Change of variables from dS to dA for surface integrals using the gradient vector? In class, we learned that: $$
\iint \vec{F} \cdot d \vec{S} = \iint \vec{F} \cdot \hat{n} \|r_{u} \times r_{v}\| dA = \iint \vec{F} \cdot \hat{n} \|r_{u} \times r_{v}\| dudv
$$
where the normal unit vector is equal to $\frac{r_{u} \times r_{v}}{\|r_{u} \times r_{v}\|}$.
Moreover, $$\|r_{u} \times r_{v}\|
$$ is the Jacobian matrix, which is the scale factor for converting dS to dA. I understand this.
However, in some solutions for my math problems. This Jacobian was completely bypassed, and the gradient was used instead.
For example:

I'm confused about the solution:

I understand how you can use any normal unit vector, but the Jacobian scale factor still needs to come into play in order to convert the dS into dA, and then to dxdy. It seems they bypassed that using the gradient vector?

*

*How did this happen? Is there a gradient scale factor? How does the gradient come into play for surface integrals?

*How would you do this question using the Jacobian scale factor and a parametrization of two variables instead of the gradient?

 A: Suppose you wish to calculate the surface integral of some function $\vec{F}(x,y,z)$ over a  surface $G(x,y,z)=0$ $($In your example, $\vec{F}(x,y,z)=\langle x, y, 1\rangle$ and $G(x,y,z)=z+x^2+y^2-1=0)$. To evaluate this, we need to first find a parameterization of the surface. Using the implicit function theorem, we can find a parameterization of the surface at the point $(x,y)$ to be of the form $(x,y,z) \mapsto (x,y,f(x,y))$ $($or more familiarly, $\vec{r}(x,y) = x \hat{i} + y\hat{j} + f(x,y)\hat{k}$. Now given this parameterization, notice that:
$$\vec{r}_x \times \vec{r}_y = \left|\begin{matrix}\hat{i} & \hat{j} & \hat{k} \\
1 & 0 & f_x \\
0 & 1 & f_y\end{matrix}\right| = \langle -f_x, -f_y, 1 \rangle$$
But remember that our level surface $G(x,y,z) = z-f(x,y)=0$ tells us that $f_x = -\frac{G_x}{G_z}$ and $f_y=-\frac{G_y}{G_z}$. Substituting these gives us:
$$\vec{r}_x \times \vec{r}_y = \langle -f_x, -f_y, 1 \rangle=\left\langle\frac{G_y}{G_z},\frac{G_y}{G_z},1\right\rangle= \frac{1}{G_z}\langle G_x,G_y,G_z\rangle = \frac{\nabla G}{\nabla G \cdot \hat{k}}$$
Therefore we have:
$$\iint_S \vec{F}\cdot \vec{n}\;dS = \iint_S \vec{F}\cdot \frac{\vec{r}_x\times \vec{r}_y}{|\vec{r}_x \times \vec{r}_y|} |\vec{r}_x \times \vec{r}_y| \;dx dy = \iint_S \vec{F}\cdot \vec{r}_x\times \vec{r}_y \;dx dy$$
$$=\iint_S \vec{F}\cdot \frac{\nabla G}{\nabla G \cdot \hat{k}} \;dx dy$$
