calculus and analysis – Parametric Surface Integral Boundary Code?

Use the definition of surface integral.

x(u_, v_) = u Cos(v);
y(u_, v_) = u Sin(v);
z(u_, v_) = v;
r(u_, v_) = {x(u, v), y(u, v), z(u, v)};
Integrate(
 y(u, v)*Norm@Cross(D(r(u, v), u), D(r(u, v), v)), {u, 0, 1}, {v, 
  0, π})
(* Integrate(
 y(u, v) Sqrt(# . #) &@Cross(D(r(u, v), u), D(r(u, v), v)), {u, 0, 
  1}, {v, 0, π})*)

2/3 (-1 + 2 Sqrt(2))

Reply the second question.

I believe that $y$ should be $u^2-v^2$ since if $y=u^3-v^2$ then the region is not a simple surface and the integral became so complicated.

x(u_, v_) = 2 u*v;
y(u_, v_) = u^2 - v^2;
z(u_, v_) = u^2 + v^2;
r(u_, v_) = {x(u, v), y(u, v), z(u, v)};
Integrate((x(u, v) + y(u, v) + z(u, v)) Sqrt(# . #) &@
  Cross(D(r(u, v), u), D(r(u, v), v)), {u, v} ∈ Disk())

(4 Sqrt(2) π)/3

x(u_, v_) = 2 u v;
y(u_, v_) = u^2 - v^2;
z(u_, v_) = u^2 + v^2;
r(u_, v_) = {x(u, v), y(u, v), z(u, v)};
reg = ParametricRegion({r(u, v), u^2 + v^2 <= 1}, {u, v});
Integrate(x + y + z, {x, y, z} ∈ reg)

(2 Sqrt(2) π)/3

Clear(x, y, z, u, v, r, θ);
x = 2 u*v /. {u -> r*Cos(θ), v -> r*Sin(θ)};
y = u^2 - v^2 /. {u -> r*Cos(θ), v -> r*Sin(θ)};
z = u^2 + v^2 /. {u -> r*Cos(θ), v -> r*Sin(θ)};
Integrate((x + y + z)*Sqrt(# . #) &@
  Cross(D({x, y, z}, r), D({x, y, z}, θ)), {r, 0, 
  1}, {θ, 0, 2 π})

(4 Sqrt(2) π)/3