Find the area of the surface z = 2/3(x^(3/2) + y^(3/2))
0 <= x <=1 , 0 <= y <=1
Well i know dz/dx = x^1/2 and dz/dy = y^1/2.
Then putting it into the formula sqrt(1 + (dz/dx)^2 + (dz/dy)^2) dA
would be double integral of sqrt(1 + x + y) dA
Then I think it would be double integrals both from 0 to 1 sqrt(1 + cos(theta) + sin(theta)) r dr dtheta
Is this correct so far? I think I did something wrong since that gets really hard to integrate.