# Calculate $\int\limits_{\Omega} \nabla \cdot F \,\mathrm dx$

$\def\d{\mathrm{d}}$During some new lecture series I started to attend I was given the following exercise. It involves vector calculus with which I am rather weak. Therefore, it might be elementary and trivial to some people here.

The exercise goes as below:

Let $h \in C^{1}([0, \infty))$ with $h(0) = 0$ and $$h'(t) > 0. \quad \forall t \in [0,\infty).$$Furthermore, let $$\Psi: \mathbb{R}^3 \to \mathbb{R}, \quad \Psi(x_1, x_2, x_3) = h(x_1^2 + x_2^2 + x_3^2)$$ and $\Omega := \Psi^{-1}([0,T])$, $T > 0$.

Let $$F: \mathbb{R}^3 \to \mathbb{R}^3, \quad F(x_1, x_2, x_3) = g(x_1^2 + x_2^2 + x_3^2)(x_1, x_2, x_3)^{\text{T}},$$ where $g\in C^{1}(\mathbb{R})$. Now calculate $$\int_{\Omega} \nabla\cdot F\,\d x.$$

My thoughts are as follows.

First of all, I am new to the notation $\nabla \cdot F$. I suppose that "$\cdot$" is the dot product and $$\nabla \cdot F = \frac{\partial F_1}{\partial x_1} + \frac{\partial F_2}{\partial x_2} + \frac{\partial F_3}{\partial x_3},$$ that is, divergence. I just know it as $\text{div}\,F$. If this is the case, I suppose it involves the Gaussian divergence theorem which says $$\int_{\Omega} \mathrm{div}\, F\,\d\Omega = \int_{\partial \Omega} F\cdot n \,\d(\partial\Omega).$$ The points where I am more or less stuck now are

1. How do I find the boundary $\partial \Omega$?
2. How do I find the normal vector $n$ on this boundary?

With regard to the second question, I know that I can find a (unitary) normal by considering the gradient of some function but I don't know which function to consider.

• The edit of the almost one year old question was really not necessary, Alex Francisco. You did not change anything substantial and the answer was already given. – Taufi Feb 12 '18 at 19:34

Luckily, I got my hands on the solution. It goes as follows.

The conditions $h(0) = 0$ and $h'(t) > 0$ imply that there is either one point $t_0 \in [0, \infty]$ for which it holds that $h(t_0) = T$ or none. Let $A$ denote the first case and $B$ the second.

In general, since $\Psi : \mathbb{R}^3 \to \mathbb{R}, \Psi(x) = h(\lvert x\rvert_2^2)$, it is $\Omega := \Psi^{-1}([0, T]) = \{x \in \mathbb{R}\;\vert \;h(\lvert x\rvert^2) \leq T\}$. In case A, there exists exactly one $t_0 \in [0, \infty)$: $h(t_0) = T$, therefore $$\Omega = \{ x\, \vert \, h(\lvert x\rvert^2 \leq h(t_0)\} = \{x \, \vert \, \lvert x \rvert \leq \sqrt{t_0}\} = B_{\sqrt{t_0}}(0)$$

In case B, it is $\Omega =\mathbb{R}^3$ since $h(t) < T \, \forall t \in [0, \infty)$.

So, after we found the appropriate set $\Omega$, we can integrate relatively easy. The usual relation is $$\int\limits_{\Omega} \nabla\cdot F\, dx = \int\limits_{\partial \Omega} F \nu \, dS$$ In case A, $\partial \Omega = \partial B_{\sqrt{t_0}}(0)$ and $\nu = \frac{1}{\sqrt{t_0}} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}$. It then follows that

$$\int\limits_{\partial \Omega} F \nu ds = \int\limits_{\partial \Omega} g(\lvert x\rvert^2)\frac{1}{R}\cdot \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} dS = g(R^2)\cdot R\int\limits_{\partial \Omega} ds = 4\pi R^3 g(R^2)$$

Funnily, in case B you cannot apply the Gaussian theorem since $\mathbb{R}^3$ is not compact. I guess the one who posed the exercise missed that.