TITLE: vector calculus - Integral over vector
QUESTION [5 upvotes]: Our physics prof wrote the following equation:
$\int\frac{\vec{r}}{r^3}d\vec{r} = \int\frac{1}{r^2}dr$
This is logical as long as I argue that $\vec{r}$ and $d\vec{r}$ are parallel, which is why the dot product evaluates as $|\vec{r}||d\vec{r}| = r dr$ However then i tried to do it by hand:
$\vec{r}d\vec{r} = \left(\begin{array}{c}x\\y\\z\\ \end{array}\right)\left(\begin{array}{c}dx\\dy\\dz\\ \end{array}\right) = xdx + ydy + zdz$
but this is nowhere near
$rdr = \sqrt{x^2+y^2+z^2}\sqrt{dx^2 + dy^2 + dz^2}$
which is why I would like to ask you what i am doing wrong. 
Thanks in advance
ftiaronsem

REPLY [3 votes]: Maybe the following helps (it seems to be sloppy notation of your prof):
The integral $$\int_{\mathbf{r}_a}^{\mathbf{r}_b}\frac{\vec{r} \cdot d\vec{r}}{r^3}$$ is a line-integral. The vector field 
$$\vec{E}(\vec{r})=\frac{\vec{r}}{r^3}$$
is the gradient of a scalar field $\phi(\vec{r}) = -r^{-1}$, i.e.,
$$\vec{E}(\vec{r}) = \vec{\nabla} \phi(\vec{r}).$$ Therefore, the line-integral is path independent and the result is given by
$$\int_{\mathbf{r}_a}^{\mathbf{r}_b} \vec{E}(\vec{r}) \cdot d\vec{r} = 
\int_{\mathbf{r}_a}^{\mathbf{r}_b} \vec{\nabla} \phi(\vec{r}) \cdot d\vec{r}
= \phi(\vec{r}_b) - \phi(\vec{r}_a) = r_a^{-1} - r_b^{-1},$$
which coincides with the result of your prof.