Apply the divergence theorem for the vector field F

In summary, the conversation discusses the use of the divergence theorem to calculate surface integrals. The theorem states that the surface integral over the boundary of a three dimensional region is equal to the triple integral of the divergence of the vector field over that region. The conversation also touches on the direction of the normal vector on the surface, which is determined by the direction the vector points away from the region. This is important when determining the sign of the normal vector in the calculation. The conversation ends with a request for further clarification on the topic.
  • #1
mathmari
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Hey! :eek:

Apply the divergence theorem over the region $1 \leq x^2+y^2+z^2 \leq 4$ for the vector field $\overrightarrow{F}=-\frac{\hat{i}x+\hat{j}y+\hat{k}z}{p^3}$, where $p=(x^2+y^2+z^2)^\frac{1}{2}$.

$\bigtriangledown \overrightarrow{F}=\frac{3}{p^5}\frac{x^2+y^2+z^2}{p^2}-\frac{3}{p^3}=\frac{3}{p^3}-\frac{3}{p^3}=0$

So $\int \int \int_D {\bigtriangledown \overrightarrow{F}}dV=0$To calculate $\int \int_S {\overrightarrow{F} \cdot \hat{n}}d \sigma$ we have to calculate first the $\hat{n}$.

Isn't it as followed?
$\hat{n}=\frac{\bigtriangledown f}{|\bigtriangledown f|}=\frac{\hat{i}x+\hat{j}y+\hat{k}z}{p}$

In my textbook it is: $\hat{n}=\pm \frac{\bigtriangledown f}{|\bigtriangledown f|}=\pm \frac{\hat{i}x+\hat{j}y+\hat{k}z}{p}$
$"+": p=2$
$"-" \text{ for } p=1$
Why is it so?
 
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  • #2
The divergence theorem say that [tex]\int\int \vec{f}\cdot \vec{n}dS= \int\int\int \nabla\cdot\vec{f} dV[/tex] where the surface integral is taken over the boundary of the three dimensional region.

But a surface has two sides. Which way does the normal vector point? The rule for the divergence theorem is that the normal vector points outward, away from the three dimensional region. Here that region is defined by "[tex]1\le x^2+ y^2+ z^2\le 4[/tex]". That is, it is inside a sphere of radius 2 but outside a sphere of radius 1. The normal vector on the sphere of radius 2 points away from the region so "outward", away from the origin. The normal vector on the sphere of radius 1 also points away from the region which means it points "inward", toward the origin.
 
  • #3
HallsofIvy said:
The divergence theorem say that [tex]\int\int \vec{f}\cdot \vec{n}dS= \int\int\int \nabla\cdot\vec{f} dV[/tex] where the surface integral is taken over the boundary of the three dimensional region.

But a surface has two sides. Which way does the normal vector point? The rule for the divergence theorem is that the normal vector points outward, away from the three dimensional region. Here that region is defined by "[tex]1\le x^2+ y^2+ z^2\le 4[/tex]". That is, it is inside a sphere of radius 2 but outside a sphere of radius 1. The normal vector on the sphere of radius 2 points away from the region so "outward", away from the origin. The normal vector on the sphere of radius 1 also points away from the region which means it points "inward", toward the origin.

Could you explain it further to me? I got stuck... (Worried)
 
  • #4
mathmari said:
Could you explain it further to me? I got stuck... (Worried)

Where are you stuck?
What do you understand and what do you not understand? (Wondering)
 

FAQ: Apply the divergence theorem for the vector field F

What is the divergence theorem?

The divergence theorem, also known as Gauss's theorem, is a fundamental theorem in vector calculus. It states that the flux of a vector field through a closed surface is equal to the volume integral of the divergence of the vector field over the region enclosed by the surface.

How is the divergence theorem applied to a vector field?

To apply the divergence theorem, the vector field must be defined in a three-dimensional space and have a continuous derivative. The surface must also be closed and have a smooth boundary. The flux of the vector field through the surface is then calculated and compared to the volume integral of the divergence of the vector field over the enclosed region.

What is the significance of the divergence theorem?

The divergence theorem is an important tool in vector calculus and has many applications in physics and engineering. It allows for the calculation of flux and volume integrals in a simpler and more efficient manner. It also provides a deeper understanding of the relationship between vector fields and their divergence.

Can the divergence theorem be applied to any vector field?

No, the divergence theorem has certain requirements that must be met in order to be applied. The vector field must be defined in three dimensions and have a continuous derivative. The surface must also be closed and have a smooth boundary. If these conditions are not met, the divergence theorem cannot be applied.

What are some practical uses of the divergence theorem?

The divergence theorem has many practical uses in various fields such as fluid mechanics, electromagnetism, and heat transfer. It is used to calculate flux and volume integrals, as well as to solve problems involving flow and divergence of vector fields. It is also used in the derivation of important equations such as the Navier-Stokes equations and the continuity equation.

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