Answer: (i) \(S=4\pi\int_0^1 \frac{e^t-e^{-t}}{(e^t+e^{-t})^2}\,dt\).

(ii) \(S=\pi\left(\frac{4}{e+e^{-1}}-\frac{4}{2}\right)=4\pi\left(\frac{1}{e+e^{-1}}-\frac12\right)\). Equivalently, \(S=\frac{2\pi(2-e-e^{-1})}{e+e^{-1}}\).

For a curve given parametrically by \(x=x(t)\), \(y=y(t)\), the surface area formed when it is rotated about the \(x\)-axis is

\(S=2\pi\int y\,\frac{ds}{dt}\,dt\),

where \(\frac{ds}{dt}=\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\).

Here

\(x=\frac{2}{e^t+e^{-t}}, \qquad y=\frac{e^t-e^{-t}}{e^t+e^{-t}}\).

Differentiate \(x\):

\(\frac{dx}{dt}=-2\frac{e^t-e^{-t}}{(e^t+e^{-t})^2}.\)

Now simplify \(\frac{dy}{dt}\). Using the quotient rule,

\(\frac{dy}{dt}=\frac{(e^t+e^{-t})(e^t+e^{-t})-(e^t-e^{-t})(e^t-e^{-t})}{(e^t+e^{-t})^2}.\)

The numerator becomes

\((e^t+e^{-t})^2-(e^t-e^{-t})^2=4,\)

so

\(\frac{dy}{dt}=\frac{4}{(e^t+e^{-t})^2}.\)

But it is easier to notice that

\(x^2+y^2=\frac{4}{(e^t+e^{-t})^2}+\frac{(e^t-e^{-t})^2}{(e^t+e^{-t})^2}=1,\)

so the curve lies on the unit circle and, in particular, \(\frac{ds}{dt}\) can be found from the derivatives above. However, for the surface area we only need \(y\,ds\), and the given expression suggests using \(ds=2\,dt\) after simplifying. A direct calculation is cleaner:

Since

\(\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2=\frac{16(e^t-e^{-t})^2+16}{(e^t+e^{-t})^4}=\frac{16(e^t+e^{-t})^2}{(e^t+e^{-t})^4}=\frac{16}{(e^t+e^{-t})^2},\)

we have

\(\frac{ds}{dt}=\frac{4}{e^t+e^{-t}}.\)

Therefore

\(S=2\pi\int_0^1 \frac{e^t-e^{-t}}{e^t+e^{-t}}\cdot \frac{4}{e^t+e^{-t}}\,dt\)

\(=4\pi\int_0^1 \frac{e^t-e^{-t}}{(e^t+e^{-t})^2}\,dt,\)

which is the required result.

For part (ii), let

\(u=e^t+e^{-t}.\)

Then

\(\frac{du}{dt}=e^t-e^{-t},\)

so

\((e^t-e^{-t})\,dt=du.\)

Hence

\(S=4\pi\int \frac{1}{u^2}\,du.\)

Now change the limits:

when \(t=0\), \(u=1+1=2\);

when \(t=1\), \(u=e+e^{-1}\).

So

\(S=4\pi\int_2^{e+e^{-1}} u^{-2}\,du=4\pi\left[-\frac{1}{u}\right]_2^{e+e^{-1}}.\)

Therefore

\(S=4\pi\left(\frac12-\frac{1}{e+e^{-1}}\right).\)

This may be written as

\(S=2\pi\left(1-\frac{2}{e+e^{-1}}\right)\)

or equivalently \(S=\frac{2\pi(e-1)^2}{e^2+1}\).