Let $\Gamma :X=\gamma(t),a\leq t\leq b$ be a rectifiable parameterized curve in the $(x,z)$-plane of $R^3$, which means $\gamma:[a,b]\to R^3$ is a $C^1$-mapping with $\gamma(t)=(x(t),0,z(t))^T$ and $\dot{\gamma}(t)\neq 0$ for all $t$.
a) Show that the set $S$, which arises from the rotation of $\Gamma$ around the z-axis in $R^3$, is a rectifiable parameterized surface. Determine the parameterization of S.
b) Show that the area is
$A(S)=2\pi\int_a^bx(t)\sqrt{\dot{x}(t)^2+\dot{z}(t)^2}dt$.
c) Determine the area of the $Torus$
$T:X=F(u,v)=\begin{pmatrix} (a+bcos(v))cos(u)\\(a+bcos(v))sin(u)\\bsin(v)\end{pmatrix}$
with $a>b>0,~ 0\leq u,~v\leq 2\pi$.
To a): I can imagine that S is a rectifiable parameterized surface, since $\Gamma$ is one and the rotation around the z-axis doesn't really change $\gamma(t)$ that much, asidesthat $x(t)$ is different for the same t. But I don't know how to show that mathematically, especially how to determine a paramterization of S.
To b): I assume that the teacher talks about the area in this case, because the original sheet only said 'Show that $A(S)=...$'.
I looked up the general formula for the area under a parametric curve, which is $A=\int_a^bF(f(t))f'(t)dt$ and I can't seem to get to the desired equation from that.
To c): I was thinking of using the equation given in b), but I am unsure about the correctness of the equation in this case because, for one the y-component is not zero and two, it depends on u and v. But I guess I would need the chain rule in this case.
For a), it is sufficent to observe that a rotation along the z-axe of angle $\theta$, is given by $$\begin{pmatrix}x'\\ y'\\ z'\end{pmatrix}=\begin{pmatrix}\cos{\theta} & \sin{\theta} & 0\\ -\sin{\theta} & \cos{\theta}& 0\\ 0& 0& 1\end{pmatrix}\begin{pmatrix}x\\ y\\ z\end{pmatrix}.$$ Soo if you consider that rotation as an application $G_{\theta}(x)$, for $x\in\mathbb{R}^{3}$, then it´s easy to see that you surface parametrization is given by $$\phi(u,t)=G_{u}(\gamma(t))=(x(t)\cos{u},x(t)\sin{u},z(t))\ \ u\in[0,2\pi].$$
For b) you have to see what is the metric asociated to this surface, or to see which is the first fundamental form of $S$. Soo you have to consider the tangent vectors $$\phi_{t}(u,t)=(x'(t)\cos{u},x'(t)\sin{u},z'(t)),$$ $$\phi_{u}(u,t)=(-x(t)\sin{u},x(t)\cos{u},z(t)),$$ and from this you have that your metric $g$ is given by $$g(u,t)=\begin{pmatrix}E & F\\ F& G\end{pmatrix},$$ where $E=\phi_{t}\cdot\phi_{t}=(x')^2+(z')^2$, $F=\phi_{t}\cdot\phi_{u}=0$ and $G=\phi_{u}\cdot\phi_{u}=x^{2}$. Soo then it is easy to see that the area elemnt of the $S$ surface is given by $det(g)$, and the are is given by $$A(S)=\int_{a}^{b}\int_{0}^{2\pi}|det(S)|dudt=2\pi\int_{a}^{b}x(t)\sqrt{(x')^2+(z')^2}dt.$$ And from these it´s easy to obtain c).