$$\iint\limits_\sum f \ d \sigma = \iiint\limits_S \operatorname{div} \textbf{f} \ dV$$
$$\operatorname{div} \textbf{f}=1+2+3=6$$
After this, we could multiply $6$ by the volume of the sphere $\frac{4}{3} \pi (3)^3$ to get $216 \pi$.
Shouldn't computing the integral give the same answer?
$$6\int_0^{2\pi} \int_0^{2\pi} \int_0^3 \ dr \ d \theta \ d \phi$$
$$=6(3)(2\pi)(2\pi)$$
$$=72 \pi^2$$
Is something wrong in the integral?
Thanks.
Several things wrong: in spherical coordinates, $$\mathrm{d}V=r^2\sin\phi\,\mathrm{d}r\,\mathrm{d}\phi\,\mathrm{d}\theta,$$ and the ball centered at $O$ with radius $3$ is represented, in spherical coordinates by $$\Delta=\bigl\{(r,\phi,\theta)\in\mathbb{R}^3\;\bigm\vert\;0\leq r\leq3,\ 0\leq\phi\leq\pi,\ 0\leq\theta\leq2\pi\bigr\}$$ so that your integral should read as: $$\iiint_\Delta r^2\sin\phi\,\mathrm{d}r\,\mathrm{d}\phi\,\mathrm{d}\theta=\int_0^3r^2\,\mathrm{d}r\int_0^\pi\sin\phi\,\mathrm{d}\phi\int_0^{2\pi}\mathrm{d}\theta=9\times2\times2\pi=36\pi.$$ Now, $6\times36\pi=216\pi$ and you're done.
