Surface to choose for the Stokes' theorem for intersection of sphere and plane. 
Use Stoke's theorem to evaluate $\int_C[y\;dx+y^2\;dy+(x+2z)\;dz]$ where $C$ is the curve of intersection of the
      sphere $x^2+y^2+z^2=a^2$ and the plane $y+z=a$, oriented counterclockwise as viewed from above.

I have found that the intersection of plane and the sphere is an ellipse
$$x^2+2\bigg(y-\frac{a}{2}\bigg)^2=\frac{a^2}{2}$$
What surface should I choose to minimize the calculations? I think this ellipse is a circle in the given plane so that might be a better choice. But how do I show that?
 A: First calculate the curl of the given vector field, then see if it has a nice form in one particular surface.
The vector field is $F=(y,y^2,x+2z)$, so $\nabla\times F = (0,-1,-1)$. Since $\nabla\times F$ is constant, integrating across a plane is the best choice, since if $S$ is the surface bounded by $C$ in the plane, and $n$ is the normal vector to the plane, then $\iint_S \nabla\times F \cdot dS=\iint_S (0,-1,-1)\cdot n\,dA=A((0,-1,-1)\cdot n)$. Since the plane is $y+z=a$, $n=\frac{1}{\sqrt{2}}(0,1,1)$, so $(0,-1,-1)\cdot n = -\sqrt{2}$. 
I'm not quite sure what you mean by the intersection of the plane and the sphere being an ellipse. The intersection of a plane and a 
sphere should always be a circle (or a point, or empty). 
Proof: we can assume the sphere is centered at the origin. Then the plane is given by $n\cdot x = d$, where $n$ is the unit normal to the plane, and $d$ is a constant, and $x$ is the variable. Let $r$ be the radius of the sphere. Then I claim that the intersection of the plane and the sphere will be a circle in the plane $n\cdot x=d$ centered at $dn$ of radius $\sqrt{r^2-d^2}$. This follows, since if $x\cdot x = r^2$, and $n\cdot x = d$, then $(x-dn)\cdot (x-dn)=x\cdot x -2dx\cdot n +d^2 n\cdot n = r^2-d^2$, so all points in the intersection lie in this circle, and conversely, if $n\cdot x=d$, and $(x-dn\cdot x-dn)=r^2-d^2$, then expanding as above and solving for $x\cdot x$, we get $x\cdot x=r^2$, so $x$ is on the sphere as well.
Now we can solve for the area of the intersection of the plane with the sphere. It's $\pi(r^2-d^2)$ (using the variables above), so we just need to find $r$ and $d$ in our case. It turns out that they are $r=a$, $d=\frac{a}{\sqrt{2}}$, so the area is $\pi\frac{a^2}{2}$.
Thus the total flux is $-\sqrt{2}\pi\frac{a^2}{2}=\frac{-\pi a^2}{\sqrt{2}}$, so $$\int_C F\cdot dr=-\frac{\pi a^2}{\sqrt{2}}.$$
