Question

Surface area of a catenoid When the catenary $y=a\cosh \frac{x}{a}$ is revolved about the $x$ -axis, it sweeps out a surface of revolution called a catenoid. Find the area of the surface generated when $y=\cosh x$ on $[-\ln 2,\ln 2]$ is rotated about the $x$ -axis.

Short answer

The surface area of the catenoid is given by the formula: $$A=\pi (\frac{1}{2}\sinh 2\ln 2+\ln 2)$$

Step-by-step solution

Identify the given function and interval

In this case, the function $f(x)=\cosh x$ and the interval for $x$ is $[-\ln 2,\ln 2]$.

Calculate the derivative of the function

We need to find $f^{\prime }(x)$ which is the derivative of $f(x)=\cosh x$. The derivative of the hyperbolic cosine function is the hyperbolic sine function: $$f^{\prime }(x)=\frac{d}{dx}\cosh x=\sinh x$$

Calculate ${[f^{\prime }(x)]}^2$

Now, we need to find the square of the derivative: ${[f^{\prime }(x)]}^2$ $${[f^{\prime }(x)]}^2=(\sinh x{)}^2$$

Calculate the integrand with the formula

Use the formula for the surface area of a solid of revolution: $$A=2\pi {\int }_a^{b}f(x)\sqrt{1+{[f^{\prime }(x)]}^2}dx$$ Replace $f(x)=\cosh x$ and ${[f^{\prime }(x)]}^2=(\sinh x{)}^2$: $$A=2\pi {\int }_{-\ln 2}^{\ln 2}\cosh x\sqrt{1+(\sinh x{)}^2}dx$$

Simplify the integrand

Use the identity ${\cosh }^2(x)-{\sinh }^2(x)=1$ to simplify the expression inside the square root: $$A=2\pi {\int }_{-\ln 2}^{\ln 2}\cosh x\sqrt{1+(\cosh x{)}^2-1}dx$$ $$A=2\pi {\int }_{-\ln 2}^{\ln 2}\cosh x\sqrt{(\cosh x{)}^2}dx$$ Since square root of a square is the absolute value of the original number, and $\cosh (x)\ge 1$ for all $x$, we get: $$A=2\pi {\int }_{-\ln 2}^{\ln 2}\cosh x(\cosh x)dx$$ $$A=2\pi {\int }_{-\ln 2}^{\ln 2}(\cosh x{)}^{2}dx$$

Evaluate the integral

Now, we need to find the integral of $(\cosh x{)}^2$ with respect to $x$: $$A=2\pi {\int }_{-\ln 2}^{\ln 2}(\cosh x{)}^{2}dx$$ To do this, we can use the identity $(\cosh x{)}^2=\frac{1}{2}(\cosh 2x+1)$: $$A=2\pi {\int }_{-\ln 2}^{\ln 2}\frac{1}{2}(\cosh 2x+1)dx$$ Now, we integrate the expression: $$A=2\pi [\frac{1}{2}{\int }_{-\ln 2}^{\ln 2}\cosh 2xdx+\frac{1}{2}{\int }_{-\ln 2}^{\ln 2}dx]$$ When we carry out the integration, $$A=\pi [\frac{1}{2}(\frac{1}{2}\sinh 2x){|}_{-\ln 2}^{\ln 2}+\frac{1}{2}x{|}_{-\ln 2}^{\ln 2}]$$ By evaluating the expression inside the brackets, we get: $$A=\pi [\frac{1}{2}(\frac{1}{2}\sinh 2\ln 2-\frac{1}{2}\sinh -2\ln 2)+\frac{1}{2}(\ln 2-(-\ln 2))]$$ Then, utilize the identity $\sinh (-x)=-\sinh (x)$ to simplify further: $$A=\pi [\frac{1}{2}(\frac{1}{2}\sinh 2\ln 2+\frac{1}{2}\sinh 2\ln 2)+\ln 2]$$ $$A=\pi (\frac{1}{2}\sinh 2\ln 2+\ln 2)$$

Write the final answer

The surface area of the catenoid generated when $y=\cosh x$ on $[-\ln 2,\ln 2]$ is rotated about the $x$ -axis is: $$A=\pi (\frac{1}{2}\sinh 2\ln 2+\ln 2)$$

Key concepts

Catenary

The concept of a catenary is deeply rooted in the natural phenomenon of hanging chains. When a chain hangs freely under its own weight, it adopts a specific curve shape known as a catenary. This shape can be described mathematically by the equation $y=a\cosh \left(\frac{x}{a}\right)$, where $a$ is a constant. The function $\cosh$, which stands for hyperbolic cosine, measures the curve as it pulls downward due to gravity.
When this curve is revolved around an axis, it creates a three-dimensional surface known as a catenoid. These catenoids have interesting properties and are seen in architectural structures like bridges and arches. They also minimize surface area, helping in the study of minimal surfaces.
Catenaries are more than mathematical curiosities; they find real-life applications. Engineers and architects leverage these properties for constructions that need to support weight or optimize materials used.

Surface of Revolution

A surface of revolution emerges when a curve is revolved around a line, typically an axis. This motion generates a three-dimensional surface. In our exercise, the curve described by the function $y=\cosh x$ is revolved around the $x$-axis. This creates a catenoid, which is a unique surface of revolution.
Visualizing this can be likened to spinning a ribbon around a stick, where the 'stick' represents the axis, and the 'ribbon' is the rotating curve. Calculating the surface area of this resultant shape involves integration techniques. These allow for the precise measurement of complex shapes by summing infinitely small pieces of the surface.
Understanding surfaces of revolution is crucial for fields such as physics and engineering, where these concepts aid in designing products ranging from automobile parts to aerodynamic structures.

Hyperbolic Functions

Hyperbolic functions, akin to trigonometric functions, are deeply embedded in calculus and analysis. They include the hyperbolic sine $\sinh$ and hyperbolic cosine $\cosh$, which characterize the shape of catenaries.
In our exercise, $y=\cosh x$ describes the catenary's curve. Its derivative, $\sinh x$, is used to find changes along this curve. This showcases the utility of hyperbolic functions in describing real-world phenomena like hanging cables or soap films between two circular wires.
Several identities relate hyperbolic functions to trigonometric functions, such as ${\cosh }^2(x)-{\sinh }^2(x)=1$, similar to the Pythagorean trigonometric identity. These identities simplify calculations and reveal underlying mathematical symmetry. Beyond pure mathematics, hyperbolic functions model growth patterns in economics and natural sciences.

Integration Techniques

Integration is fundamental to calculating the surface area of a catenoid. To evaluate such an area, a specific integration technique is required. This includes setting up an integral using the formula for the surface area of a solid of revolution:
$$A=2\pi {\int }_a^{b}f(x)\sqrt{1+{[f^{\prime }(x)]}^2}dx$$
This involves integrating over the interval specific to the problem, which in this exercise is $[-\ln 2,\ln 2]$. The process begins by finding the derivative $f^{\prime }(x)$, squaring it, and placing it into the formula.
One of the simplifying tricks involves the identity ${\cosh }^2(x)-{\sinh }^2(x)=1$ to reduce the expression under the square root, easing the integral computation.
Mastery of integration techniques is critical for students and professionals tackling problems in calculus and applied mathematics. These techniques simplify complex problems and make them manageable, driving solutions in various scientific domains.