Integration of Vectors and improper integrals

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 $$0$$.