# Find the flux of the vector field

$\mathbf F(x,y,z) = (0, \frac{y}{z(1+z^2)},0)$ through the surface $x^2+y^2 = z^2$ between $z=0$ and $z=2$.

My workings:

\begin{align} \phi(r,\theta) &= (r\cos\theta, r\sin\theta, r) \\ \implies \frac{\partial \phi}{\partial r} &= (\cos\theta, \sin\theta, 1)\quad\text{and}\quad\frac{\partial \phi}{\partial \theta} = (-r\sin\theta, r\cos\theta, 0) \\ \end{align}

So, $\frac{\partial \phi}{\partial r} \times \frac{\partial \phi}{\partial \theta} = (-r\cos\theta,-r\sin\theta, r)$

Also $\mathbf F(\phi) = (0, \frac{\sin\theta}{r(1+r^2)}, 0)$

Computing the surface integral gives

$$A = \int_0^{2}\int_0^{2\pi} \frac{-\sin^2\theta}{1+r^2} d\theta dr = -\pi \tan^{-1}(2)$$

But the correct answer is given by $\pi \tan^{-1}(2)$. Im not sure where I've gone wrong, i guess its to do with the orientation of the cross product, but i dont get why that would change because surely $r>0.$

To illustrate, the surface is a cone. The $\color{#f60}{\frac{\partial\phi}{\partial r}}$ vector points upward along the cone, and the $\color{#00f}{\frac{\partial\phi}{\partial\theta}}$ vector points counterclockwise (into the page, in the figure above), so their cross product $\color{#090}{\frac{\partial\phi}{\partial r}\times\frac{\partial\phi}{\partial\theta}}$ will point inside the cone, thus the extra negative sign.