# Obtaining the surface area element geometrically

My text book derives a parametric formula for the surface area element $dS$ (which I follow and can understand). However it then goes on to remark that the surface area element can also be obtained geometrically as $$dS=\bigg|\frac{1}{\cos\theta}\bigg|dxdy=\frac{|\textbf{n}|}{|\textbf{n} \bullet\textbf{k}|}dxdy$$ where $\cos\theta$ is the angle between the normal vector n to the surface, and the unit vector k in the direction of the positive z-axis.

I am confuse with the $\bigg|\frac{1}{\cos\theta}\bigg|$ part? I thought that $\cos\theta$ might come from $\textbf{n} \bullet\textbf{k}=|\textbf{n}||\textbf{k}|\cos\theta$, where $|\textbf{n}|=|\textbf{k}|=1$ but I think this assumption is wrong as $\textbf{n}$ is not a unit vector nor does it explain why the fraction $\frac{1}{\cos\theta}$?

• That formula holds for the dot product whether or not they are unit vectors. – Paul Jul 13 '17 at 11:17
• Sorry I still don't follow. Appreciate it if you could possibly expand your explanation. – Eiraus Jul 13 '17 at 11:31
• You have $|\mathbf n|$ in the numerator on the right-hand side, which takes care of $\mathbf n$ not necessarily being a unit vector. Substitute your expansion of $\mathbf n\cdot\mathbf k$ into this expression and simplify. – amd Jul 13 '17 at 20:58