Integrate over all values of a vector

[Ionophore]

Hi,

This is probably a really simple question, but I think that I am getting lost in notation. I want to integrate the following over all values of the (2-dimensional) vector $$\overline{r}$$:

$$\int_{\overline{r}} \frac{\delta(\abs{\overline{r}-L})}{2\pi L} \overline{r} d\overline{r}$$

Basically, I want to integrate over all space. I think that the way to proceed is to convert to spherical polar coordinates but I'm not really sure how.

Edit
... delta is a dirac delta function. Sorry I didn't specify that before... maybe it's adding unnecessary complexity to my question. All i really want to know is how to deal with the $$\overline{r}$$ out front.

Thank you,

-ben

 

[CompuChip]

Your notation is also sloppy. For example, is L a vector, or did you mean
$$\delta(|\vec r| - L)$$
(the delta restricting the length of r?)

Also, should the integral produce again a vector v? Then basically, it says$$v_i = \int_{-\infty}^\infty \int_{-\infty}^\infty \frac{\delta(r_1^2 + r_2^2 - L)}{2\pi L} r_i \, dr_1 dr_2$$

where $v_i$ denotes the i-th component.

 

[Ionophore]

Whoops, I'm sorry, I did mean magnitude:

$$\int_{\vec{r}} \frac{\delta(\abs{|\vec{r}|-L})}{2\pi L} \vec{r} d\vec{r}$$

So, then, I am to understand that the above is really a double integral? What is meant by $$r_i$$, $$dr_1$$ and $$dr_2$$?

 

[Hurkyl]

So, then, I am to understand that the above is really a double integral?

One of the basic theorems of multivariable calculus is that a single integral over a two-dimensional region can be computed by a double integral, each over one-dimensional regions. (And similarly for higher-dimensional regions) Surely you've done such things before? e.g. integrating density over a three-dimensional region to calculate mass?

I assume the same ideas would hold for multivariate distributions. (so that it would be valid with a Dirac delta distribution)

Incidentally, it looks much easier in Cartesian coordinates. By the way, did you mean $\delta^2$ instead of $\delta$?

 

[Ionophore]

So:

$$\int_{\vec{r}} \frac{\delta(\abs{|\vec{r}|-L})}{2\pi L} \vec{r} d\vec{r} = \int_{r_{x}} \int_{r_{y}} \frac{\delta(\abs{|\vec{r}|-L})}{2\pi L} r_{x}r_{y}dr_{x}dr_{y}$$

Where $$r_{x}$$ and $$r_{y}$$ are the components of $$\vec{r}$$, and the integrations extend from -infinity -> infinity. I think I see.

Thanks everyone,

-ben

Edit:
And yes, of course, I have "done" this before, but that intermediate step is never shown, and so I just sort of memorized the general idea without really understanding it.

 

[CompuChip]

Almost. Note that there is a vector $\vec r$ in the integrand. So either you will calculate a vector, that is:
$$\vec I = \int_{r_{x}} \int_{r_{y}} \frac{\delta(\abs{|\vec{r}|-L})}{2\pi L} \vec r dr_{x}dr_{y}$$
meaning
$$I_x = \int_{r_{x}} \int_{r_{y}} \frac{\delta(\abs{|\vec{r}|-L})}{2\pi L} r_x dr_{x}dr_{y}$$
and
$$I_y = \int_{r_{x}} \int_{r_{y}} \frac{\delta(\abs{|\vec{r}|-L})}{2\pi L} r_y dr_{x}dr_{y}$$.
Or, you are still forgetting some absolute value, and you actually mean
$$\int_{r_{x}} \int_{r_{y}} \frac{\delta(\abs{|\vec{r}|-L})}{2\pi L} |\vec r| dr_{x}dr_{y} = \int_{r_{x}} \int_{r_{y}} \frac{\delta(\abs{|\vec{r}|-L})}{2\pi L} \sqrt{r_x^2 + r_y^2} dr_{x}dr_{y}$$.

Also, since you mentioned something about polar coordinates: note that this is just one way to write the integral. This is what we usually do, because we are used to think about (x, y) Cartesian coordinates in the plane. However, you can also go to polar coordinates $(r, \theta)$, then your integral would read (assuming you meant |r| instead of r-vector in the integrand):
$$\int_{-\infty}^\infty \int_0^{2\pi} \frac{\delta(\abs{r-L})}{2\pi L} r \times r \, dr d\theta$$
(check that!). It looks different, but note that there are still two integrals which need to be done. Which coordinates you want to use is up to you, and is often dictated by some (usually physical) principle such as a symmetry; for example if you have rotation invariance (the integrand only depends on $|\vec r| = \sqrt{x^2 + y^2}$ it is much easier to use polar coordinates), if your integrand is $x^2$ you would be better of using Cartesian coordinates.