= Solution
{c}
Express the <position vector> in the suggested <orthonormal basis> as $\mathbf r=X\mathbf e_1+Y\mathbf e_2+Z\mathbf e_3$. Then
$$
x=\frac{-X+Z}{\sqrt2},\qquad
y=\frac{-X-Z}{\sqrt2},\qquad z=Y.
$$
This <orthogonal transformation> preserves $x^2+y^2+z^2=X^2+Y^2+Z^2$ and has <Jacobian determinant> of absolute value one, so $dV=dX\,dY\,dZ$. The basis actually reverses <orientation>; the absolute value of the <Jacobian determinant> is why no minus sign enters the <volume integral>. Also $x-y=\sqrt2 Z$, so the <half-space> becomes $Z\geq1/2$.
At each fixed $Z\in[1/2,1]$, the section is the <disk> $X^2+Y^2\leq1-Z^2$, of <area> $\pi(1-Z^2)$. Its <reflection symmetry> in $X$ makes the <integral> of $X$ vanish. Hence the required <triple integral> is
$$
\begin{aligned}
I&=-\frac1{\sqrt2}\int_{1/2}^1
\iint_{X^2+Y^2\leq1-Z^2}(X+Z)\,dX\,dY\,dZ\\
&=-\frac\pi{\sqrt2}\int_{1/2}^1Z(1-Z^2)\,dZ
=-\frac\pi{\sqrt2}\left[\frac{Z^2}2-\frac{Z^4}4\right]_{1/2}^1
=\boxed{-\frac{9\pi}{64\sqrt2}}.
\end{aligned}
$$
The negative sign is also consistent with the cap lying in the direction $(1,-1,0)$, where its average $y$ coordinate is negative.
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