# Equivalence of notation in Divergence Theorem

Normally I read the Divergence Theorem written as (\oiint doesn't exist here):

\begin{align} \oint_{\partial \Omega} \vec{F} \cdot \hat{n} \; dS = \iiint_{\Omega} \nabla \cdot \vec{F} \; dV \end{align}

Where $\partial \Omega$ is the boundary of region $\Omega$ and $dS$ are infinitesimal surface of $\partial \Omega$ and $dV$ infinitesimal volume of $\Omega$.

But reading a paper CASTILLO and GRONE, A matrix analysis approach to higher order approximations for divergence and gradients satisfying global conservation law, it states the Divergence Theorem as:

\begin{align} \int_{\partial \Omega} f \, \vec{v} \cdot \hat{n} \; dS = \int_{\Omega} \nabla \cdot \vec{v} \, f \; dV + \int_{\Omega} \vec{v} \, \nabla f \; dV \end{align}

Are those equivalent? if so how can I correlate the RHS of both equations?

I really think that he is abusing notation using a single integral instead a triple one and not directly specifying the loop integral in left hand side, but even with that I cannot correlate, maybe I'm missing some definition here.

The only error here is the second integral on the RHS should have the dot product as the integrand:

$$\int_{\Omega} \vec{v} \cdot \, \nabla f \; dV,$$

since $\nabla \cdot (f \,\vec{v}) = f \, \nabla \cdot \vec{v} + \nabla f \cdot \vec{v}$.

Aside from appearance, given an integrable function $f:\Omega \subset \mathbb{R}^3 \to \mathbb{R}$ there is no difference between

$$\int_\Omega f, \quad \int_\Omega f \, dV, \quad\text{and } \quad \int\int\int_\Omega f \, dV$$

There is difference between a multiple Riemann integral and an iterated Riemann integral, for example,

$$\int_{[0,1]^3} f \quad \text{and} \quad\int_0^1 \left(\int_0^1 \left(\int_0^1 f(x,y,z) \, dz \right)\, dy \right) \, dx$$

where the object on the right may exist but the object on the left may not. However, if $|f|$ is integrable on $[0,1]^3$, then the two must be equal, as one consequence of Fubini's theorem.

• I double checked the paper, the $\cdot$ really doesn't exist. Maybe a mistype. – Lin Jun 29 '18 at 3:04
• Definitely a mistype because $\vec{v}\, \nabla f$ is a vector. – RRL Jun 29 '18 at 3:06
• Just look up the vector identity for $\nabla \cdot (f \, \vec{v})$ in Wikipedia if you want confirmation. – RRL Jun 29 '18 at 3:08
• Is $f \vec{v}$ equivalent to $\vec{F}$? since $f$ is a scalar valued function and $\vec{v}$ a vector. (of course I could write $\vec{f}$ but I'm also using the capital here to show up the vector function). – Lin Jun 29 '18 at 3:09
• Yes, they are equivalent. You can regard $\vec{v}$ as carrying all the information about the direction of the vector field at every point, while $f$ just rescales $\vec{v}$ so that it matches the magnitude of $\vec{F}$. This is acomplished by setting $$f := \sqrt{\vec{F}\cdot\vec{F}}$$ and $$\vec{v} := \frac{1}{f}\vec{F}$$. Then $\vec{F} = f\vec{v}$ (quite obviously). – Jackozee Hakkiuz Jun 29 '18 at 3:42