The base of a fence perpendicular to plane $xy$ is located on the line $\sigma(t)=(\cos t,\sin t,0),t\in [0,\pi]$, and the height of the fence at the point $\sigma (t)$ is $t^2$. Calculate the surface of the fence.
I have $f(\sigma (t))=t^2$ and $\sigma'(t)=(-\sin t,\cos t,0)$
Then $|\sigma'(t)|=\sqrt{(-\sin t)^2+(\cos t)^2+0^2}=1$
Hence, what I am looking for is $\int_0^{\pi} f(\sigma (t))|\sigma'(t)|dt=\int_0^{\pi} t^2dt$
Is the setup of the problem's solution correct?
Yes, your setup is correct. Note that the base of your fence is a circle, but the height of the fence is not periodic: at $0$, the height of the fence has a jump from 0 to $(2\pi)^2$, which does not make it a very realistic example.
After the edit, this last remark is no longer true: the base is now just a half-circle.