Find the surface integral of $\mathbf{u}\cdot\mathbf{n}$ over $S$ where $S$ is the part of the surface $z=x+y^2$ with $z<0$ and $x>-1$, $\mathbf{u}$ is the vector field $(2y+x,\,-1,\,0)$ and $\mathbf{n}$ has negative $z$-component.
I have the surface written parametrically as $$(x,\,y,\,x+y^2).$$ In my book the authors claim two parallel vectors to this surface are $$(1,\,0,\,1)\quad\text{and}\quad(0,\,1,\,2y).$$ Since they don't show how this was achieved, I'm assuming there is some quick trick they used to find them. Anyone have a guess what that trick might be?
Your surface is the graph of the function $$ f(x,y) = x + y^2 $$ Therefore for any vector $v = v_x \partial_x + v_y \partial_y$ in $\mathbb{R}^2$, based at the point $(x,y)$, you have a tangent vector to the graph given by $$ (v_x, v_y, v\cdot\nabla f) $$ based at $(x,y,f(x,y))$.
Now evaluate the above using first $v = \partial_x$ gives
$$ (v_x, v_y, v\cdot\nabla f) = (1,0,\partial_x f) = (1,0,1) $$
independent of $(x,y)$, the base point. Evaluating using $v = \partial_y$ gives
$$ (v_x, v_y, v\cdot\nabla f) = (0,1,\partial_y f) = (0,1,2y) $$
at the base point $(x,y)$.