# Integrating a vector valued function

1. Sep 27, 2009

### osc

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

Calculate

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

where

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

3. 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

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

can it?

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

I could have understood something like

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

but now I'm lost.

2. Sep 27, 2009

### 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

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

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)

3. Sep 28, 2009

### osc

Ok, that makes sense.

Thank you!