Question i cannot work out. I assume you need to get both sides in terms of u and v (parameterized), but im getting pretty confused after completing the first few steps.
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You may use geometry to evaluate the left side triple integral. Simply multiply the divergence by the volume of sphere of radius $3$ – AgentS Oct 14 '14 at 11:22
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For emphasis, ganeshie8's argument here requires that $\nabla \cdot \mathbf{F}$ is constant (which is indeed to the case here). – Travis Willse Oct 14 '14 at 11:24
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The left-hand size is over the solid ball $V$, whereas the right-hand side is over just its boundary, the sphere of radius $3$ centered at the origin. The fact that we can translate an integral in two dimensions into one in three dimensions (which may be easier) is what makes the Divergence Theorem a powerful tool.
For the left-hand side, we could change variables (spherical coordinates would work well here), but computing gives $\nabla \cdot \mathbf{F} = 2$, so the left hand side becomes $$2 \iiint_V dx\,dy\,dz,$$ which can be evaluated without calculus.
On the right-hand side, we need to pick a parameterization $\bf r$ of the sphere $S$ with some coordinates $(u, v)$, and use the parameterization formula $$\iint_S \mathbf{F} \cdot \mathbf{\hat{n}} \,dS = \int_{\mathbf{r}^{-1}(S)} \mathbf{F}(\mathbf{r}(u, v)) \cdot (\mathbf{r}_u \times \mathbf{r}_v) \,du \,dv.$$
To be clear, $\mathbf{r}^{-1}(S)$ is simply the domain of the parameterization.
Travis Willse
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