z=xy,\(x^{2}\)+\(y^{2}\)＜\(a^{2}\)
∂z/∂x=y,∂z/∂y=x
∬dxdy/\(\sqrt{\quad}\)(1+(∂z/∂x\()^{2}\)+(∂z/∂y\()^{2}\))=∬rdrdθ/\(\sqrt{\quad}\)(1+\(r^{2}\))
ここで、t=1+\(r^{2}\)とおくと、
dt=2rdr,1＜t＜1+\(a^{2}\)となり、積分は
∫[0,2π]dθ∫(\(\frac{1}{2}\))dt/\(\sqrt{\quad}\)t
=2π[t^(\(\frac{1}{2}\))]
=2π(\(\sqrt{\quad}\)(1+\(a^{2}\))-\(\sqrt{\quad}\)1)
=2π(\(\sqrt{\quad}\)(1+\(a^{2}\))-1)