Evaluate the integral $\int_\Gamma(y^2-z^2)dx+(z^2-x^2)dy+(x^2-y^2)dz$ I need to calculate the integral
$$\int_\Gamma(y^2-z^2)dx+(z^2-x^2)dy+(x^2-y^2)dz$$
being $\Gamma=S_1\cap S_2$, given:


*

*$S_1=\{(x,y,z)\in\mathbb{R}:2x+2y+z=3\}$

*$S_2=\{(x,y,z)\in\mathbb{R}:z=9-x^2-y^2\}$
In my problem i'm asked to solve it using both direct integration and the Stokes Theorem. I've started trying the Stokes part, calculating the rotational of $F$:
$$F=((y^2-z^2),(z^2-x^2),(x^2-y^2))\Longrightarrow \text{rot}(F)=-2(y+z,x+z,x+y)$$
Now, I know that the normal vector $N$ is $(2,2,1)$ (because $S_1\subset\Gamma$ and that's $S_1$'s normal vector at any point).
So now I using Stokes I have that (after simplifying)
$$\int_\Gamma(y^2-z^2)dx+(z^2-x^2)dy+(x^2-y^2)dz=\frac{-2}{3}\iint_S(3x+3y+4z)d\sigma$$
I'm stucked here. I don't get what $S$ am I supposed to use in the double integral (I guess  I must do some variable change).
For the direct integration part, i don't know where to start. I will thank any help.
Edit: I need two different solutions, one using direct integration, and another using Stokes Theorem.

Direct approach:
Let's study the intersection by equating the $z$:
$$3-2x-2y=9-x^2-y^2$$
$$x^2+y^2-2x-2y=9-3=6$$
$$(x-1)^2+(y-1)^2=(2\sqrt2)^2$$
Hence I would use the parametrization $x=1+2\sqrt2\cos \theta, y = 1+2\sqrt2\sin \theta, z=-1-4\sqrt2\sin \theta - 4\sqrt2 \cos \theta.$
$$\frac{dx}{d\theta} = -2\sqrt2 \sin \theta, \frac{dy}{d\theta}= 2\sqrt2\cos \theta, \frac{dz}{d\theta}= -4 \sqrt2\cos \theta + 4\sqrt2\sin \theta.$$
\begin{align}
&\int_{\Gamma} (y^2-z^2) \, dx + (z^2-x^2) \, dy + (x^2-y^2) \, dz \\
&= \int_0^{2\pi}y^2\left( \frac{dx}{d\theta} - \frac{dz}{d\theta} \right) +z^2\left( \frac{dy}{d\theta} - \frac{dx}{d\theta} \right) + x^2 \left( \frac{dz}{d\theta} - \frac{dy}{d\theta} \right)\, d\theta \\
&= \int_0^{2\pi} \left( 1+4\sqrt2\sin \theta + 8\sin^2 \theta\right)\left( -6\sqrt2\sin \theta + 4\sqrt2\cos \theta\right) \\
&+\left( 1+32 \sin^2 \theta + 32 \cos^2 \theta + 8\sqrt2 \sin \theta + 8\sqrt2 \cos \theta + 32 \sin \theta \cos \theta\right)\left( 2\sqrt2\cos \theta + 2\sqrt2\sin \theta\right) \\
&+\left( 1+4\sqrt2\cos \theta + 8\cos^2 \theta\right)\left( -6\sqrt2\cos \theta + 4\sqrt2\sin\theta\right) \, d\theta \\
&= \int_0^{2\pi} -16 \sin^2 \theta -16 \cos^2 \theta \, d\theta \\
&=\int_0^{2\pi} -16  \, d\theta \\
&= -32 \pi
\end{align}
Remark: There is no need to work out the details in the middle as we know that for $\{\sin^3 \theta, \cos^3 \theta, \sin^2 \theta \cos \theta, \sin \theta \cos^2 \theta, \sin \theta \cos \theta, \sin \theta, \cos \theta\}$, they vanish when we integrate form $0$ to $2 \pi$. We just have to keep track of $\sin^2 \theta$ and $\cos^2 \theta$.

Alternatively, we can use the Stoke's theorem,
$$curl(F)=-2(y+z, x+z, x+y)$$
The normal vector is $(2,2,1)$.
The dot product is $-2(3x+3y+4z)=-2(3x+3y+4(3-2x-2y))=-2(12-5x-5y)$.
Let's integrate using the parametrization $x=1+r\cos \theta, y = 1+r \sin \theta, 0 \le \theta \le 2\pi, 0 \le r \le 2\sqrt2$:
\begin{align}
&\int_{\Gamma} (y^2-z^2) \, dx + (z^2-x^2) \, dy + (x^2-y^2) \, dz \\
&= -2\int_0^{2\sqrt2} \int_0^{2\pi}  (12-5(1+r\cos \theta)-5(1+r\sin \theta))r\,  d\theta \,dr \\
&= -2 \int_0^{2\sqrt2} \int_0^{2\pi}  (2r-5r^2\cos \theta-5r^2\sin \theta)\,  d\theta \,dr \\
&= -2 \int_0^{2\sqrt2} \int_0^{2\pi}  (2r)\,  d\theta \,dr \\
&= -4\pi \int_0^{2\sqrt2} 2r \, dr \\
&= -32 \pi
\end{align}
