The surface area $S$ of a curve $y = f(x)$ revolved around the x-axis from $x = a$ to $x = b$ is given by the formula:

$S = \int_a^b 2\pi y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \,dx$

For the given catenary:

$y = c \cosh(x/a)$

Step 1: Compute $dy/dx$

The derivative of $y$ is:

$\frac{dy}{dx} = c \sinh(x/a) \cdot \frac{1}{a} = \frac{c}{a} \sinh(x/a)$

Thus,

$\left(\frac{dy}{dx}\right)^2 = \left(\frac{c}{a} \sinh(x/a)\right)^2 = \frac{c^2}{a^2} \sinh^2(x/a)$

Step 2: Compute $\sqrt{1 + (dy/dx)^2}$

$\sqrt{1 + \frac{c^2}{a^2} \sinh^2(x/a)}$

Using the identity:

$1 + \sinh^2(x/a) = \cosh^2(x/a)$

we get:

$\sqrt{1 + \frac{c^2}{a^2} \sinh^2(x/a)} = \sqrt{\frac{a^2 + c^2 \sinh^2(x/a)}{a^2}} = \frac{c}{a} \cosh(x/a)$

Step 3: Compute the Surface Area Integral

$S = \int_0^c 2\pi c \cosh(x/a) \cdot \frac{c}{a} \cosh(x/a) \,dx$

$S = \frac{2\pi c^2}{a} \int_0^c \cosh^2(x/a) \,dx$

Using the identity:

$\cosh^2 u = \frac{1 + \cosh 2u}{2}$

we rewrite the integral:

$\int \cosh^2(x/a) \,dx = \frac{1}{2} \int (1 + \cosh(2x/a)) \,dx$

$= \frac{1}{2} \left( x + \frac{a}{2} \sinh(2x/a) \right)$

Evaluating from $0$ to $c$:

$\int_0^c \cosh^2(x/a) \,dx = \frac{1}{2} \left( c + \frac{a}{2} \sinh(2c/a) - 0 - \frac{a}{2} \sinh 0 \right)$

$= \frac{1}{2} \left( c + \frac{a}{2} \sinh(2c/a) \right)$

Step 4: Compute the Final Expression for $S$

$S = \frac{2\pi c^2}{a} \times \frac{1}{2} \left( c + \frac{a}{2} \sinh(2c/a) \right)$

$= \frac{\pi c^2}{a} \left( c + \frac{a}{2} \sinh(2c/a) \right)$

$= \pi c^2 \left(\frac{c}{a} + \frac{1}{2} \sinh(2c/a) \right)$

Thus, the surface area is:

$S = \pi c^2 \left(\frac{c}{a} + \frac{1}{2} \sinh(2c/a) \right)$

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Would you like me to simplify it further or provide numerical evaluations for specific values of $a$ and $c$?

Related Questions:

1. How do you derive the arc length of a catenary curve?
2. What is the volume of the solid formed by revolving the same catenary around the x-axis?
3. How does the surface area change as $c \to \infty$?
4. What happens to the surface area if the catenary is rotated about the y-axis instead?
5. Can we generalize this result for a catenary of the form $y = A \cosh(x/B)$?

Math Tip:

Hyperbolic functions, like $\cosh(x)$ and $\sinh(x)$, often simplify problems involving catenary curves, as they have identities similar to trigonometric functions but with exponential properties.