- #1
osc
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Homework Statement
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]
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.