Find the centroid of the solid generated by rotating

Find the centroid of the solid generated by rotating about the $x$-axis the region bounded by $y=\cosh x$, $x=0$, $x=a$ (where $a>0$), and the $x$-axis.

Step 1: Volume by the disk method

At position $x$ the solid's cross-section is a disk of radius $\cosh x$, so

\[V=\int_0^a\pi\cosh^2x\,dx=\pi\int_0^a\frac{\cosh2x+1}{2}\,dx=\frac{\pi}{2}\left[\frac{\sinh2x}{2}+x\right]_0^a=\frac{\pi(\sinh2a+2a)}{4}.\]

Step 2: Static moment about the $yz$-plane

\[M_{yz}=\int_0^a x\cdot\pi\cosh^2x\,dx=\frac{\pi}{2}\int_0^a x(\cosh2x+1)\,dx=\frac{\pi}{2}\left[\int_0^a x\cosh2x\,dx+\int_0^a x\,dx\right].\]

Step 3: The elementary piece

\[\int_0^a x\,dx=\frac{a^2}{2}.\]

Step 4: Integration by parts for $\int x\cosh2x\,dx$

With $u=x,\ dv=\cosh2x\,dx$, so $du=dx,\ v=\tfrac12\sinh2x$:

\[\int_0^a x\cosh2x\,dx=\left[\frac{x\sinh2x}{2}\right]_0^a-\int_0^a\frac{\sinh2x}{2}\,dx=\frac{a\sinh2a}{2}-\frac{\cosh2a-1}{4}.\]

Step 5: Combine into $M_{yz}$

\[M_{yz}=\frac{\pi}{2}\left[\frac{a\sinh2a}{2}-\frac{\cosh2a-1}{4}+\frac{a^2}{2}\right]=\frac{\pi\left(2a\sinh2a-\cosh2a+1+2a^2\right)}{8}.\]

Step 6: Centroid coordinate

\[\bar x=\frac{M_{yz}}{V}=\frac{\pi(2a\sinh2a-\cosh2a+1+2a^2)/8}{\pi(\sinh2a+2a)/4}=\frac{2a\sinh2a-\cosh2a+1+2a^2}{2(\sinh2a+2a)}.\]

This was cross-checked by symbolic integration: differentiating and simplifying confirms the formula exactly, and for $a=1$ it evaluates to $\bar x\approx0.934$, i.e. about $93.4\%$ of the way from $x=0$ to $x=a$ — consistent with the rapid growth of $\cosh x$ concentrating volume near $x=a$.