= Solution
{c}
The <partial derivatives> of the <parametrization> are
$$
\begin{aligned}
\mathbf r_\theta&=(-\sin\theta/\sqrt2,\cos\theta\cos\phi,\cos\theta\sin\phi),\\
\mathbf r_\phi&=(0,-\sin\theta\sin\phi,\sin\theta\cos\phi).
\end{aligned}
$$
Their <cross product> is
$$
\mathbf r_\theta\wedge\mathbf r_\phi
=\left(\sin\theta\cos\theta,\frac{\sin^2\theta\cos\phi}{\sqrt2},\frac{\sin^2\theta\sin\phi}{\sqrt2}\right).
$$
Taking its <euclidean norm> gives the <surface area element>
$$
dA=\frac{\sin\theta}{\sqrt2}\sqrt{1+\cos^2\theta}\,d\theta\,d\phi.
$$
The two poles and duplicated seam have zero <surface area>, so integrating over the stated <parameter domain> counts the <surface area> once. With the <change of variables> $u=\cos\theta$ and the supplied <definite integral>,
$$
\begin{aligned}
\operatorname{area}(S)
&=\frac{2\pi}{\sqrt2}\int_0^\pi\sin\theta\sqrt{1+\cos^2\theta}\,d\theta\\
&=\sqrt2\pi\int_{-1}^1\sqrt{1+u^2}\,du\\
&=\boxed{2\pi+\sqrt2\pi\ln(1+\sqrt2).}
\end{aligned}
$$
Here $-\ln(\sqrt2-1)=\ln(\sqrt2+1)$ because $(\sqrt2-1)(\sqrt2+1)=1$.
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