= Solution
{c}
Use <cylindrical coordinates>. The real square root restricts the <projection> to $0\leq r\leq1$, and there $r-1\leq0\leq\sqrt{1-r^2}$. Since $\nabla x=\mathbf e_x$, the <volume integral> is
$$
\begin{aligned}
\iiint_V\nabla x\,dV
&=2\pi\mathbf e_x\int_0^1r\bigl(\sqrt{1-r^2}-r+1\bigr)\,dr\\
&=2\pi\mathbf e_x\left(\frac13-\frac13+\frac12\right)
=\boxed{\pi\mathbf e_x}.
\end{aligned}
$$
For an independent boundary calculation, split $\partial V$ into an upper <hemisphere> $S_+$ and a lower <right circular conical surface> $S_-$. They meet in the <circle> $r=1,z=0$ and there is no additional planar face.
On $S_+$, use the <parametrization>
$$
\mathbf R(\alpha,\theta)
=(\sin\alpha\cos\theta,\sin\alpha\sin\theta,\cos\alpha),
\qquad 0\leq\alpha\leq\pi/2,\quad0\leq\theta\leq2\pi.
$$
The outward <unit normal vector> is $\mathbf R$, and the <surface area element> is $\sin\alpha\,d\alpha\,d\theta$. The $y$ and $z$ components of $\iint_{S_+}x\mathbf n\,dS$ vanish by the angular <integrals> of $\sin\theta\cos\theta$ and $\cos\theta$. The $x$ component is
$$
\int_0^{2\pi}\cos^2\theta\,d\theta\int_0^{\pi/2}\sin^3\alpha\,d\alpha
=\pi\cdot\frac23=\frac{2\pi}{3}.
$$
On $S_-$, the <parametrization> $(r\cos\theta,r\sin\theta,r-1)$ has outward <vector area element>
$$
\mathbf n\,dS=(\cos\theta,\sin\theta,-1)r\,dr\,d\theta.
$$
Its negative vertical component points out below the solid. Multiplying by $x=r\cos\theta$ and using the same angular <symmetry in integration> gives
$$
\iint_{S_-}x\mathbf n\,dS
=\mathbf e_x\int_0^1r^2\,dr\int_0^{2\pi}\cos^2\theta\,d\theta
=\frac\pi3\mathbf e_x.
$$
Therefore the independently evaluated <surface integral> is
$$
\boxed{\iint_{\partial V}x\mathbf n\,dS
=\left(\frac{2\pi}{3}+\frac\pi3\right)\mathbf e_x
=\pi\mathbf e_x,}
$$
in agreement with the <volume integral>. The isolated apex of the <cone> does not contribute to the <surface integral>.
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