length \(L_{k}\) of the tangent line segment in the \(k\) th subinterval is \(L_{k}=\sqrt{\left(\Delta x_{k}\right)^{2}+\left(f^{\prime}\left(m_{k}\right) \Delta x_{k}\right)^{2}}\) Graph cannot copy c. Show that the lateral surface area of the frustum of the cone swept out by the tangent line segment as it revolves about the \(x\) -axis is 2\(\pi f\left(m_{k}\right) \sqrt{1+\left(f^{\prime}\left(m_{k}\right)\right)^{2}} \Delta x_{k}\) d. Show that the area of the surface generated by revolving \(y=f(x)\) about the \(x\) -axis over \([a, b]\) is $$ \lim _{n \rightarrow \infty} \sum_{k=1}^{n}\left(\begin{array}{l}{\text { lateral surface area }} \\ {\text { of } k \text { th frustum }}\end{array}\right)=\int_{a}^{b} 2 \pi f(x) \sqrt{1+\left(f^{\prime}(x)\right)^{2}} d x $$

Step-by-Step Solution

Answer

The surface area is \( \int_a^b 2\pi f(x) \sqrt{1 + (f'(x))^2} \mathrm{d}x \).

Step 1: Express Radii for Tangent Line Midpoints

For the first part, a tangent line equation at point \(m_k\) is constructed. The endpoints of this line are expressed as \(r_1\) and \(r_2\). These endpoints represent the values at the midpoint minus and plus the change over the subinterval. So, \(r_1\) is calculated as \(f(m_k) - f'(m_k) \frac{\Delta x_k}{2}\), which is the value at the midpoint minus half the tangent's slope times the width of the interval, and \(r_2 = f(m_k) + f'(m_k) \frac{\Delta x_k}{2}\) is similarly constructed by adding the same term.

Step 2: Calculate Length of Tangent Line Segment

To find the length \(L_k\) of the tangent line on the \(k\)-th subinterval, we apply the Pythagorean theorem to the horizontal and vertical components. The horizontal difference is \(\Delta x_k\), and the vertical difference, considering the slope of the tangent \(f'(m_k)\), is \(f'(m_k) \Delta x_k\). The length is thus \(L_k = \sqrt{(\Delta x_k)^2 + (f'(m_k) \Delta x_k)^2}\).

Step 3: Determine Lateral Surface Area of Frustum

The lateral surface area of a frustum formed by revolving the tangent line segment around the \(x\)-axis is determined by \(2\pi r L\), where \(r\) is the average radius \(f(m_k)\) and \(L\) is the length of the tangent. This results in \(2\pi f(m_k) \times \sqrt{1 + (f'(m_k))^2} \Delta x_k\).

Step 4: Derive Integral Expression for Surface Area

The total surface area over the interval \([a, b]\) is obtained by summing the lateral areas of each frustum and taking the limit as the number of partitions \(n\) approaches infinity. This is expressed as \( \int_a^b 2\pi f(x) \sqrt{1 + (f'(x))^2} \mathrm{d}x \), which is reached by applying the definition of the integral as the limit of a Riemann sum.

Key Concepts

Surface of Revolution

When we study the surface of revolution, we are focusing on a surface generated by rotating a curve around a given axis. In the context of the exercise, consider a function \( y = f(x) \) that describes a curve on a graph. To form a surface of revolution, we revolve this curve around the \( x \)-axis. The fascinating outcome is a smooth, often symmetrical, 3D shape.
For instance, imagine revolving a simple straight line around the \( x \)-axis; you would get a cone-like surface. The crucial concept here is that every point on the curve generates a circular slice at a particular position along the \( x \)-axis.
To calculate the surface area of such a 3D solid, we decompose it into tiny frustums—small sections between two radii—which resemble truncated cones. By considering these mini-cones, we approximate the surface area, refining our estimates by increasing the number of these pieces, thereby approaching the real surface area, as explained in the method of integration.

Integration Techniques

Integration plays a vital role when calculating the surface area of a surface generated by revolution. It allows us to sum up very small elements over a continuous interval, providing an accurate measure of the whole. In essence, we utilize integration to handle the infinite sum of differential elements.
For our exercise, we're given a function \( f(x) \) and its derivative \( f'(x) \). The process involves creating a Riemann sum, where each term is tied to the lateral surface area of one frustum. As the number of partitions \( n \) becomes larger, the sum approaches the integral.

• We use the formula for the lateral surface area of a frustum, which is \( 2\pi f(m_k) \sqrt{1 + (f'(m_k))^2} \Delta x_k \).
• When finding the full surface area, this is expressed integrally as \( \int_a^b 2\pi f(x) \sqrt{1+(f'(x))^2} \, dx \).

This integral expression leverages the limit of a Riemann sum, encapsulating the concept of differential calculus and integral calculus working in tandem to handle continuous functions.

Tangent Line Analysis

Tangent line analysis is key to understanding how to approximate areas for these revolution surfaces. By examining the curve closely, a tangent line offers an immediate snapshot of the curve's slope at a particular point. This line is pivotal for constructing each frustum.

• The tangent at any point on the curve \( f \) is described by its slope \( f'(m_k) \) at the midpoint \( m_k \).
• The line cuts through points \( r_1 \) and \( r_2 \), calculated by subtracting and adding half the slope multiplied by the width of the subinterval, \( \Delta x_k \).
• Subsequently, the length of this tangent line segment \( L_k \) is determined using the Pythagorean Theorem, producing \( L_k = \sqrt{(\Delta x_k)^2 + (f'(m_k) \Delta x_k)^2} \) due to its horizontal and vertical components.

By effectively representing the curve's behavior locally, tangent lines simplify the analysis necessary for extending these into three-dimensional segments of the surface.

Calculus Concepts

When addressing this exercise, several foundational calculus concepts come into play. Understanding these is crucial for grasping the entirety of the surface area computation.

• Differential Calculus: This involves understanding the derivative, which provides the slope of the tangent line at any given point on a curve. Here, we use \( f'(m_k) \) to denote this slope at midpoint \( m_k \).
• Integral Calculus: The integral helps in accumulating quantities over a continuum. In our case, it allows computation of the net surface area across sub-intervals from \( a \) to \( b \).
• Geometric Interpretation: Calculus provides a geometrical lens, allowing us to visualize how curves are stretched to create surfaces by examining the rotational aspect of geometry.

Each of these concepts builds upon one another, forming the mathematical backbone that makes the calculation of surface areas engendered by revolving curves feasible and reliable.