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Integrating a vector valued function

  1. Sep 27, 2009 #1

    osc

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

    Calculate

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

    where

    [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]

    so
    [tex]r=\sqrt{x^2+y^2+z^2}[/tex]
    and
    [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
    [/tex]

    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

    lanedance

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    Homework Helper

    it depends on cicurmstance / notation

    first
    [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

    osc

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    Ok, that makes sense.

    Thank you!
     
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