= Solution
For an arbitrary constant vector $\mathbf c$, the <product rule for divergence> gives
$$
\nabla\mathbin{\cdot}(\phi\mathbf c)=\mathbf c\mathbin{\cdot}\nabla\phi.
$$
The <divergence theorem> therefore implies
$$
\mathbf c\mathbin{\cdot}\int_V\nabla\phi\,dV
=\int_V\nabla\mathbin{\cdot}(\phi\mathbf c)\,dV
=\int_S\phi\,\mathbf c\mathbin{\cdot}d\mathbf S
=\mathbf c\mathbin{\cdot}\int_S\phi\,d\mathbf S.
$$
Since this holds for every $\mathbf c$,
$$
\boxed{\int_V\nabla\phi\,dV=\int_S\phi\,d\mathbf S}.
$$
For $\phi=x+y$, $\nabla\phi=(1,1,0)$. The ball of radius $a$ has volume $4\pi a^3/3$, so
$$
\int_V\nabla\phi\,dV=\frac{4\pi a^3}{3}(1,1,0).
$$
On the sphere, use $x=a\sin\theta\cos\varphi$, $y=a\sin\theta\sin\varphi$ and the stated outward <oriented surface element>. The first component of the surface integral is
$$
a^3\int_0^\pi\sin^3\theta\,d\theta
\int_0^{2\pi}(\cos^2\varphi+\sin\varphi\cos\varphi)\,d\varphi
=a^3\frac43\pi,
$$
and symmetry gives the same second component and a zero third component. The two sides agree.
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