Find the volume of the solid generated by rotating

Find the volume of the solid generated by rotating the region bounded by the curves $y = a\sinh\frac{x}{a}$, $x = \pm a$, and $y = 0$ about the $x$-axis.

We use the disk method together with the standard hyperbolic power-reduction identity.

Step 1: Set up the disk method

Since $\sinh$ is odd, $y=a\sinh\frac xa$ takes both signs on $[-a,a]$, but $y^2$ (and hence the disk area) only needs $|y|$:

\[A(x) = \pi y^2 = \pi a^2\sinh^2\frac xa\]

Step 2: Use the power-reduction identity

From $\cosh2u = 2\sinh^2u+1$: $\sinh^2u = \dfrac{\cosh2u-1}{2}$, so:

\[\sinh^2\frac xa = \frac{\cosh\frac{2x}{a}-1}{2}\]

Step 3: Set up the integral using even symmetry

\[V = \pi a^2\int_{-a}^{a}\sinh^2\frac xa\,dx = \pi a^2\int_0^{a}\left(\cosh\frac{2x}{a}-1\right)dx\]

Step 4: Integrate

\[\int\cosh\frac{2x}{a}\,dx = \frac{a}{2}\sinh\frac{2x}{a}+C\]

\[V = \pi a^2\left[\frac a2\sinh\frac{2x}{a}-x\right]_0^{a} = \pi a^2\left(\frac a2\sinh2-a\right) = \pi a^3\left(\frac{\sinh2}{2}-1\right)\]

\[\boxed{V = \pi a^3\left(\frac{\sinh2}{2}-1\right) \text{ cubic units}}\]

Step 5: Equivalent exponential form

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

\[V = \frac{\pi a^3(e^2-e^{-2}-4)}{4}\]

Step 6: Numerical value

$\sinh2\approx3.627$, so $\dfrac{\sinh2}{2}-1\approx0.814$, giving:

\[V \approx 0.814\pi a^3 \approx 2.557a^3 \text{ cubic units}\]

Summary

\[\boxed{V = \pi a^3\left(\frac{\sinh2}{2}-1\right) = \frac{\pi a^3(e^2-e^{-2}-4)}{4} \approx 2.557a^3 \text{ cubic units}}\]

Key insight: The identity linking $\sinh^2$ to $\cosh(2u)$ is the hyperbolic mirror of the familiar trigonometric power-reduction formula, and it converts what looks like an intractable $\sinh^2$ integral into an elementary one, producing a volume expressed naturally in terms of $\sinh2$.