Divergence of the Radial Field F = (x, y)

Differentiate each component with respect to its own variable

Divergence eats a vector field and returns a number at each point. You differentiate the first component with respect to $x$, the second with respect to $y$, and add. Only those two diagonal pairings appear.

Here $P(x,y) = x$ and $Q(x,y) = y$, so both derivatives are as easy as derivatives get.

Adding gives $\nabla \cdot \mathbf{F} = 2$ at every point of the plane, the origin included.

The mistake: adding the wrong two partials

This field has four first partials, and only two of them belong to the divergence.

• $\frac{\partial P}{\partial x} = 1$ and $\frac{\partial Q}{\partial y} = 1$: these two are the divergence.
• $\frac{\partial P}{\partial y} = 0$ and $\frac{\partial Q}{\partial x} = 0$: these two are the curl.
• Swap the pairs and you report a divergence of $0$ when the true value is $2$: you have read the curl's two partials and put the divergence's name on them.

The other habit to break is stacking instead of summing. If your answer is $\langle 1, 1 \rangle$, you have built a vector, which is what the gradient does. Divergence of a plane field is one number.

Why a constant answer is the right shape here

Picture the arrows: at each point the arrow points straight away from the origin, with length equal to the distance from the origin. Far out the arrows are long, near the origin they are short. Divergence does not measure that length. It measures how fast nearby arrows spread apart.

That spreading rate is the same everywhere, which is what the constant $2$ records. Read as a fluid velocity, a small blob carried by this flow grows its area at a relative rate of $2$ per unit time no matter where it starts.

The pattern generalises: in $n$ dimensions the field $\langle x_1, x_2, \dots, x_n \rangle$ has divergence $n$, because each of the $n$ diagonal partials contributes exactly $1$.