what is the surface integral of graphs

Question

The surface integral of a vector field $F$ on $\mathbb R^3$ over a parametrized surface $\Phi:D\to \mathbb R^3$ is the number $$ \int \int _{S} F\cdot dS = \int \int _{D} F(\Phi (u,v)) \cdot (\Phi_u \times \Phi_v) du dv $$

(Here, $S:=\Phi(D)$)

---

My Question is: How to prove the following assertion: "If $F= Pi + Qj + Rk$ is a vector field in space and $S$ is the surface $z= f(x,y)$ where $f$ is a function on $D,$ the surface integral of $F$ over $S$ is: $$ \int \int_{S} F\cdot S = \int \int_{D} (-Pf_x-Qf_y+R) dx dy. $$"

Answer 1

You can always parameterize a graph in the obvious way: $$ \Phi(x,y) = (x,y,f(x,y)) $$

Then you have $\Phi_x = (1,0,f_x)$ and $\Phi_y=(0,1,f_y)$, with cross product $$ \Phi_x \times \Phi_y = (-f_x,-f_y,1) $$

Taking the dot product with $F=(P,Q,R)$, you get the integrand: $(-Pf_x - Qf_y + R)$.