I am struggling to understand the concept of an area element in a surface integral.
This is the problem.
"Consider a surface integral of the form $I = \int_{S} \boldsymbol{B} \cdot d \boldsymbol{\sigma}$ over the surface of a tetrahedron whose vertices are at the origin and at the points (1,0,0), (0,1,0), and (0,0,1), with $\boldsymbol{B}=...$" and $d\boldsymbol{\sigma} = \hat{\boldsymbol{n}}dA$.
I want to focus on a triangle area element $dA$ formed by the three coordinates (1,0,0), (0,1,0), and (0,0,1).
They go on to say the following:
"The width of the triangle at height $z$ is $\sqrt{2}(1-z)$, and a change $dz$ in $z$ produces a displacement $dz\sqrt{3/2}$."
In result, $dA=\frac{\sqrt{2}(1-z)dz\sqrt{3/2}}{2}$.
However, the way I solved it is get the area $A$ and take the derivative with respect to $z$.
In that case, I get $A=\frac{bh}{2}=\frac{(\sqrt{2}(1-z))(z\sqrt{3/2})}{2}=\frac{\sqrt{3}}{2}z(1-z)$.
Lastly, taking the derivative results in : $dA=\frac{\sqrt{3}}{2}(1-2z)dz$.
Simply put, I think it should be $dA=dbdh/2$, but the example says $dA=bdh/2$.
It seems that the base width is also changing w.r.t the height $z$, so I don't know what I am wrong conceptually.
I will really appreciate any comments.
This is an example from "Mathematical Methods for Physicists" by Arfken, Weber, and Harris. 7th Edition, Example 3.7.2 "A Surface Integral", Pg. 161-162