Integration using Stokes theorem

Question

I do not fully understand how to use Stokes theorem to compute the following integral could someone show me in this example.

$$\int_C (x+2y)dx+(2z+2x)dy+(z+y)dz\,$$ $C$ is intersection of sphere $x^2 +y^2 +z^2 =1$ and plane $y=z$

Answer 1

Stokes theorem says $\oint f(r(t))\cdot dr = \iint \nabla\times f(x,y,z) dS$

Where S is any surface bounded by that contour.

$\nabla\times f(x,y,z) = (-1,0,0)$

Why don't we use disk on the plane y=z as our surface. The normal to the plane is $(0,1,-1)$

and we can stop right there, as $\nabla\times f(x,y,z) \cdot n = 0$

Answer 2

Using differential 1-form

$$\omega = \left( {x + 2y} \right)dx + \left( {2z + 2x} \right)dy + \left( {z + y} \right)dz$$

we get for $z = y$

$$\omega = \left( {x + 2y} \right)dx + \left( {2x + 4y} \right)dy$$

For exterior derivative, that is applying operator $d$, it follows

$$d\omega = 2dy \wedge dx + 2dx \wedge dy$$

$$d\omega = - 2dx \wedge dy + 2dx \wedge dy = 0$$.

But know, from Stoke's theorem

$$\int\limits_{\partial S} \omega = \int\limits_S {d\omega } = 0$$