Should I use the unit normal to calculate the surface integral?

Question

So I am supposed to find the flux through the surface $x^2+y^2 \leq4$ in the xy-plane. So I find the parameterization of the disk:

$r(u,v)=(R cos(\theta), Rsin(\theta),0)$

Then I find the crossproduct $\frac{\partial r}{\partial R} \times \frac{\partial r}{\partial u}=(0,0,R)$

So the surface integral is

$ \iint_S F\bullet(\frac{\partial r}{\partial R} \times \frac{\partial r}{\partial \theta})r\,dr\,d \theta$

But in the solution they use $(0,0,1)$, and therefore we don't get the same answer. My book tells me however that when you take surface integrals over vector fields, you don't need the unit normal. You only need to find the unit normal if you take the surface integral over a scalar function. So why I am I wrong?

Answer 1

When you parametrize the surface as $ \phi (u, v)$, the surface integral is,

$ \displaystyle \iint_D \vec F \cdot (\phi_u \times \phi_v) ~ du ~ dv \tag1$

$[$Depending on orientation, the normal vector could alternatively be $(\phi_v \times \phi_u)$$]$

In this case, it is a disk in xy-plane and the parametrization of the surface is,

$\phi (r, t) = (r \cos t, r \sin t, 0), 0 \leq r \leq 2, 0 \leq t \leq 2 \pi$

$\phi_r \times \phi_t = (0, 0, r)$

So following $(1)$, the surface integral is,

$\displaystyle \int_0^{2\pi} \int_0^2 \vec F \cdot (0, 0, r) ~ dr ~ dt$

The book instead uses the fact that upward unit normal vector to any surface in xy-plane is $(0, 0, 1)$. That leads to surface integral as,

$\displaystyle \iint_{x^2 + y^2 \leq 4} \vec F \cdot (0, 0, 1) ~ dx ~ dy$

Now converting to polar coordinates, the surface integral is

$\displaystyle \int_0^{2\pi} \int_0^2 \vec F \cdot (0, 0, 1) ~ r ~ dr ~ dt$