To calculate the surface area of a parametrized surface, we approximated each small segment of the surface as a parallelogram spanned by $\displaystyle \pdiff{\dlsp}{\spfv}(\spfv,\spsv)\Delta \spfv$ and $\displaystyle \pdiff{\dlsp}{\spsv}(\spfv,\spsv) \Delta \spsv$. To find the area of the parallelogram, we took the magnitude of their cross product.

One feature of the cross product is that its magnitude is the area of the parallelogram spanned by the two vectors. The other feature of the cross product is that it is perpendicular to both of the vectors. It turns out that the vectors $\pdiff{\dlsp}{\spfv}(\spfv,\spsv)$ and $\pdiff{\dlsp}{\spsv}(\spfv,\spsv)$ are both tangent to the surface. Consequently, their cross product \begin{align*} \pdiff{\dlsp}{\spfv}(\spfv,\spsv) \times \pdiff{\dlsp}{\spsv}(\spfv,\spsv) \end{align*} is perpendicular to the surface and therefore is a normal vector to the surface. We frequently want a unit normal vector, meaning a normal vector with length one. To obtain a unit normal vector, we just divide by its magnitude: \begin{align*} \vc{n} = \frac{\displaystyle \pdiff{\dlsp}{\spfv}(\spfv,\spsv) \times \pdiff{\dlsp}{\spsv}(\spfv,\spsv)}{\displaystyle \left\| \pdiff{\dlsp}{\spfv}(\spfv,\spsv) \times \pdiff{\dlsp}{\spsv}(\spfv,\spsv) \right\|}. \end{align*}