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Assume $0\le a<R$. In cylindrical coordinates, the sphere imposes $r^2+z^2\le R^2$, and the volume element is $rdrd\varphi dz$. The required triple integral isHenceIt tends to zero as $a\to 0$ and to the full sphere volume $4\pi R^3/3$ as $a\to R$, checking both limits.

Express the position vector in the suggested orthonormal basis as $\mathbf{r}=X{\mathbf{e}}_1+Y{\mathbf{e}}_2+Z{\mathbf{e}}_3$. ThenThis orthogonal transformation preserves $x^2+y^2+z^2=X^2+Y^2+Z^2$ and has Jacobian determinant of absolute value one, so $dV=dXdYdZ$. The basis actually reverses orientation; the absolute value of the Jacobian determinant is why no minus sign enters the volume integral. Also $x-y=\sqrt{2}Z$, so the half-space becomes $Z\ge 1/2$.

At each fixed $Z\in [1/2,1]$, the section is the disk $X^2+Y^2\le 1-Z^2$, of area $\pi (1-Z^2)$. Its reflection symmetry in $X$ makes the integral of $X$ vanish. Hence the required triple integral isThe negative sign is also consistent with the cap lying in the direction $(1,-1,0)$, where its average $y$ coordinate is negative.

The liquid occupies $0\le z\le h$ and $0\le r\le \sqrt{az}$. A horizontal slice is a disk of area $\pi az$. Integrating these areas givesEquivalently, the triple integral is ${\int }_0^{2\pi }{\int }_0^h{\int }_0^{\sqrt{az}}rdrdzd\theta$, including the cylindrical Jacobian determinant factor $r$.