Calculate the surface area generated by rotating the arc

Calculate the surface area generated by rotating the arc of the catenary $y = a\cosh\dfrac{x}{a}$ about the $x$-axis between $x=0$ and $x=a$ (where $a>0$).

Step 1: Differentiate

\[y'=\sinh\frac xa \implies 1+(y')^2=1+\sinh^2\frac xa=\cosh^2\frac xa \implies \sqrt{1+(y')^2}=\cosh\frac xa\]

Step 2: Set up the integral

\[S=2\pi\int_0^a a\cosh\frac xa\cdot\cosh\frac xa\,dx = 2\pi a\int_0^a\cosh^2\frac xa\,dx\]

Step 3: Use the double-angle identity

\[S=2\pi a\int_0^a\frac{\cosh\frac{2x}a+1}2\,dx = \pi a\left[\frac a2\sinh\frac{2x}a+x\right]_0^a\]

Step 4: Evaluate

\[S=\pi a\left(\frac a2\sinh2+a\right)=\frac{\pi a^2}2(\sinh2+2)\]

\[\boxed{S=\frac{\pi a^2}2(\sinh2+2)}\]

Step 5: In exponential form

Since $\sinh2=\frac{e^2-e^{-2}}2$:

\[S=\frac{\pi a^2(e^2-e^{-2}+4)}4\]

Step 6: Numerical value

With $\sinh2\approx3.627$: $S\approx2.814\pi a^2\approx8.84a^2$.