1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Integrating a vector valued function

  1. Sep 27, 2009 #1


    User Avatar

    1. The problem statement, all variables and given/known data


    [tex]\int _{\mathbb{R}^{3+}} V(\textbf{r} ) d\textbf{r}[/tex]


    [tex]V(\textbf{r})=\frac{1}{r},\ \ r=||\textbf{r}||[/tex]

    3. The attempt at a solution

    I'm guessing
    [tex]\textbf{r}=x \textbf{i} + y \textbf{j} + z \textbf{k}[/tex]

    [tex]d\textbf{r}= \textbf{i} dx + \textbf{j} dy + \textbf{k} dz[/tex]

    But this can't mean that

    [tex]\int _{\mathbb{R}^{3+}} V(\textbf{r} ) d\textbf{r}=
    \int _{\mathbb{R}^{3+}} \frac{1}{\sqrt{x^2+y^2+z^2}} (\textbf{i} dx + \textbf{j} dy + \textbf{k} dz)=
    \int _{\mathbb{R}^{+}} \frac{1}{\sqrt{x^2+y^2+z^2}} \textbf{i} dx
    +\int _{\mathbb{R}^{+}} \frac{1}{\sqrt{x^2+y^2+z^2}} \textbf{j} dy
    +\int _{\mathbb{R}^{+}} \frac{1}{\sqrt{x^2+y^2+z^2}} \textbf{k} dz

    can it?

    What would i,j,k mean in an integral?

    I could have understood something like

    [tex]\int _{\mathbb{R}^{3+}} V(\textbf{r} ) \cdot d\textbf{r}[/tex]

    but now I'm lost.
  2. jcsd
  3. Sep 27, 2009 #2


    User Avatar
    Homework Helper

    it depends on cicurmstance / notation

    [tex] V(\textbf{r}) : \mathbb{R}^3 \rightarrow \mathbb{R} [/tex]
    ie it is a scalar value function acting on R3

    that means
    [tex]\int _{\mathbb{R}^{3+}} V(\textbf{r} ) \cdot d\textbf{r}[/tex]
    makes no sense as you can only perform a dot product between two vectors

    so looking at your integral
    [tex]\int _{\mathbb{R}^{3+}} V(\textbf{r} ) d\textbf{r}[/tex]

    as the integral is over all R3 its most likely dr means a volume element, in some of the different notations floating around
    [tex] d\textbf{r} = dV = dr^3 = dx.dy.dz [/tex]

    if this is the case, its probably worth tranforming to spherical co-ordinates, as the function is spherically symmetric

    Remember to include your Jacobian term (effectively giving the volume element in spehrical coordinates)
  4. Sep 28, 2009 #3


    User Avatar

    Ok, that makes sense.

    Thank you!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook