Originally Posted by

**danny arrigo** I think what you're refering to is the following.

If your surface $\displaystyle S$ is given parametrically as

$\displaystyle \vec{r} (p,q) = <x(p,q),y(p,q),z(p,q)>$ for $\displaystyle a < p < b,\;\;\;c < q < d$

then the surface area of S is given by

$\displaystyle \int_c^d \int_a^b || \vec{r}_u \times \vec{r}_v || dp\,dq$

where

$\displaystyle \vec{r}_p = < x_p, y_p, z_p>,\;\;\;\vec{r}_q = < x_q, y_q, z_q>,$

I think in the beginning it's easier to use cartesian coordinates and then switch to polar (if necessary) then trying to use this formula in general. If so, then

$\displaystyle \vec{r} = <x, y, f(x,y)>$ (here $\displaystyle p = x, \;\;q = y$)

then

$\displaystyle \vec{r}_x = <1, 0, f_x>$ and $\displaystyle \vec{r}_y = <0,1, f_y>$

so

$\displaystyle || \vec{r}_x \times \vec{r}_y|| = \sqrt{1+f_x^2+f_y^2}$

Hence the formula I gave earlier. Are you familar with this

$\displaystyle \iint_R \sqrt{1+f_x^2+f_y^2} \,dx\,dy$?