Parametric Surface Integral Boundary Code?

Question

I am having trouble using MMA to calculate 2 parametrized surface integrals. I'd be grateful for any fixes people can offer.

I usually calculate surface integrals with the following code, which works, with me inputting the parameters for x, y, and z, the region in line 5, and the integrand in line 6:

x[u_, v_] := u + v;

y[u_, v_] := u - v;

z[u_, v_] := 1 + 2 u + v;

r[u_, v_] := {x[u, v], y[u, v], z[u, v]};

reg = ParametricRegion[r[u, v], {{u, 0, 2}, {v, 0, 1}}];

Integrate[x + y + z, {x, y, z} \[Element] reg]

But on the following problem this code unexpectedly doesn't work!

Integrand is y, with parameters as shown in the problem below and u from 0 to 1 and v from 0 to Pi.

Clear[u, v, x, y, z]
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]};
reg = ParametricRegion[r[u, v], {{u, 0, 1}, {v, 0, Pi}}];
NIntegrate[y, {x, y, z} \[Element] reg]

I have hand calculated this problem, and the correct answer (verified with teacher) is $2 /3 (2 Sqrt(2) - 1)$. I have tried Nintegrate and Integrate, but the latter returns gibberish and the former the wrong answer.

At the same time, this problem befuddles me. Using the exact same code as above, the usually works, I have a region where u and v don't go between numbers but are represented by $u^2 + v^2 <=1$. On my line where I enter the parametric region, I've always used bounded intervals, not an equation. So this code messes up. I would love help on how to rephrase the region for problems like this.

x[u_, v_] := 2 u v;
y[u_, v_] := u^3 - 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} \[Element] reg]

Answer 1

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

x = 2 u*v;
y = u^2 - v^2;
z = u^2 + v^2;
z^2 == x^2 + y^2 // Simplify
(* True *)
Clear[x, y, z, reg]
reg = ImplicitRegion[
   z == Sqrt[x^2 + y^2 ] && 0 <= z <= 1, {x, y, z}];
Integrate[x + y + z, {x, y, z} ∈ reg]

(2 Sqrt[2] π)/3