Transformation of DS and DXDY in surface integral of area

As shown in the figure , Hypothetical surface $z=f(x,y)$ On a small piece $\Delta S$ stay xoy The projection on the axis is rectangular ABCD, Because the block is small enough , You can think of it as a parallelogram . be $\Delta S$ And $\Delta x\Delta y$ The relation of can be regarded as parallelogram EFGH Area and rectangle ABCD The relationship between areas .

rectangular $AD$ The length of is $\Delta x$（x Amount of change ）,$AB$ The length of is $\Delta y$（y Amount of change ）, be
$$S_{ABCD}=\Delta x\Delta y$$

$E$ The coordinates of the points are $(x,y,f(x,y))$,$F$ The coordinates of the points are $(x,y+\Delta y,f(x,y+\Delta y))$, among $f(x,y+\Delta y)$ Linear approximation can be used , namely $f(x,y+\Delta y)\approx f(x,y)+\Delta y\cdot f_y^{\prime }$, be $F$ The coordinates of the points are $(x,y+\Delta y,f(x,y)+\Delta y\cdot f_y^{\prime })$, The same can be $H$ The coordinates of the points are $(x+\Delta x,y,f(x,y)+\Delta x\cdot f_x^{\prime })$, be
$$\overrightarrow{EF}=\left(0,\Delta y,\Delta y\cdot f_y^{\prime }\right)$$
$$\overrightarrow{EH}=\left(\Delta x,0,\Delta x\cdot f_x^{\prime }\right)$$
$$\overrightarrow{EF}\times \overrightarrow{EH}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 0& \Delta y& \Delta y\cdot f_y^{\prime }\\ \Delta x& 0& \Delta x\cdot f_x^{\prime }\end{array}\right|=\left(\Delta x\Delta y\cdot f_x^{\prime },\Delta x\Delta y\cdot f_y^{\prime },-\Delta x\Delta y\right)=\left(f_x^{\prime },f_y^{\prime },-1\right)\Delta x\Delta y$$
$$\Delta S=\left|\overrightarrow{EF}\times \overrightarrow{EH}\right|=\sqrt{{\left(f_x^{\prime }\right)}^2+{\left(f_y^{\prime }\right)}^2+1}\Delta x\Delta y$$
Take infinitesimal
$$dS=\sqrt{{\left(f_x^{\prime }\right)}^2+{\left(f_y^{\prime }\right)}^2+1}dxdy$$