5. This is a typical problem to solve conditional extremum, and Lagrange multiplier method can be used.

Let F(x, y, λ) = xy+λ (x+y-2)

F'x=y+λ=0

F'y=x+λ=0

F'λ=x+y-2=0

X=y= 1，λ =- 1。

So z = xy max =/kloc-0 /×1=1.

6.∫∫D dσ means integral area =πr? =4π

∫D dσ=∫(0，2π) dθ∫(0，2) rdr

=2π∫(0，2) rdr

=πr？ |(0,2)=4π