Given a surface defined by z=g(x,y) for (x,y) in a region R of the xy plane, then the surface area is given by:

$\displaystyle \iint_R\sqrt{1+\left(\frac{\partial{g}}{\partial{x }}\right)^2+\left(\frac{\partial{g}}{\partial{y}}\ right)^2}dA$.

For your problem, R is the rectangle bound by $\displaystyle x=X_i$, $\displaystyle x=X_f$, $\displaystyle y=Y_i$, and $\displaystyle y=Y_f$; while $\displaystyle \frac{\partial{g}}{\partial{x}}=\frac{x}{2f}$ and $\displaystyle \frac{\partial{g}}{\partial{y}}=\frac{y}{2f}$, and so you have $\displaystyle S=\int_{X_i}^{X_f}\int_{Y_i}^{Y_f}\sqrt{1+\frac{x^ 2}{4f^2}+\frac{y^2}{4f^2}}\,dy\,dx$. Like you said, substitute $\displaystyle u=\frac{x}{2f}$ and $\displaystyle v=\frac{y}{2f}$. Then we get:

$\displaystyle S=4f^2\int_{X_i/2f}^{X_f/2f}\int_{Y_i/2f}^{Y_f/2f}\sqrt{1+u^2+v^2}\,dv\,du$.

Now, a table of integrals will tell you that $\displaystyle \int\sqrt{a^2+x^2}\,dx=\frac{x\sqrt{a^2+x^2}}{2}+\ frac{a^2}{2}\ln(x+\sqrt{a^2+x^2})$ (see #8

here, for example). Replacing x with v and $\displaystyle a^2$ with $\displaystyle 1+u^2$, the above applies to our inner integral, and we have:

$\displaystyle S=4f^2\int_{X_i/2f}^{X_f/2f}\left[\frac{v\sqrt{1+u^2+v^2}}{2}+\frac{1+u^2}{2}\ln(v+\ sqrt{1+u^2+v^2})\right]_{v=Y_i/2f}^{Y_f/2f}\,du$

$\displaystyle S=2f^2\int_{X_i/2f}^{X_f/2f}\left(\frac{Y_f\sqrt{1+\frac{Y_f^2}{4f^2}+u^2}-Y_i\sqrt{1+\frac{Y_i^2}{4f^2}+u^2}}{2f}+(1+u^2)\le ft(\ln\left(\frac{Y_f}{2f}+sqrt{1+\frac{Y_f^2}{4f^ 2}+u^2}\right)-\ln\left(\frac{Y_i}{2f}+sqrt{1+\frac{Y_i^2}{4f^2}+ u^2}\right)\right)\right)\,du$.

Now this is a bit more complicated, but the integrand may be broken up into four terms, each of which may be integrated. For the terms $\displaystyle \frac{Y_f}{2f}\sqrt{1+\frac{Y_f^2}{4f^2}+u^2}$ and $\displaystyle -\frac{Y_i}{2f}\sqrt{1+\frac{Y_i^2}{4f^2}+u^2}$, we can use the form $\displaystyle \int\sqrt{a^2+x^2}\,dx=\frac{x\sqrt{a^2+x^2}}{2}+\ frac{a^2}{2}\ln(x+\sqrt{a^2+x^2})$, this time with x replaced by u, and with either $\displaystyle 1+\frac{Y_f^2}{4f^2}$ or $\displaystyle 1+\frac{Y_i^2}{4f^2}$ in place of $\displaystyle a^2$.

For the terms $\displaystyle (1+u^2)\ln\left(\frac{Y_f}{2f}+sqrt{1+\frac{Y_f^2} {4f^2}+u^2}\right)$ and $\displaystyle -(1+u^2)\ln\left(\frac{Y_i}{2f}+sqrt{1+\frac{Y_i^2} {4f^2}+u^2}\right)$, well, those may take more creative substitutions, but I'm pretty sure they can be done analytically as well. Some messy algebra later, you'll have your analytic answer.

--Kevin C.