Welcome to our community

Be a part of something great, join today!

Apply the divergence theorem for the vector field F

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,047
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?
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
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.
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,047
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)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
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)