Show with Stokes that $\oint_C (y\mathbf{i}+z\mathbf{j}+x\mathbf{k})\bullet d\mathbf{r}=\sqrt{3}\pi a^2$ when $C$ is intersection of $x^2+y^2+z^2=a^2$ and $x+y+z=0$. My work: $$z=g(x,y)=-x-y$$ $$\mathbf{N}=\frac{-\frac{\partial g}{\partial x}\mathbf{i}-\frac{\partial g}{\partial y}\mathbf{j}+\mathbf{k}}{\sqrt{1+\left(\frac{\partial g}{\partial x}\right)^2+\left(\frac{\partial g}{\partial y}\right)^2}}=\pm\frac{1}{\sqrt 3}(\mathbf{i}+\mathbf{j}+\mathbf{k})$$ $$dS=\sqrt{1+\left(\frac{\partial g}{\partial x}\right)^2+\left(\frac{\partial g}{\partial y}\right)^2}dA=\pm\sqrt 3 dA$$ So stokes says that: $$\oint_C (y\mathbf{i}+z\mathbf{j}+x\mathbf{k})\bullet d\mathbf{r}=\iint_R \nabla \times \mathbf{F}\bullet\mathbf{N}dS$$ $$\begin{align} \iint_R \nabla \times \mathbf{F}\bullet\mathbf{N}dS&=\iint_R (-\mathbf{i}-\mathbf{j}-\mathbf{k})\bullet(\pm\frac{1}{\sqrt 3}(\mathbf{i}+\mathbf{j}+\mathbf{k}))\pm\sqrt{3}dA \\ &=\pm\sqrt{3} \iint_R \pm\frac{3}{\sqrt{3}} dA \\ &=3 \iint_R 1 dA \\ &=3\int_0^{2\pi}d\phi\int_0^{a}rdr=3\pi r^2\end{align}$$
Soo... I got it wrong. Can someone explain what my mistake was?
Your double integral is over a circular region, but the intersection of the sphere and plane given is actually $x^2+xy+y^2=\frac{a^2}{2}$, an ellipse.
Edited for more detail:
Note that $\iint_R\,dA$ is exactly the area enclosed by this ellipse, so we just need to find this area and multiply it by 3.
One way to do this would be continue using calculus and convert to polar coordinates: $$r^2+r^2\cos\theta\sin\theta=\frac{a^2}{2}$$ $$r^2=\frac{a^2}{2(1+\cos\theta\sin\theta)}$$
This leads to a pretty ugly integral which I don't really want to think about right now...
Another way would be to use some linear algebra. Note that $2x^2+2xy+2y^2$ corresponds to the symmetric matrix $\begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}$. If you orthogonally diagonalize this matrix, you should get eigenvalues of 1 and 3 and corresponding orthonormal eigenvectors of $\begin{pmatrix} \frac{1}{\sqrt 2} \\ -\frac{1}{\sqrt 2}\end{pmatrix}$ and $\begin{pmatrix}\frac{1}{\sqrt 2} \\ \frac{1}{\sqrt 2}\end{pmatrix}$.
What does this tell us? We can rewrite the equation of the ellipse as $$\frac{\left(\frac{1}{\sqrt 2}x-\frac{1}{\sqrt 2}y\right)^2}{a^2}+\frac{\left(\frac{1}{\sqrt 2}x+\frac{1}{\sqrt 2}y\right)^2}{\frac{a^2}{3}}=1$$ So this ellipse has semimajor axis of length $a$ and semiminor axis of length $\frac{a}{\sqrt 3}$, and it encloses an area of $\frac{\pi a^2}{\sqrt 3}$.
Multiplying this by the 3 from above, we get the desired result.