Surface integral problem formula question

[flyingpig]

Homework Statement

My book says proves this formula

$$\iint_S f(x,y,z) dS = \iint_D f(x,y,g(x,y)) \sqrt{\left (\frac{\partial z}{\partial x} \right )^2 + \left (\frac{\partial z}{\partial y} \right )^2 + 1 } \;dA$$

Any surface with equation z = g(x,y) can be regarded as a parametric surface with parametric equations

x = x

y = y

z = g(x,y)

r_x = i + g_x k


r_y = j + g_y k


|r_x x r_y| = $$\sqrt{\left (\frac{\partial z}{\partial x} \right )^2 + \left (\frac{\partial z}{\partial y} \right )^2 + 1 }$$

Thus


$$\iint_S f(x,y,z) dS = \iint_D f(x,y,g(x,y)) \sqrt{\left (\frac{\partial z}{\partial x} \right )^2 + \left (\frac{\partial z}{\partial y} \right )^2 + 1 } \;dA$$

Question

How do they know that every parametrization falls nicely as x = x and y = y?

 

[Mark44]

In general, the surface is w = f(x, y, z), but if we're given that z is a function of x and y (i.e., z = g(x, y)), then we have w = f(x, y, g(x, y)). And instead of integrating over some surface S in three-dimensional space, we can integrate over a two-dimensional region in the x-y plane, D. I think that's all they're saying.