Evaluate a curve integral Evalute the integral $$\oint_C ((x+1)^2+(y-2)^2 \rm ds$$
C is the intersection of surface $x^2+y^2+z^2=1$ and plane $x+y+z=1$.
I know this is the first kind line integral. But I never meet this form of the line's equation.  It is expressed as intersection of two equation. Confused about this king integral...
 A: Hint: $z=1-x-y$ and $x^2 + y^2 + (1-x-y)^2=1$
A: Hint: Perhaps this works. After dropping $z$ from two equations we get
$$y^2+(x-1)y+(x^2-x)=0$$
from $b^2-4ac=3\left(\dfrac49-(x-\dfrac13)^2\right)$, I think the substitution
$$x=\dfrac13+\dfrac23\sin t$$
$$y=\dfrac13\left(1-\sin t\pm\sqrt{3}\cos t\right)$$
works! Details should be considered preciously.
A: The first surface in question is a sphere. The second is a plane. We know that intersection of a plane and a sphere is a circle. The most helpful part is the symmetry.
A few was to approach this:


*

*Eliminate $z$ via expression from the plane equation. You get an ellipse (which by symmetry you know it's at 45° in the $xy$ plane) which you can parameterize in polar coordinates, or by changing variables to $u=x+y$, $v=x-y$ to align it to the axes and use either polar/cylindrical (better) or cartesian coordinates (worse, as you need to split it in two).

*The normal of the circle is in the direction (1,1,1). Rotate the (x,y,z) coordinates to bring the normal into the (1,1,1) direction. Then, the curve is a simple centered circle in the $xy$ plane which you can easily manage with polar coordinates.

*Brute force it by going directly into spherical coordinates. This way, the first equation is immediately satisfied. The second equation becomes a rule for $\theta$ with relation to $\phi$ (somewhat ugly, but survivable).
