Calculate $\oint F\cdot dr$ on the curve which is created by the intersection of plane $z=2x$ and paraboloid $z=x^2+y^2$. Suppose the direction of the curve is anti-clockwise. Let the field be $F=\langle 0,0,y\rangle$.
First the definition of $\oint F\cdot dr$ used in my class: $$ \oint F\cdot dr=\oint F\cdot\langle dx,dy,dz\rangle $$ Now according to Stokes' theorem: $$ \oint F\cdot dr=\iint \operatorname{curl}F\cdot n\cdot dS $$ where $n$ is the normal unit vector to the surface.
As far as I understand our curve is a circle with the center at $(1,0)$ and with radius $r=1$: $$ x^2+y^2=2x\\ (x-1)^2+y^2=1 $$ so $n=\frac{\langle -2,0,1\rangle}{\sqrt 5}$ Therefore: $$ \iint_D \operatorname{curl}F\cdot n\cdot dS=\iint_D \langle 1,0,0\rangle\cdot\frac{\langle -2,0,1\rangle}{\sqrt 5}dS=\\ =-\frac{2}{\sqrt 5}\iint_D dS $$ Because our curve is the circumference of the circle so $\iint_DdS=\pi r^2=\pi$.
Thus $\oint F\cdot dr=-\frac{2\pi}{\sqrt 5}$.
But the answer is incorrect as the correct solution is $-2\pi$. I assume that if I didn't normalize $n$ my answer would be correct but $n$ should be normalized as far as I understand.
What am I doing wrong?
It is true that the boundary of the surface in the plane projects a circle of radius $1$ onto the $xy$ plane. The surface on the plane itself is not a circle nor not even a disk, to my imagination it would be an ellipse. As such it's area is not $\pi$ but rather $\sqrt{5}\pi$. This factor shows up because of tilting.
We are looking at the surface $S$ with $(x-1)^2+y^2 \leq 1$ and $z=2x$. We wish to calculate,
$$\iint_{S} dS$$
Because $z=2x$ we have $z_x=2$ and $z_y=0$, hence $\sqrt{1+z_x^2+z_y^2}=\sqrt{5}$ and $dS=\sqrt{5} dA$, so:
$$\iint_{S} dS=\sqrt{5} \iint_{(x-1)^2+y^2\ \leq 1} dA$$
$$=\sqrt{5} \pi$$