# Confused on how to apply the volume integral form of the Biot-Savart law

In Viscous evolution of point vortex equilibria (Jing, 2011) the author explains that the velocity equation for the Lamb-Oseen vortex is found by applying the 2D Biot-Savart law to the vorticity equation. The Biot-Savart law is given in the paper as $$u(x) = \dfrac{1}{2\pi} \int_{\mathbb{R}^2}{ \dfrac{ (x-\overset{\sim}{x})^\bot }{ {\left\lVert x - \overset{\sim}{x} \right\rVert}^2 } \omega(\overset{\sim}{x}) \, d\overset{\sim}{x} }$$ where $x = \mathrm{(x,y)}$ and $x^\bot = \mathrm{(-y,x)}$, and the vorticity field is given as $$\omega(x,t) = \dfrac{\Gamma}{4 \pi \nu t} \exp {(- \dfrac{ {\left\lVert x \right\rVert}^2 }{4 \nu t} )}$$ Now, due to axisymmetry I believe the velocity field can be easily determined using Ampère's law in this case: $$\oint_C {v_\theta(r,t) \cdot dl} = \int_0^{2\pi} v_\theta(r,t) r \, d\theta = 2 \pi r \, v_\theta(r,t)$$and$$\iint_S { \omega(\overset{\sim}{r},t) \cdot dS } = \int_0^{2\pi} \int_0^r \omega(\overset{\sim}{r},t) \overset{\sim}{r} \, d \overset{\sim}{r} \, d\theta = \Gamma (1 - \exp {( -\dfrac{r^2}{4 \nu t} )} )$$so that$$v_\theta(r,t) = \dfrac{\Gamma}{2 \pi r} (1 - \exp {( -\dfrac{r^2}{4 \nu t} )} )$$ and this indeed gives the desired answer. Using Biot-Savart the author gives the following expression, which gives the same result: $$v_\theta(r,t) = \dfrac{1}{2\pi} \int_0^{2\pi} \int_0^r \dfrac{\omega(\overset{\sim}{r},t)}{r} \overset{\sim}{r} \, d \overset{\sim}{r} \, d\theta = \dfrac{\Gamma}{2 \pi r} (1 - \exp {( -\dfrac{r^2}{4 \nu t} )} )$$ However, for the life of me I can't figure out how to get that expression starting from the Biot-Savart law! In particular, where did the various instances of $(x - \overset{\sim}{x})$ (or $(r - \overset{\sim}{r})$) go? Did they cancel out somehow? Why does the denominator only have $r$ in it?

I've tried visualizing the 3D (electromagnetic) analogue of the problem and filling that in, but the fact that the current density $J$ depends on $\overset{\sim}{x}$ whereas the rest depends on $x - \overset{\sim}{x}$ keeps messing me up. All the examples I've been able to find online use the line integral version of Biot-Savart with a constant current $I$, and use Ampère's law for everything else!

I actually want to apply this to a more complex (3D) case, but after scouring the Internet (it feels that way, at least) I'm pretty stuck. This is my first question, so I hope it doesn't sound too confused. Any help would be much appreciated!

Edit: The non-uniform current density was a red herring. Consider the magnetic field inside an infinitely long, straight wire at distance $r$ from the center, induced by constant current density $J$ pointing up along the $z$-axis. From Ampère's law, we know the enclosed current induces the field $$\vec{B}(r) = \{B_r,B_\theta,B_z\} = \{0,\frac{r}{2}J,0\}$$ My reading of the Biot-Savart law leads to the following expression: $$\vec{B}(r) = \frac{1}{4\pi} \int_0^{2\pi} \int_0^r \int_{-\infty}^{\infty} \frac{ \{0,0,J\} \times \{r-r',0,z-z'\} }{ \left( (r-r')^2 + (z-z')^2 \right)^{3/2} } r' dz' dr' d\theta' = \{0,-rJ,0\}$$ Where I used $z-z'$ merely for symmetry (simply $z'$ would give the same answer). As you can see this gives the wrong result. The correct result is given by $$\vec{B}(r) = \frac{1}{4\pi} \int_0^{2\pi} \int_0^r \int_{-\infty}^{\infty} \frac{ \{0,0,J\} \times \{r,0,z'\} }{ \left( r^2 + z'^2 \right)^{3/2} } r' dz' dr' d\theta' = \{0,\frac{r}{2}J,0\}$$ The analogous 2D expression matches the equation given in Jing, 2011. I still don't understand why using the static distance between source and measurement is correct here, though. The way the Biot-Savart law is written (in all the forms I've encountered), it looks like the distance should be involved in the integration!