Suppose a surface, S, is given implicitly by an equation of the form F(x; y; z) = 0.
Thus, z is dened implicitly as a function of x and y, say z = f(x; y) for (x; y)
in a region R. If Fz 6= 0, one can show that fx =

Question

Suppose a surface, S, is given implicitly by an equation of the form F(x; y; z) = 0.
Thus, z is dened implicitly as a function of x and y, say z = f(x; y) for (x; y)
in a region R. If Fz 6= 0, one can show that fx =

anonymous · Answer

$$A = \int\limits_{}^{}\int\limits_{}^{R}\sqrt{(Fx^2 + Fy^2 + Fz^2)} / |Fz|$$

anonymous · Answer

oops, i meant: $$\int\limits_{}^{}\int\limits_{R}^{}$$

jamesj · Answer

what do you mean by this: "If Fz 6= 0"

anonymous · Answer

sorry, coppied and pasted and didn't read. $$Fz \neq 0$$

jamesj · Answer

First of all, if you had a good ol' smooth function z = f(x,y), what's the integral expression for the surface area of the graph of z over a domain in the xy-plane, R?

anonymous · Answer

$$A = \int\limits_{}\int\limits_{R}^{}\sqrt{1 + f_x^2 + f_y^2}$$

jamesj · Answer

Right; so you need to show the two integrands are equal.  Check again carefully your other hypotheses because they don't look quite right to me.