Understanding d^3 vec{r} and its Role in Integrals | Helpful Explanation

[InsertName]

Homework Statement

This isn't my homework question exactly, but it will help me to do the question. We have an integral that is of the form:

$$\int$$$$\int$$$$\int$$ f($$\vec{r}$$)g($$\vec{r}$$) $$\vec{r}$$ d$$^{3}$$$$\vec{r}$$

Homework Equations

N/A

The Attempt at a Solution

I don't know what this d$$^{3}$$$$\vec{r}$$ is. I'm normally faced with dV or dr or dx or something of this form.

I know that r = sqrt (x^2 + y^2 + z^2)
$$\vec{r}$$ = xi + yj + zk
and how to represent x, y and z in terms of r, theta and phi but I don't know where to go form here.

 

[Andrew Mason]

The Attempt at a Solution

I don't know what this d$$^{3}$$$$\vec{r}$$ is. I'm normally faced with dV or dr or dx or something of this form.

If you let f(r), g(r) and r be the derivative of F(r) and G(r) R respectively with respect to a variable x, say, the expression becomes:

$$\int \int \int \frac{dF(\vec{r})}{dx}\frac{dG(\vec{r})}{dx}\frac{d\vec{R}} {dx}d^{3}\vec{r} = \int \int \int dF(\vec{r})dG(\vec{r})d\vec{R} \frac{d^{3}\vec{r}}{dx^3}$$

Does that help?

AM

 

[InsertName]

Thanks for the quick reply...

I'm not entirely sure if it does help with this integral. Perhaps it would be best to be more specific. The functions are of the form:

f($$\vec{r}$$) = e$$^{-r/a}$$

g($$\vec{r}$$) = xe$$^{-r/2a}$$

Clearly, the functions can be combined to give

xe$$^{-3r/2a}$$

P.s. I'd like to solve as much of this problem myself as possible.

 

[Andrew Mason]

Thanks for the quick reply...

I'm not entirely sure if it does help with this integral. Perhaps it would be best to be more specific. The functions are of the form:

f($$\vec{r}$$) = e$$^{-r/a}$$

g($$\vec{r}$$) = xe$$^{-r/2a}$$

Clearly, the functions can be combined to give

xe$$^{-3r/2a}$$

P.s. I'd like to solve as much of this problem myself as possible.

I was trying to show you that d^3r is related to the third derivative of r.

AM

 

[Dickfore]

$$d^{3}\vec{r} \equiv dV$$

 

[InsertName]

Hero.