Questions tagged [divergence-theorem]

Gauss Divergence Theorem relates the flux of a vector field through a closed surface to the divergence of the field in the volume enclosed.

In vector calculus, the divergence theorem, also known as Gauss's theorem or Ostrogradsky's theorem, is a result that relates the flux of a vector field through a closed surface to the divergence of the field in the volume enclosed.

More precisely, the divergence theorem states that the surface integral of a vector field over a closed surface, which is called the flux through the surface, is equal to the volume integral of the divergence over the region inside the surface. Intuitively, it states that the sum of all sources of the field in a region (with sinks regarded as negative sources) gives the net flux out of the region.

The divergence theorem is an important result for the mathematics of physics and engineering, in particular in electrostatics and fluid dynamics.

In physics and engineering, the divergence theorem is usually applied in three dimensions. However, it generalizes to any number of dimensions. In one dimension, it is equivalent to integration by parts. In two dimensions, it is equivalent to Green's theorem.

Suppose $V$ is a subset of $\mathbb {R} ^{n}$ (in the case of $n = 3$, $V$ represents a volume in three-dimensional space) which is compact and has a piecewise smooth boundary $S$ (also indicated with $\partial V = S$). If $\mathbf {F}$ is a continuously differentiable vector field defined on a neighborhood of $V$, then:

$$ \iiint _ {V}\left(\mathbf {\nabla } \cdot \mathbf {F} \right)\ dV = \oint _ S (\mathbf {F} \cdot \mathbf {n} )\ dS \text . $$

The left side is a volume integral over the volume $V$, the right side is the surface integral over the boundary of the volume $V$. The closed manifold $\partial V$ is oriented by outward-pointing normals, and $\mathbf {n}$ is the outward pointing unit normal at each point on the boundary $\partial V$. ($d\mathbf {S}$ may be used as a shorthand for $\mathbf {n} dS$.) In terms of the intuitive description above, the left-hand side of the equation represents the total of the sources in the volume $V$, and the right-hand side represents the total flow across the boundary $S$.

Source: Wikipedia

276 questions
5
votes
1 answer

Measure of the unit ball in $\mathbb{R}^n$

The unit sphere in $\mathbb{R}^n$, with respect to the Euclidean norm $\|\cdot\|$, is $$B_1(0) = \{x \in \mathbb{R}^n : \|x\| \leq1 \}.$$ Is it possible to compute the Lebesgue measure of this set using the Divergence Theorem? Can you give me a hint…
Nicola
  • 160
2
votes
1 answer

Divergence theorem(calculating flux of vector field)

I need to find flux of vector field $F=(x,y,z)$ over area $T=\{|x|+|y|+|z|\le 4|(x,y,z)\in \mathbb R^3\}$ oriented in the direction of outside normal, using:a)divergence theorem; b)directly. First, using divergence theorem i get integral $\iiint…
Trevor
  • 533
1
vote
0 answers

How can I think about a compact set with a smooth edge?

We discuss at the moment about the gaussian Formula (divergence theorem). There we have that the set on which we integrate must be compact and have a smooth edge. I know what compact means, but I can't imagine how a smooth edge looks like. We had…
user123234
  • 2,885
1
vote
0 answers

Divergence of a Swirl

What would the divergence of a swirl vector be of this image from google below? I understand that divergence is the amount of fluid flow in and out of a point. As such, I think the divergence at any point is positive since the vector field lines…
Woody
  • 181
1
vote
1 answer

A question on the divergence theorem

Let $\Omega \subseteq \mathbb{R}^n$ a bounded domain with boundary $\partial \Omega \in C^1$. Then if $w \in C^1(\Omega)$ it holds $$\int_\Omega \text{div} \ w \ dV = \int_{\partial \Omega} \ w \cdot \nu \ d\sigma$$ This is a classical formulation…
Gabrielek
  • 1,898
0
votes
1 answer

Divergence Theorem for a Parabola

Let $E$ be defined in $\mathbb{R}^3$ as the area between the parabola $z = x^2 + y^2$ and the plane $z = 2$. Given the vector $\vec{F} = \langle 2x, 2y, 0 \rangle$, find the flux integral $\iint_S \vec{F}\cdot\mathrm{d}\vec{r}$ using the divergence…
Mailbox
  • 896
0
votes
1 answer

Does the flux of a surface depend on how many surfaces the object has?

In my vector analysis class, we were given a vector field $\mathbf{F}(\mathbf{r})= \frac{1}{r^3}\mathbf{r}$ This vector field's divergence is 0, so according to the divergence theorem the flux over any closed object's surface $S_1$ within this field…
Kimple
  • 3
0
votes
1 answer

Problem to understand the divergence theorem in the exercise

For the vector field $\mathbf{F}=(xy^2,yz^2,x^2z)$ use the divergent theorem to evaluate $$\iint_S\mathbf{F}\cdot d\mathbf{S}$$ where $S$ is the sphere of radius $1$ centered at the origin. We use the divergence theorem and get that the divergence…
-1
votes
1 answer

Why DIVERGENCE THEOREM is not valid on open surface?

I have been reading proof of divergence theorem , but there has been a question that I just cannot solve for my self , and I searched a lot , but not much was out there , mathematically what is wrong with open surfaces in here? (why nobody even…