Integrating a vector valued function

[osc]

Homework Statement

Calculate

$$\int _{\mathbb{R}^{3+}} V(\textbf{r} ) d\textbf{r}$$

where

$$V(\textbf{r})=\frac{1}{r},\ \ r=||\textbf{r}||$$

The Attempt at a Solution

I'm guessing
$$\textbf{r}=x \textbf{i} + y \textbf{j} + z \textbf{k}$$

so
$$r=\sqrt{x^2+y^2+z^2}$$
and
$$d\textbf{r}= \textbf{i} dx + \textbf{j} dy + \textbf{k} dz$$

But this can't mean that

$$\int _{\mathbb{R}^{3+}} V(\textbf{r} ) d\textbf{r}=<br /> \int _{\mathbb{R}^{3+}} \frac{1}{\sqrt{x^2+y^2+z^2}} (\textbf{i} dx + \textbf{j} dy + \textbf{k} dz)=<br /> \int _{\mathbb{R}^{+}} \frac{1}{\sqrt{x^2+y^2+z^2}} \textbf{i} dx<br /> +\int _{\mathbb{R}^{+}} \frac{1}{\sqrt{x^2+y^2+z^2}} \textbf{j} dy<br /> +\int _{\mathbb{R}^{+}} \frac{1}{\sqrt{x^2+y^2+z^2}} \textbf{k} dz$$

can it?

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

I could have understood something like

$$\int _{\mathbb{R}^{3+}} V(\textbf{r} ) \cdot d\textbf{r}$$

but now I'm lost.

 

[lanedance]

it depends on cicurmstance / notation

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

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

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

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

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)

 

[osc]

Ok, that makes sense.

Thank you!