Area of projected surface Suppose I have a surface given by $z=f(x,t)$, $a\le x \le b$ and $c\le t \le d$. How can I find the area of the surface projected onto the $(x,z)$ plane?
We may assume appropriate monotonicity conditions to have a nice projection.
 A: Consider the parameterization $R(x,y) = (x,y, f(x,y))$ for $a \leq x \leq b$ and $c \leq t \leq d$. The normal to the surface at $R(x,y)$ is given by $N(x,y)=R_x \times R_y = \langle 1,0, f_x \rangle \times \langle 0,1, f_y \rangle = \langle -f_x,-f_y, 1 \rangle$. The infinitesimal area element is then given by $d\vec{S} =\langle -f_x,-f_y, 1 \rangle \, dx \, dy$. If we were to integrate $|| \vec{dS} ||$ over the surface then this would give the area of the surface. However, you wish to find the area of the projection. Naturally, we project $\vec{dS}$ onto the $xz$-plane as follows:
$$ dA_{xz} = |\vec{dS} \cdot \langle 0,1,0 \rangle | $$
To see this formula checks in an obvious case, if $\vec{dS} = (dS) \langle 0,1,0 \rangle$ then we ought to have area of the surface and its projection match. Likewise, if the area element has no $y$-component then it means the surface only has a line as a projection hence the area of the projection ought to be zero. Thus,
$$ A_{projected}= \int dA_{xz} = \int_{a}^b \int_{c}^d |f_y| dy \, dx. $$
As you predicted, we probably should require $f_y \neq 0$ or we'll need a more detailed analysis.
To be safe, let's check my claim in a verifiable case. Consider the cylinder $z=y^2$ for $0 \leq x \leq 1$ and $0 \leq y \leq 2$. Geometrically, this projects $(0,2,4)$ to $(0,0,4)$ and $(0,0,0)$ to $(0,0,0)$ and the other two points are the same with $x=1$; $(1,0,4)$ and $(1,0,0)$ this gives a $1 \times 4$ rectangle of area 4. On the other hand, $f(x,y)=y^2$ hence $f_y=2y$ and $\int_{0}^1 \int_{0}^2 |2y| dy \, dx = 4$.
My example $z=x^2+y^2$ needs some work, I hope to fix it later tonight. 
