(a) Revolve the curve \(y=x^{-1}\), from \(x=1\) to \(x=\infty\), about the \(x\) axis to create a surface and a volume. Write integrals for the surface area and the volume. Find the volume, and show that the surface area is infinite. Hint: 'The surface area integral is not easy to evaluate, but you can casily show that it is greater than \(\int_{1}^{\infty} x^{-1} d x\) which you can evaluate. (b) 'The following question is a challenge to your ability to fit together your mathematical calculations and physical facts: In (a) you found a finite volume and an infinite area. Suppose you fill the finite volume with a finite amount of paint and then pour off the excess leaving what sticks to the surface. Apparently you have painted an infinite area with a finite amount of paint! What is wrong? (Compare Problem \(15.31 \mathrm{c}\) of Chapter 1.)

When we revolving a curve around an axis to form a 3D object, the volume of that object can be computed using a volume integral. For our problem, we revolve the curve defined by the equation \(y = x^{-1}\) (or \(y = \frac{1}{x}\)) around the x-axis from \(x = 1\) to \(x = \infty\). The volume integral is set up using the disc method. This method slices the solid into thin discs and sums the volume of each disc.
The formula to find the volume is:
\[ V = \int_{a}^{b} \pi \, [f(x)]^2 \, dx \]
For our specific curve, this translates to:
\[ V = \pi \, \int_{1}^{\infty} \left( \frac{1}{x} \right)^2 \, dx \]
Simplifying this integral gives:
\[ V = \pi \, \int_{1}^{\infty} \frac{1}{x^2} \, dx \]
This integral evaluates to \(\pi\), giving us a finite volume of \(\pi\) cubic units.

The surface area of a solid of revolution can be computed with a surface area integral. When we revolve the curve \(y = x^{-1}\) about the x-axis, we need a slightly more complex formula than for volume. The formula for calculating the surface area is:
\[ A = 2\pi \, \int_{a}^{b} f(x) \sqrt{1 + [f'(x)]^2} \, dx \]
Here, \(f(x) = \frac{1}{x}\), and the derivative \(f'(x) = -x^{-2}\). Plugging these into our formula, we get:
\[ A = 2\pi \, \int_{1}^{\infty} \frac{1}{x} \, \sqrt{1 + \left(-\frac{1}{x^2}\right)^2} \, dx \]
As shown, simplifying under the square root gives us:
\[ A = 2\pi \, \int_{1}^{\infty} \frac{1}{x} \, \sqrt{1 + \frac{1}{x^4}} \, dx \]
The function \(\frac{1}{x} \sqrt{1 + \frac{1}{x^4}}\) is always greater than \(\frac{1}{x}\), leading us to realize that \(\int_{1}^{\infty} \frac{1}{x} \, dx\) diverges, resulting in an infinite surface area.

Understanding the idea of infinite surface areas can be tricky. In our context, it means that as we revolve the curve \(y = x^{-1}\), the resulting surface extends infinitely even though it forms a finite volume.
When we look at the integral:
\[ \int_{1}^{\infty} \frac{1}{x} \, dx = [\ln|x|]_{1}^{\infty} \]
This diverges, meaning our surface area does not converge to a finite number.
The counterintuitive part is that a finite volume can have an infinite surface area. This happens because as \(x\) approaches infinity, the surface becomes exceedingly long and thin, contributing to an infinite area while maintaining a finite enclosed volume.

The disc method is essential for finding the volume of solids of revolution. This technique involves slicing the solid perpendicular to the axis of rotation into thin discs.
Each disc’s volume is approximated by the volume of a cylinder: \(V = \pi \, r^2 \, h\). Here:

• \(r\) is the radius of the disc. It is given by the function \(f(x)\).
• \(h\) is the thickness of each slice. It corresponds to the differential \(dx\), or a very small width in the x-direction.

The total volume is the sum of the volumes of all the discs, which is computed by integrating from the lower to the upper bounds of \(x\). In our case, this gives the integral:
\[ V = \pi \, \int_{1}^{\infty} \frac{1}{x^2} \, dx \]
Through integration, we get a finite value representing the solid’s volume.

Calculus is a fundamental branch of mathematics that helps us understand and quantify changes. It's essential for tackling problems involving rates, areas, volumes, and much more.
In this exercise, calculus allows us to:

• Set up integrals for volume and surface area
• Evaluate those integrals
• Understand concepts like divergence and convergence

Key techniques include:

• Integration: Used for summing infinitesimally small quantities to find areas under curves and volumes.
• Differentiation: Helps in finding rates of change, slopes, and understanding how functions behave.

Understanding and using these tools allows us to solve more complex real-world problems and understand deeper mathematical concepts.