Yes, that was perfect. I used the surface z=g(x,y)=y^2-x, and got a normal vector from <-\frac{\partial g} {\partial x}, -\frac{\partial g}{\partial y}, 1> . Did my curl and dot product calculations to get
\int\int_d <1,-2y,1>\cdot<-1,0,-x>dA=\int\int_d -x-1dA
I converted to polar using...