Calculate the surface area generated by rotating

Calculate the surface area generated by rotating the hyperbola $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ about the $y$-axis for $-h \lt y \lt h$.

Step 1: Set up

$x=\dfrac ab\sqrt{b^2+y^2}$, so $x'=\dfrac{ay}{b\sqrt{b^2+y^2}}$, and

\[1+(x')^2=\frac{b^4+(a^2+b^2)y^2}{b^2(b^2+y^2)} \implies \sqrt{1+(x')^2}=\frac{\sqrt{b^4+(a^2+b^2)y^2}}{b\sqrt{b^2+y^2}}\]

Step 2: Set up the integral

By symmetry about $y=0$, with $c=\sqrt{a^2+b^2}$:

\[S=\frac{4\pi a}{b^2}\int_0^h\sqrt{b^4+c^2y^2}\,dy\]

Step 3: Substitute $u=\dfrac{cy}{b^2}$

\[S=\frac{4\pi ab^2}c\int_0^{ch/b^2}\sqrt{1+u^2}\,du\]

Step 4: Evaluate with the standard antiderivative

With $\alpha=\dfrac{ch}{b^2}=\dfrac{h\sqrt{a^2+b^2}}{b^2}$:

\[S=\frac{2\pi ab^2}c\left[\alpha\sqrt{1+\alpha^2}+\ln\left(\alpha+\sqrt{1+\alpha^2}\right)\right]\]

Step 5: Result

\[\boxed{S=\frac{2\pi a}{\sqrt{a^2+b^2}}\left[h\sqrt{b^4+(a^2+b^2)h^2}+b^2\ln\left(\frac{h\sqrt{a^2+b^2}+\sqrt{b^4+(a^2+b^2)h^2}}{b^2}\right)\right]}\]

This is a hyperboloid of revolution (one sheet), with waist radius $a$ at $y=0$, widening as $|y|$ increases.