Question

Find surface area of the catenoid \(y=\cosh (x)\) from \(x=-1\) to \(x=1\) that is created by rotating this curve around the \(x\) -axis.

Short answer

The surface area is given by \(2\pi (1 + \sinh(2))\).

Step-by-step solution

Understanding the Surface Area Formula

To find the surface area of a surface of revolution, we use the formula:\[A = 2\pi \int_{a}^{b} y \sqrt{1 + \left( \frac{dy}{dx} \right)^2 } \, dx\]where \( y = \cosh(x) \) and the interval is from \( x = -1 \) to \( x = 1 \).

Calculate the Derivative

Compute \( \frac{dy}{dx} \) for \( y = \cosh(x) \). The derivative is:\[\frac{dy}{dx} = \sinh(x)\]This follows from the identity: \( \frac{d}{dx}[ \cosh(x) ] = \sinh(x) \).

Substitute into the Formula

Substitute \( y = \cosh(x) \) and \( \frac{dy}{dx} = \sinh(x) \) into the surface area formula:\[A = 2\pi \int_{-1}^{1} \cosh(x) \sqrt{1 + (\sinh(x))^2} \, dx\]

Simplify the Integrand

Use the identity \( 1 + \sinh^2(x) = \cosh^2(x) \) to simplify the square root:\[\sqrt{1 + \sinh^2(x)} = \cosh(x)\]Thus, the integrand becomes: \( \cosh(x) \cdot \cosh(x) = \cosh^2(x) \).

Compute the Integral

Now, compute the integral:\[A = 2\pi \int_{-1}^{1} \cosh^2(x) \, dx\]Using the identity \( \cosh^2(x) = \frac{1 + \cosh(2x)}{2} \), we have:\[A = 2\pi \int_{-1}^{1} \frac{1 + \cosh(2x)}{2} \, dx\]This simplifies to:\[A = \pi \left[ x + \frac{1}{2} \sinh(2x) \right]_{-1}^{1}\]

Evaluate the Integral

Evaluate the integral by substituting the limits:\[A = \pi \left( \left[ 1 + \frac{1}{2} \sinh(2 \cdot 1) \right] - \left[ -1 + \frac{1}{2} \sinh(2 \cdot -1) \right] \right)\]After simplifying, we find:\[A = 2\pi \left( 1 + \sinh(2) \right)\]

Final Result

The surface area is calculated as:\[A = 2\pi (1 + \sinh(2))\]Using \( \sinh(2) = e^2/2 - 1/e^2 \), compute the numerical value for the final answer.

Key concepts

Catenoid

The catenoid is a fascinating mathematical shape that looks like a soap film stretched between two rings. It's formed when a catenary curve, described by the hyperbolic cosine function, is rotated around an axis. This creates a surface of revolution that has minimal surface properties. This means the catenoid has the least possible surface area for a given boundary, which makes it very efficient and elegant.

In the exercise, the curve given by the function \(y = \cosh(x)\) is rotated around the \(x\)-axis from \(x = -1\) to \(x = 1\). This operation forms a catenoid. The term "cosh(x)" refers to the hyperbolic cosine, which is important in generating this shape.

Understanding the properties of the catenary and its rotational counterpart, the catenoid, helps us appreciate the beauty and applications of geometry in areas like architecture and physics.

Hyperbolic Functions

Hyperbolic functions, analogous to trigonometric functions, are essential in describing hyperbolas in mathematics. Instead of circles, trigonometric functions model, hyperbolic functions model phenomena with hyperbolas. This makes them invaluable in solving certain calculus problems, especially those involving hyperbolas and exponential growth.

The hyperbolic cosine and hyperbolic sine functions, \(\cosh(x)\) and \(\sinh(x)\), are often used together. The identity \(1 + \sinh^2(x) = \cosh^2(x)\) is crucial in simplifying complex expressions and integrals. In the given problem, calculating the derivative of \(\cosh(x)\) yields \(\sinh(x)\), showcasing these functions' interconnected nature.

These functions appear frequently in physics, particularly in problems involving forces and potential energies. Their distinctive roles complement those of their trigonometric counterparts, expanding our mathematical toolbox for analyzing different scenarios.

Integral Calculus

Integral calculus is a branch of mathematics centered around finding areas under curves. It is used to calculate accumulated quantities, such as area, volume, and surface area. By performing an integral, one sums an infinite number of infinitesimally small factors.

In this problem, integral calculus was crucial for finding the surface area of the catenoid. The formula used involves rotation about an axis and requires calculating an integral that accounts for the shape's curvature. To solve it, we used the identity \(\cosh^2(x) = \frac{1 + \cosh(2x)}{2}\), transforming a more complex integral into something easier to manage.

Real-world engineering and scientific problems often rely on integral calculus to model and predict phenomena. Understanding how to compute integrals of various forms is a vital skill in many fields, from engineering to economics.