Right code for solving Stoke's Theorem problem

Question

In solving Stoke's Theorem problems, I have developed a step by step method. I'm now trying to write a MMA script to help me solve these problems. But I'm a little stuck.

Here's my step by step method, which works well if tediously by hand.

So imagine this problem:

Using Stoke's Theorem, solve the surface integral $F(x,y,z) = Cos(z)i + x^2j + 2yk$ where $C$ is the intersection of the plane $z = 2-x$ and $x^2 + y^2 = 4$. The answer is 8$pi$.

I started my step by step with this code, but bogged down in how to make each vector element from the curl into the $P, Q, R$ of Step 2 and how then to multiply the partial derivatives by $P,Q,R$ in Step 5.

F = {Cos[z],  x^2, 2 y}
C = Curl[F, {x, y, z}] (*Step 1*)

g = 2 - x (*Step 3 solving for z of the curve*)
g1 = D[g, x] )(*Step 4*)
g2 = D[g, y]

Thanks for any help.

UPDATE:

Daniel was kind enough to supply some code, which looks great. But I'm trying to reconcile answers from books with how this code solves problems. So his code gives -8pi to a problem where the book says 8 pi. That could be an orientation issue. But here's an example from a UPenn math class (see 1example 1 where Daniel's code says the answer is -4pi, but the professor says "pi." Here's the code I used to evaluate Example 1. Perhaps I misunderstood the application.

f[x_, y_, z_] = {y^2 z, x z, x^2 y^2};
curl[x_, y_] = Curl[f, {x, y, z}];
surf[x_, y_] = {x, y, x^2 + y^2};
surfelem[x_, y_] = 
  Cross[D[surf[x0, y], x0], D[surf[x, y0], y0]] /. {x0 -> x, 
    y0 -> y};
reg = ImplicitRegion[x^2 + y^2 == 1, {x, y}];
Integrate[curl[x, y].surfelem[x, y], {x, y} \[Element] reg]

Answer 1

We first define the vector field and the curl (note, use lower case symbols, as uppercase are used by MMA):

f[x_,y_,z_] = {Cos[z], x^2, 2 y};
curl[x_, y_] = Curl[f[x,y,z], {x, y, z}];

Then we define the surface and the surface element, that we get by the cross product between the derivative of the surface in x and y direction:

surf[x_, y_] = {x, y, 2 - x};
surfelem[x_, y_] = 
  Cross[D[surf[x0, y], x0], D[surf[x, y0], y0]] /. {x0 -> x, y0 -> y};

The we need the region over which to integrate:

reg = ImplicitRegion[x^2 + y^2 == 4, {x, y}];

And now the integral is the dot product of the curl and the surface element integrated over the region:

Integrate[curl[x, y].surfelem[x, y], {x, y} \[Element] reg]

Answer 2

Reply the updated of the question.

We use the parametric form of the surface z=x^2 +y^2,that is

f[r_, θ_] := {r*Cos[θ], r*Sin[θ], r^2}

Assume the vector field is F[x_, y_, z_] := {y^2*z, x*z, x^2*y^2}; and then we substitute it into the Curl[F[x, y, z], {x, y, z}]

(Curl[F[x, y, z], {x, y, z}] /. Thread[{x, y, z} -> f[r, θ]]) 

All the code as below.

Clear[F,f];
F[x_, y_, z_] := {y^2*z, x*z, x^2*y^2};
f[r_, θ_] := {r*Cos[θ], r*Sin[θ], r^2};
Integrate[(Curl[F[x, y, z], {x, y, z}] /. 
    Thread[{x, y, z} -> f[r, θ]]) . 
  Cross[D[f[r, θ], r], D[f[r, θ], θ]], {r, 0, 
  1}, {θ, 0, 2 π}]

π

Reply the original of the question.

Use the same method,we can also calculate the original question.

The parametric surface of z=2-x, 0<= x^2+y^2<=4is

f[r_, θ_] := {2 r*Cos[θ], 2 r*Sin[θ], 2 - 2 r*Cos[θ]};

Clear[F, f];
F[x_, y_, z_] := {Cos[z], x^2, 2 y};
f[r_, θ_] := {2 r*Cos[θ], 2 r*Sin[θ], 2 - 2 r*Cos[θ]};
Integrate[(Curl[F[x, y, z], {x, y, z}] /. 
    Thread[{x, y, z} -> f[r, θ]]) . 
  Cross[D[f[r, θ], r], D[f[r, θ], θ]], {r, 0, 1}, {θ, 0, 2 π}]

8 π

Use x,y as parametric

Clear[f, F];
F[x_, y_, z_] := {y^2*z, x*z, x^2*y^2};
f[x_, y_] := {x, y, z} /. z -> x^2 + y^2;
Integrate[(Curl[F[x, y, z], {x, y, z}] /. {z -> x^2 + y^2}) . 
  Cross[D[f[x, y], x], D[f[x, y], y]], {x, y} ∈ 
  Disk[{0, 0}, 1]]

π

Clear[F, f];
F[x_, y_, z_] := {Cos[z], x^2, 2 y};
f[x_, y_] := {x, y, z} /. z -> 2 - x;
Integrate[(Curl[F[x, y, z], {x, y, z}] /. z -> 2 - x) . 
  Cross[D[f[x, y], x], D[f[x, y], y]], {x, y} ∈ 
  Disk[{0, 0}, 2]]

8 π