9. (12 pts) Set up the surface integral $\iint_{S}(z-x) d S$, where $S$ is the portion of the surface $z=x+y^{2}$, where $0 \leq x \leq y, 0 \leq y \leq 1$. DO NOT EVALUATE.

Step 1 ：The surface integral is given by \(\iint_{S}(z-x) d S\), where \(S\) is the portion of the surface \(z=x+y^{2}\), where \(0 \leq x \leq y, 0 \leq y \leq 1\).

Step 2 ：The surface \(S\) is given by \(z = x + y^2\). So, we can write \(z - x = y^2\).

Step 3 ：The surface integral can be expressed as \(\iint_{S} y^2 dS\).

Step 4 ：The differential area element \(dS\) in Cartesian coordinates is given by \(dS = dx dy\).

Step 5 ：So, the surface integral becomes \(\iint_{S} y^2 dx dy\).

Step 6 ：The limits of integration for \(x\) are from \(0\) to \(y\) and for \(y\) are from \(0\) to \(1\).

Step 7 ：So, the surface integral is \(\int_{0}^{1} \int_{0}^{y} y^2 dx dy\).

Step 8 ：Final Answer: The surface integral is set up as \(\boxed{\int_{0}^{1} \int_{0}^{y} y^2 dx dy}\).