Volume integral of grad Could someone explain to me what it means to do a volume integral over a vector field. It doesn't seem to make sense! I have attached the question but don't understand what the last part "means". Any help with understanding and the final part of the question itself would be appreciated. > **$\bf 10C\,\,\,\,$Vector Calculus** > > $\quad$ State the divergence theorem for a vector field $\mathbf u(\mathbf x)$ in a region $V$ bounded by a piecewise smooth surface $S$ with outward normal $\bf n$. > $\quad$ Show, by suitable choice of $\bf u$, that $$\int_V\nabla f\,\mathrm dV=\int_S f\,\mathrm d\mathbf S\tag{$*$}$$ for a scalar field $f(\mathbf x).$ > $\quad$ Let $V$ be the paraboloidal region given by $z\geqslant 0$ and $x^2+y^2+cz\leqslant a^2$, where $a$ and $c$ are positive constants. Verify that $(*)$ holds for the scalar field $f=xz$.

One way to calculate a volume integral of a vector field $\mathbf{u}(\mathbf{x})$ is to expand $\mathbf{u}$ into its Cartisian components: $\mathbf{u}(\mathbf{x})=\mathbf{\hat e}_xf(\mathbf{x})+\mathbf{\hat e}_yg(\mathbf{x})+\mathbf{\hat e}_zh(\mathbf{x})$, where $f,g,h$ are scalar fields. Then by linearity (the unit vectors are constants),

$$\int_V\mathbf{u}(\mathbf{x})dV=\int_V\left(\mathbf{\hat e}_xf(\mathbf{x})+\mathbf{\hat e}_yg(\mathbf{x})+\mathbf{\hat e}_zh(\mathbf{x})\right)dV\\\ =\mathbf{\hat e}_x\int_Vf(\mathbf{x})dV+\mathbf{\hat e}_y\int_Vg(\mathbf{x})dV+\mathbf{\hat e}_z\int_Vh(\mathbf{x})dV.$$

So the volume integral of a vector field is just the vector whose components are scalar field volume integrals.