## Area of Surface Given in General Cartesian Form

Theorem
The general equation for the surface area of a surface given in Cartesian coordinates
$x,y,z$
, so for a surface with equation of the form
$f(x,y,z)=0$
the surface area is

$A = \int_{xy} \frac{ \sqrt{(\frac{\partial f}{\partial x })^2 + (\frac{\partial f}{\partial y})^2 + (\frac{\partial f}{\partial z})^2}}{\frac{\partial f}{\partial z}} dxdy$

Proof
If
$\mathbf{n}$
is the unit normal at the centre of surface element
$d \mathbf{S}$
then
$\frac{d \mathbf{S}}{dxdy} = \frac{\mathbf{n}}{| \mathbf{n} \cdot \mathbf{k}} \rightarrow d \mathbf{S} = \frac{\mathbf{n}}{| \mathbf{n} \cdot \mathbf{k}} dxdy \rightarrow dS = \frac{ |\mathbf{n} |}{| \mathbf{n} \cdot \mathbf{k}} dxdy = \frac{1}{| \mathbf{n} \cdot \mathbf{k}} dxdy$

We can take
$\mathbf{n} = \frac{\frac{df}{dx} \mathbf{i} + \frac{df}{dy} \mathbf{j} + \frac{df}{dz} \mathbf{k} }{\sqrt{(\frac{\partial f}{\partial x })^2 + (\frac{\partial f}{\partial y})^2 + (\frac{\partial f}{\partial z})^2}}$

Then
$\mathbf{n} \cdot \mathbf{k} = \frac{\frac{\partial f}{\partial z}}{{\sqrt{(\frac{\partial f}{\partial x })^2 + (\frac{\partial f}{\partial y})^2 + (\frac{\partial f}{\partial z})^2}}}$

Hence
$A = \int_{xy} \frac{1}{\frac{\frac{\partial f}{\partial z}}{{\sqrt{(\frac{\partial f}{\partial x })^2 + (\frac{\partial f}{\partial y})^2 + (\frac{\partial f}{\partial z})^2}}}} dxdy= \int_{xy} \frac{\sqrt{ (\frac{\partial f}{\partial x })^2 + (\frac{\partial f}{\partial y})^2 + (\frac{\partial f}{\partial z})^2}}{\frac{\partial f}{\partial z}} dxdy$