Problem 61 - Calculus of a Single Variable

Question

Volume and Surface Area Let \(R\) be the region bounded by \(y=1 / x,\) the \(x\) -axis, \(x=1,\) and \(x=b,\) where \(b>1 .\) Let \(D\) be the solid formed when \(R\) is revolved about the \(x\) -axis. (a) Find the volume \(V\) of \(D\) (b) Write the surface area \(S\) as an integral. (c) Show that \(V\) approaches a finite limit as \(b \rightarrow \infty\) . (d) Show that \(S \rightarrow \infty\) as \(b \rightarrow \infty\) .

Step-by-Step Solution

Answer

The volume \(V\) of the solid \(D\) is \(2\pi (b-1)\). The surface area \(S\) of \(D\) is given by the integral \(2\pi \int_{1}^{b} \frac{1}{x} \sqrt{1+\frac{1}{x^4}}dx\). As \(b \rightarrow \infty\), volume \(V\) approaches infinity and the surface area \(S\) also approaches infinity.

Step 1: Calculating the Volume \(V\)

To calculate the volume \(V\), use the method of cylindrical shells. The formula for this method is \(V = 2\pi \int_{a}^{b}yxdx\). Here \(a=1\), \(b\) is a variable, \(y=1/x\), and \(x\) is also a variable. So, the integral becomes \(V = 2\pi \int_{1}^{b}1dx = 2\pi [x]_{1}^{b}= 2\pi (b-1)\).

Step 2: Writing the Surface Area \(S\) as an Integral

The formula for the surface area of revolution is \(S = 2\pi \int_{a}^{b} y \sqrt{1+(dy/dx)^2} dx\). Differentiating \(y = 1/x\) gives \(dy/dx = -1/x^2\). Replacing the differentials and original function in the formula gives \(S = 2\pi \int_{1}^{b} 1/x \sqrt{1+1/x^4}dx\).

Step 3: Checking the Limit of \(V\) as \(b \rightarrow \infty\)

As per the calculated formula \(V=2\pi (b-1)\), the limit of \(V\) as \(b \rightarrow \infty\) is \(V=\lim_{b \to \infty} 2\pi (b - 1) = \infty - 2\pi = \infty\).

Step 4: Checking the Limit of \(S\) as \(b \rightarrow \infty\)

The limit of \(S\) as \(b \rightarrow \infty\) is not easy to compute directly, but it can be intuitively understood. The function inside the integral, \(1/x \sqrt{1+1/x^4}\), never reaches zero, which implies that as \(b \rightarrow \infty\), the integral (surface area \(S\)) will tend to infinity.

Key Concepts

Integral Calculus

Integral calculus is a branch of mathematics that focuses on finding the total size or value, often referred to as the 'integral', from the infinitesimal pieces that compose it. In practice, it allows us to calculate areas, volumes, and total accumulated values. The integral is denoted by the symbol \( \int \), with limits of integration specifying the interval we're interested in.

For example, to find the volume of a solid of revolution, we would set up a definite integral with lower limit 'a' and upper limit 'b', which represent the bounds on the x-axis. The function that we integrate, such as \( y = 1/x \) in our exercise, represents a curve that, when revolved around an axis, generates the solid whose volume we want to find. Integral calculus employs various techniques to calculate these integrals, and in the context of the solid formed by revolving a function, we use methods such as discs, washers, or cylindrical shells.

Method of Cylindrical Shells

The method of cylindrical shells is an integral calculus technique for finding volumes of solids of revolution. Instead of slicing the solid perpendicular to the axis of revolution—which results in discs or washers—we slice parallel to the axis, creating cylindrical elements. Imagine rolling up a thin rectangle into a cylinder to visualize how we use shells.

The formula for a volume using cylindrical shells is \(V = 2\pi \int_{a}^{b} f(x) g(x)dx\), where \(f(x)\) represents the radius of the shell (the distance from the axis of rotation to the shell's midpoint), and \(g(x)\) represents the height of the shell. The integral sums the infinite number of these cylindrical shells between 'a' and 'b' to give the total volume. In our exercise, \(y = 1/x\) is revolved about the x-axis, and by using this method, we found \(V = 2\pi (b-1)\) for a given value of 'b'.

Limits in Calculus

Limits are a fundamental concept in calculus that deals with the behavior of a function as its inputs get infinitely close to a specific value. Limits help economists, scientists, and mathematicians make sense of things that change, and they are essential for defining both the derivative and the integral.

In the context of our exercise, we're interested in what happens to the volume \(V\) and surface area \(S\) of the solid as 'b' tends to infinity. By evaluating \(\lim_{{b \to \infty}} V\) and \(\lim_{{b \to \infty}} S\), we gain insight into whether these quantities approach a finite value (converge) or grow without bound (diverge). This exercise shows us that while \(V\) diverges as \(b \to \infty\), indicating an infinitely large volume, the situation is different for \(S\), which we have to investigate differently.

Surface Area of Revolution

When a curve is revolved around an axis, the resulting shape is akin to a pottery wheel's creation—a three-dimensional object whose surface area we can find using calculus. The formula to determine the surface area of a solid of revolution is \(S = 2\pi \int_{a}^{b} y \sqrt{1 + (dy/dx)^2} dx\), where \(y\) is the radius function, and \(dy/dx\) is its derivative. This formula accounts for both the distance around the slice of the solid (circumference) and the 'smoothness' of the surface—it's essentially the product of the curve length for one full rotation about the axis.

In our exercise, we used this formula to express the surface area of the solid created by revolving \(y = 1/x\) around the x-axis. The challenge comes in evaluating the integral as \(b\) approaches infinity. Unlike volume, the surface area does not converge to a finite number as \(b \to \infty\), instead it increases without bound—indicating that even though a solid can have an infinite volume, its surface area can too expand infinitely, much like an ever-lengthening spiral or an unbounded cylindrical helix.