Answer: (a) \(2\sinh^2 A=\cosh 2A-1\).

(b) \(S=\frac{1}{32}\pi\left(\frac{820}{81}-\ln 3\right)\).

(a) Using the definitions

\(\sinh A=\frac{1}{2}(e^A-e^{-A})\) and \(\cosh 2A=\frac{1}{2}(e^{2A}+e^{-2A})\).

Then

\(2\sinh^2A=2\left(\frac{1}{2}(e^A-e^{-A})\right)^2\)

\(\qquad=\frac{1}{2}(e^A-e^{-A})^2\)

\(\qquad=\frac{1}{2}(e^{2A}-2+e^{-2A})\)

\(\qquad=\frac{1}{2}(e^{2A}+e^{-2A})-1\)

\(\qquad=\cosh 2A-1\).

So \(2\sinh^2A=\cosh 2A-1\).

(b) For a curve rotated about the \(x\)-axis, the surface area is

\(S=\int_0^{2/3}2\pi y\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx\).

Here \(y=x^2\), so \(\frac{dy}{dx}=2x\). Hence

\(S=2\pi\int_0^{2/3}x^2\sqrt{1+4x^2}\,dx\).

Use the substitution \(x=\frac{1}{2}\sinh u\). Then

\(dx=\frac{1}{2}\cosh u\,du,\qquad x^2=\frac{1}{4}\sinh^2u,\qquad \sqrt{1+4x^2}=\sqrt{1+\sinh^2u}=\cosh u.\)

When \(x=0\), \(u=0\). When \(x=\frac{2}{3}\), \(\sinh u=\frac{4}{3}\), so \(u=\sinh^{-1}\!\left(\frac{4}{3}\right)\).

Therefore

\(S=2\pi\int_0^{\sinh^{-1}(4/3)}\left(\frac{1}{4}\sinh^2u\right)(\cosh u)\left(\frac{1}{2}\cosh u\right)du\)

\(\qquad=\frac{1}{4}\pi\int_0^{\sinh^{-1}(4/3)}\sinh^2u\cosh^2u\,du\).

Using \(\sinh 2u=2\sinh u\cosh u\),

\(\sinh^2u\cosh^2u=\frac{1}{4}\sinh^22u\),

so

\(S=\frac{1}{16}\pi\int_0^{\sinh^{-1}(4/3)}\sinh^22u\,du\).

From part (a), \(2\sinh^2A=\cosh 2A-1\), hence

\(\sinh^22u=\frac{1}{2}(\cosh 4u-1).\)

So

\(S=\frac{1}{32}\pi\int_0^{\sinh^{-1}(4/3)}(\cosh 4u-1)\,du\)

\(\qquad=\frac{1}{32}\pi\left[\frac{1}{4}\sinh 4u-u\right]_0^{\sinh^{-1}(4/3)}.\)

Now

\(\sinh^{-1}\!\left(\frac{4}{3}\right)=\ln\left(\frac{4}{3}+\sqrt{1+\frac{16}{9}}\right)=\ln\left(\frac{4}{3}+\frac{5}{3}\right)=\ln 3.\)

Let \(\alpha=\ln 3\). Then

\(\sinh 4\alpha=\frac{1}{2}(e^{4\alpha}-e^{-4\alpha})=\frac{1}{2}\left(e^{\ln 81}-e^{-\ln 81}\right)=\frac{1}{2}\left(81-\frac{1}{81}\right).\)

Hence

\(\frac{1}{4}\sinh 4\alpha=\frac{1}{8}\left(81-\frac{1}{81}\right)=\frac{820}{81}.\)

Therefore

\(S=\frac{1}{32}\pi\left(\frac{820}{81}-\ln 3\right).\)