Question

(Continuation of Exploration 2\()\) Let \(x=g(y)>0\) have a continuous first derivative on \([c, d] .\) Show the area of the surface generated by revolving the curve \(x=g(y)\) about the \(y\) -axis is $$S=\int_{c}^{d} 2 \pi g(y) \sqrt{1+\left(g^{\prime}(y)\right)^{2}} d y$$

Short answer

The formula \(S = \int_{c}^{d} 2 \pi g(y) \sqrt{1 + (g'(y))^2} dy\) for the surface area of the surface of revolution generated by the curve \(x = g(y)\) about the y-axis, where the function \(x = g(y) > 0\) has a continuous first derivative on the interval \([c, d]\), is derived using the concept of the surface area of a small strip of the surface, approximated as a circular strip, and integrating this expression over the given interval. The proof involves understanding of calculus, especially the concept of derivative and indefinite integration.

Step-by-step solution

Understand the problem

The problem is asking to prove the given formula which calculates the surface area generated by revolving the curve \(x=g(y)\) about the y-axis. Here, \(x = g(y) > 0\) is a continuous function with first derivative on the interval \([c, d]\). The formula to prove is: \[S = \int_{c}^{d} 2 \pi g(y) \sqrt{1 + (g'(y))^2} dy\]

Define the surface of revolution

When a curve is revolved around the y-axis, it sweeps out a surface of revolution. The surface area \(dS\) of a small strip of this surface, at \(y\), with width \(dy\), is approximately the circumference of the strip times its width. The strip is like a circle of radius \(g(y)\) and width \(\sqrt{1 + (g'(y))^2} dy\) (using the Pythagorean theorem). So, \[dS \approx 2 \pi g(y) \sqrt{1 + (g'(y))^2} dy\]

Prove the formula by integrating

To find the total surface area \(S\), integrate \(dS\) from \(c\) to \(d\):\[S = \int_{c}^{d} dS\]Substitute \(dS\) from Step 2:\[S = \int_{c}^{d} 2 \pi g(y) \sqrt{1 + (g'(y))^2} dy\]This proves the given formula.

Key concepts

Calculus

Calculus is a branch of mathematics focused on limits, functions, derivatives, integrals, and infinite series. It's the mathematical study of continuous change, and it has two main branches: differential calculus and integral calculus. Differential calculus concerns the concept of a derivative, which allows us to understand how a function changes at any given point. Integral calculus, on the other hand, involves the concept of an integral, which allows us to accumulate quantities over intervals. Together, these concepts allow us to solve complex problems in engineering, physics, economics, and beyond, including finding the surface area of objects formed by revolving a curve around an axis—the very issue addressed in our textbook exercise.

Calculus provides the tools needed for solving real-world problems that require the calculation of rates at which one quantity changes relative to another or total values from continuously varying quantities, making it a fundamental discipline for many fields of science and engineering.

Definite Integrals

Definite integrals are a cornerstone of integral calculus. They are used to calculate the accumulated total of a quantity that can vary continuously, like area, volume, and in our case, surface area. A definite integral has upper and lower bounds, which in the exercise are represented by the limits of integration, \(c\) and \(d\).

The definite integral of a function over a specified interval gives us a precise measurement of the total accumulation of the function's value as we move along the interval from the start point to the end point. In terms of surface area, the definite integral adds up infinitely small pieces of the surface to reach a finite total surface area. Mathematically, the formula provided in the exercise utilizes a definite integral to sum up the surface area of each infinitesimally thin strip as the curve revolves around the y-axis.

Surface Area Calculations

Surface area calculations are often encountered in geometry and calculus. In calculus, when dealing with a three-dimensional object, especially one that is generated by rotating a curve around an axis, calculating the surface area can become complex. The formula in the exercise encapsulates this concept perfectly.

In finding the surface area of the shape formed by rotating a curve around an axis, we slice the shape into infinitesimally thin rings or strips. We then calculate the area of each strip and sum these areas from one end of the object to the other, a process that is achieved through integration. This is why, in the exercise, we see the integral of the circumference of each infinitesimal strip—given by \(2 \pi g(y)\)—multiplied by its width, which includes the Pythagorean increment to account for the slope of the function we're rotating.

Derivative Applications

Derivatives represent the rate at which a function is changing at any given point and are a major part of differential calculus. They're crucial in calculating things like velocity, acceleration, and slopes of curves.

In the context of the surface area of revolution problem, derivatives help account for the change in the radius of the surface area strips as we move along the curve. As the curve is revolved around the y-axis, its slope \(g'(y)\) affects how much each strip deviates from being a perfect cylinder. Without accounting for this slope, we would underestimate the surface area. The application of the derivative in this manner allows us to find a more accurate representation of the surface area of the rotated curve, showcasing one of the many practical applications of derivatives in calculus.