In integrating $(x,y,z)$ around the circle centred at $ (0,0,1)$ passing through $(0,0,2)$ and $(0,1,1)$, the solution I'm seeing starts by saying:
It's a circle of radius $1$ centered at $(0,0,2)$, lying in the $yz$ plane, described by $x=0$ and $y^2+(z-1)^2=1$
(It then prametrises it and concludes that the answer is $0$). The problem states that the circle is centred at $(0,0,1)$ and the solution starts by saying that it's centred at $(0,0,2)$. So which is it?
How I tried to solve it:
Let $\displaystyle I= \int_{C} (x,y,z)\cdot d\overline{r}$. The circle can be described by $x^2+y^2+(z-1)^2=1$. I parametrised it via: $x= \cos{t},~y=\sin{t},~z=t$ for $t \in [0, 2\pi)$, getting the apparently wrong answer of $I = \int_0^{2\pi}t d{t} = 2\pi^2.$
The correct equation for the circle is $y^2+(z-1)^2=1$ indeed by the givens the circle can't be centered in (0,0,2) since the distances with $(0,0,1)$ and $(0,1,1)$ are different.