# How can a volume integral yield a vector field?

1. Jan 12, 2016

### rcmps772

I'm using the textbook Electricity and Magnetism by Purcell. In the section about continuous charge distributions I found the following formula

$$\mathbf{E}(x,y,z)= \frac{1}{4\pi\epsilon_0 } \int \frac{\rho(x',y',z')\boldsymbol{\hat r} dx'dy'dz'}{r^{2}}$$.

It's stated that (x,y,z) is fixed while we let the variables x', y' and z' range over the domain of integration. What puzzles me is that radial unit vector, which is supposed to point from (x', y', z') to (x,y,z), making the integrand a vector valued function.
What am I missing?

2. Jan 12, 2016

### blue_leaf77

Yes, it is the radial unit vector which gives the vector nature to the total E field.

3. Jan 12, 2016

### rcmps772

I understand the idea, but as a mathematical object it looks like nonsense to me. There are no examples in the textbook that use this formula directly. They use symmetry arguments such that only the magnitude of one direction is calculated.

4. Jan 12, 2016

### blue_leaf77

In which way does it look nonsense to you?
Well, the symmetry argument derives automatically from the vector summation, doesn't it? Without the presence of the radial unit vector in the integrand, there can't be such thing as cancellation of one or more of the field's components. The examples in your textbook do use that formula, but most of the times, examples are a group of problems which can be solved and understood easily. That's why a physical situation which is facilitated by the symmetry argument is commonly found in examples.

5. Jan 12, 2016

### rcmps772

In the same way that this looks
$$\int (r_x,r_y,r_z) dx'dy'dz'$$
This is essentially the same as what's up there, just removed clutter. I don't understand how this works.

6. Jan 12, 2016

### PeroK

The integral of a vector is, by definition, the vector formed by integrating each of the components.

7. Jan 12, 2016

### rcmps772

Well, makes sense. I'm guessing that it's defined just as easily through Riemann sums.Thank you for the help.

8. Jan 12, 2016

### PeroK

Yes, exactly. Although, once you have proved the properties of a scalar integral, the properties of a vector integral follow.

9. Jan 12, 2016

### Staff: Mentor

If it weren't a continuous charge distribution, but instead a finite number of discrete point charges, you'd calculate the electric field at a given point by summing the contributions from each point charge. Each contribution $\vec{E_i}$ would be a vector, so you'd write it as the product of its magnitude ($Q_i/r_i^2$ where $Q_i$ and $r_i$ are the charge of and distance to the $i$'th point charge) and a unit vector in the direction of that charge. These vectors point in different directions, but that doesn't stop you from adding them.

The integral you're looking at is the same thing except extended to an infinite number of infinitesimally small charges represented by $\rho$. The integrand is the vector-valued contribution to the total field produced by the infinitesimal charge in the infinitesimal volume dx'dy'dz at a distance $r$ in the direction $\hat{r}$; the integral is the sum of all of these.