I am being asked to integrate the vector field $F(x,y,z)=(0,2z(a-1),0)$ over the section $S$ of the sphere $x^2 + y^2 + z^2 = 4$ that has $0\leq x\leq1$ and $0\leq y\leq1$. Parameterising this surface seems like it would be overcomplicating things, so because the sphere is just a level set, I can just use:
$\int_S\vec{F}.d\vec{A}=\int_A\frac{\vec{F}.\vec{\nabla} f}{{\vec{e_3}}.\vec{\nabla} f}dxdy$
where $\vec{\nabla}f=(2x,2y,2z)$ is the gradient of $f(\vec{x})=x^2 + y^2 + z^2$.
After a few calculations, this simplifies $\frac{\vec{F}.\vec{\nabla} f}{{\vec{e_3}}.\vec{\nabla} f}=2y(a-1)$. But here is where I am unsure. Is it correct to say that
$\int_A2y(a-1)dxdy=\int_0^1dx\int_0^12y(a-1)dy$,
which is simple enough to calculate, or should I be parameterising $x$ and $y$ in some way? $A$ is described as the 'area of the surface S projected onto the $xy$- plane' in the notes I am reading. But which part of the surface is the 'projection' onto the $xy$-plane? Why would it be correct to choose the 'square' part of $S$ as the projection, and not the 'circular' part (i.e. where $|z|>\sqrt2$)? Perhaps I am overthinking but my notes do not give much guidance on this.
Yes, you are correct. As, You are to find the value of $\iint_{S} \vec{F}\cdot \hat{n} dS$. Since $\hat{n}=\frac{\vec{\nabla}f}{|\vec{\nabla}f|}=\frac{x\hat{i}+y\hat{j}+z\hat{k}}{2}$, therefore $\vec{F}\cdot \hat{n}=yz(a-1)$. Now, we take projection of the surface $S$ on $xy$-plane and we get: $$ dS= \sqrt{1+{z_x}^2+{z_y}^2} dx dy=\frac{2}{|z|}dx dy=\frac{2}{z}dx dy.$$ Note that our $S$ tells us that $z>0$. Now, the surface integral is reduced as:
You don't have to always parameterize the surface until or unless the question is specifically asked to solve by parametrization.