This is a difficult question that I don't understand how to answer but I tried and I hope someone could tell me if my answers are right or wrong. I managed to do a and b only. For c I need some guidance on how to answer it.
The circle with centre (r, 0, r) and radius r in the x-z-plane is rotated through an angle α as shown in the sketch above. We let K denote this rotated circle.
a) Find the parametric representation of the original circle in the x-z-plane.
$\begin{pmatrix}r+rcos\left(t\right)\\ 0\\ r+rsin\left(t\right)\end{pmatrix}$
b) Find the parametric representation of the rotated circle K.
$\begin{pmatrix}r+rcos\left(t\right)\cdot cos\left(\theta \right)\\ r+rcos\left(t\right)\cdot sin\left(\theta\right)\\r+rsin\left(t\right)\end{pmatrix}$
c) Compute the path integral$\int _c\vec{F}d\vec{x}\:with\:\vec{F}\left(\vec{x}\right)=\begin{pmatrix}x-rcos\left(\alpha \right)\\ y-rsin\left(\alpha \right)\\ z-r\end{pmatrix}\:$
I understand that I need to find the derivative of x(t)? Is that right? And then I have to insert it into Fx to get F(x(t))and then integrate it? How do I insert x(t) into F(x)?
You did very well for part a. Note that this is not the only parametrization you could have chosen.
You made a small error for part b. From part a, you indicated a set of positions that produce the circle. For part b, you essentially want to rotate these points an angle $\alpha$ about the $z$-axis. You need to have the parametrization of $K$ with position vector $\vec{p}$ as follows:
$ \vec{p}(t)= \begin{pmatrix} x(t) \\ y(t) \\ z(t) \end{pmatrix} = \begin{pmatrix} [r+r\cos(t)]\cos(\alpha) \\ [r+r\cos(t)]\sin(\alpha) \\ r+r\sin(t) \end{pmatrix} = \begin{pmatrix} r\cos(\alpha)+r\cos(t)\cos(\alpha) \\ r\sin(\alpha)+r\cos(t)\sin(\alpha) \\ r+r\sin(t) \end{pmatrix} $
This comes directly from the multiplication of the original parametrization of the position vector with a CCW rotation matrix about the $z$-axis. For part c, I think there might be an error because $x$ isn't a vector in the context of the problem. I think you meant $d\vec{x}$ as a differential path element which is $d\vec{p}$. Similarly, it should be $\vec{F}(\vec{p})$.
You need to establish the function in terms of your parameter $t$ by substituting in for $x$, $y$, and $z$ from your parametrization of $K$. Then you need to account for $d\vec{p}$ by recognizing that $d\vec{p} = \frac{d\vec{p}}{dt}dt$.
The function vector w.r.t. the parametrization is as follows:
$\vec{F}(\vec{p}) = \begin{pmatrix} x - r\cos(\alpha) \\ y - r\sin(\alpha) \\ z - r \end{pmatrix} = \begin{pmatrix} r\cos(\alpha)+r\cos(t)\cos(\alpha) - r\cos(\alpha) \\ r\sin(\alpha)+r\cos(t)\sin(\alpha) - r\sin(\alpha) \\ r+r\sin(t) - r \end{pmatrix} = \begin{pmatrix} r\cos(t)\cos(\alpha) \\ r\cos(t)\sin(\alpha) \\ r\sin(t) \end{pmatrix} $
The differential path vector is as follows:
$d\vec{p} = \frac{d\vec{p}}{dt}dt = \begin{pmatrix} -r\sin(t)\cos(\alpha) \\ -r\sin(t)\sin(\alpha) \\ r\cos(t) \end{pmatrix}dt$
Therefore, the following holds:
$\vec{F}(\vec{p}) \cdot \frac{d\vec{p}}{dt} = \begin{pmatrix} r\cos(t)\cos(\alpha) \\ r\cos(t)\sin(\alpha) \\ r\sin(t) \end{pmatrix} \cdot \begin{pmatrix} -r\sin(t)\cos(\alpha) \\ -r\sin(t)\sin(\alpha) \\ r\cos(t) \end{pmatrix} \\ = -r^2\sin(t)\cos(t)\cos^2(\alpha)-r^2\sin(t)\cos(t)\sin^2(\alpha) + r^2\sin(t)\cos(t)\\ = -r^2\sin(t)\cos(t) + r^2\sin(t)\cos(t) = 0 $
The function vector is orthogonal to the path for all $t$, so the path integral is simply $0$.