<s>[INST] Evaluate $\int ydx + zdy + xdz$ using Stokes' Theorem? 
Evaluate $\int ydx + zdy + xdz$ where $C $ is intersection of $x+y=2$
  and $x^2+y^2+z^2=2(x+y) $ traversed counterclockwise as viewed from
  origin

I am using Stokes' theorem to solve this question so 
We want $\int \int curl F.N \; dS$ where $N$ is the normal unit vector to surface S, where S is a surface bounded by $C$ 
$F = \langle  y,z,x\rangle$
$curl  F  = \langle -1,-1,-1 \rangle $
I take $S $  on the plane $x+y = 2$
$\nabla (x+y) = \langle 1,1,0 \rangle  = A(say)$
Then unit normal vector $N =  \langle -1/\sqrt2,-1/\sqrt2, 0\rangle $ {Multiplied by $-1$ because we are viewing it from origin }
$curl F.N= \sqrt(2)$ 
Now intersection of $x+y=2 $ and $x^2+y^2+z^2 = 2(x+y) $ gives
$x^2 + y^2 + z^2 = 4$
To get projection onto $xy$ plane $z=0$ we get
$x^2 +y^2 =4$
Now I am stuck in finding $dS$
How do I get dS = $\sqrt{z_x^2 +z_y^2 +1}dA$  where $A:x^2 +y^2 =4$
This is because my $S:x+y=2$ has no $z$ term
 [/INST] Since $\nabla\times F\cdot \vec{N}=\sqrt{2}$, you only need to evaluate 
$$\sqrt{2}\iint_S dS$$
which is $\sqrt{2}$ multiplied by the area of the disk. (Note that the intersection is a disk, and it passes through the center of the original ball.) The original ball can be written as
$$(x-1)^2+(y-1)^2+z^2=2$$
whose center is $(1,1,0)$. The plane $x+y=2$ also passes through the center $(1,1,0)$. Hence the area of the disk is $\pi r^2=2\pi$, and the answer for the integral is $2\sqrt{2}\pi$.
Alternatively, if we project this surface to $xz$-plane, (we cannot project it to $xy$-plane since it is perpendicular to the $xy$-plane) we substituting $y=2-x$ into the other equation to obtain
$$x^2+(2-x)^2+z^2=4\implies (x-1)^2+\frac{z^2}{2}=1.$$
Now setting $y=2-x$, we have
$$\sqrt{1+y_x^2+y_z^2}=\sqrt{2}.$$
Computing 
$$\sqrt{2}\cdot\sqrt{2}\iint_A dA$$
where $A$ is region enclosed by the ellipse, we obtain $2\sqrt{2}\pi$, same as above.
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