The calculation of the surface area of a parametrized surface closely mirrors the calculation of the arc length of a parametrized curve. We estimated the arc length of a parametrized curve by chopping up its domain $[a,b]$ into small segments and approximating the corresponding segments of the curve as straight line segments. We could easily calculate the length of the straight line segment approximation to the curve, from which we derived the integral giving the arc length of the curve.

The procedure for calculating surface area is similar. A function $\dlsp: \R^2 \to \R^3$ (confused?) maps a region $\dlr$ in the plane onto a surface. For example, the function \begin{align*} \dlsp(\spfv,\spsv) = (\spfv\cos \spsv, \spfv\sin \spsv, \spsv). \end{align*} parametrizes a helicoid for $(\spfv,\spsv) \in \dlr = [0,1] \times [0, 2\pi]$. Similar to the arc length calculation, the first step in estimating surface area is to chop up the region $\dlr$ into a bunch of small rectangles, as illustrated below.

*A parametrized helicoid with domain chopped into small rectangles.* The function $\dlsp(\spfv,\spsv) = (\spfv\cos \spsv, \spfv\sin \spsv, \spsv)$ parametrizes a helicoid when $(\spfv,\spsv) \in \dlr$, where $\dlr$ is the rectangle $[0,1] \times [0, 2\pi]$. The region $\dlr$ is shown as the green rectangle floating above the helicoid, and it is divided into small rectangles. You can drag the green point in $\dlr$ to specify both $\spfv$ and $\spsv$. You cannot directly move the red point on the surface as it is moves with $\spfv$ and $\spsv$ to be at the point $\dlsp(\spfv,\spsv)$.

The small rectangles form a grid on $\dlr$. The rectangles are mapped by $\dlsp$ onto the helicoid, forming a grid on the surface. In fact, the grid on the surface was shown in all the helicoid applets because we just display an approximation to the true helicoid using such a grid.

The function $\dlsp$ maps each small rectangle in $\dlr$ onto
the surface. Below, you can see below how each rectangle in $\dlr$ (outlined in
green) maps onto a small part of the surface, called the
**image** of the rectangle (outlined in red).

*A parametrized helicoid with surface area elements.* The function $\dlsp(\spfv,\spsv) = (\spfv\cos \spsv, \spfv\sin \spsv, \spsv)$ parametrizes a helicoid when $(\spfv,\spsv) \in \dlr$, where $\dlr$ is the rectangle $[0,1] \times [0, 2\pi]$. The region $\dlr$ is shown as the green rectangle floating above the helicoid, and it is divided into small rectangles. As you drag the green point in $\dlr$ to specify both $\spfv$ and $\spsv$, a surrounding small rectangle is outlined in green. This rectangle is mapped into the small region of the helicoid that is outlined in red and surrounds the red point $\dlsp(\spfv,\spsv)$. The area of the red region on the helicoid depends on how $\dlsp$ stretches or shrinks the small green rectangle as it maps it on the surface.

In reality, the image of each rectangle is some “curvy rectangle” on the surface. (In the above applets, the image of each rectangle appears to have straight edges, but what is shown is only an approximation of the actual helicoid.)

The next step to calculate the surface area is to estimate the area of each of the curvy rectangles. We use exactly the same procedure we did to calculate the “area expansion factor” for a change of variables in double integrals. (In fact, the change of variable function $\cvarf: \R^2 \to \R^2$ can be viewed as a special case of the function $\dlsp: \R^2 \to \R^3$; the surface that $\cvarf$ parametrizes just happens to lie in the plane.)

Let the size of each small rectangle in $\dlr$ be $\Delta\spfv \times \Delta\spsv$. As with changing variables, we approximate each curvy rectangle as a parallelogram spanned by the vectors $\pdiff{\dlsp}{\spfv}\Delta\spfv$ and $\pdiff{\dlsp}{\spsv}\Delta\spsv$. The area of the parallelogram is simply the magnitude of the cross product of those two vectors $$\Delta A= \left\| \pdiff{\dlsp}{\spfv} \times \pdiff{\dlsp}{\spsv}\right\|\Delta\spfv \Delta\spsv.$$

The total surface area is approximated by a Riemann sum of such terms. If we let the rectangles shrink so that $\Delta\spfv$ and $\Delta\spsv$ go to zero, we would see that the total surface area is the double integral \begin{align*} A = \iint_\dlr \left\| \pdiff{\dlsp}{\spfv}(\spfv,\spsv) \times \pdiff{\dlsp}{\spsv}(\spfv,\spsv) \right\| d\spfv\,d\spsv. \end{align*}

The expression for surface area is also similar to the expression we derived for the length of a parametrized curve, \begin{align*} L=\int_a^b \| \dllp\,'(t) \| dt. \end{align*} For surface area, instead of integrating over an interval $[a,b]$, we integrate over the region $\dlr$. In the formula for path length, the expression $\| \dllp\,'(t) \|$ estimates how much $\dllp$ is stretching the interval $[a,b]$ as it maps $[a,b]$ onto the path (or, we could think of it as determining how fast a particle with position $\dllp(t)$ is moving). The expression \begin{align*} \left\| \pdiff{\dlsp}{\spfv}(\spfv,\spsv) \times \pdiff{\dlsp}{\spsv}(\spfv,\spsv) \right\| \end{align*} is analogous to $\| \dllp\,'(t) \|$. It indicates how much $\dlsp$ is stretching $\dlr$ as it maps $\dlr$ onto the surface.

You can read an example here.