Area of Surface Given in General Cartesian Form
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
\[A = \int_{xy} \frac{1}{\frac{\frac{\partial f}{\partial z}}{{\sqrt{(\frac{\partial f}{\partial x })^2 + (\frac{\partial f}{\partial y})^2 + (\frac{\partial f}{\partial z})^2}}}} dxdy= \int_{xy} \frac{\sqrt{ (\frac{\partial f}{\partial x })^2 + (\frac{\partial f}{\partial y})^2 + (\frac{\partial f}{\partial z})^2}}{\frac{\partial f}{\partial z}} dxdy \]
