Calculate
$$ \int_\ell \vec{F}(x) \cdot\vec{T}(x)$$
Where $\vec{F} = (y^2, x, z^2)$ and $\ell : \vec{v}(t) = (\cos t, \sin t, 1);~~ 0 \leq t \leq 2\pi$
I'm not too deep into vector calculus so I'm not sure how to tackle this integral. I'm not sure if I have to use Stokes's or Gauss's Theorem. Tips are very welcome. Thanks.
It appears that the question is written with really confusing notation. This looks like a straightforward line integral where $\vec{F}(x)$ is some field and $\vec{T}(x)$ is the derivative of the curve $\vec{v}(t)$. Honestly, a better more consistent way to write this would be to write:
$$ \int_{\mathscr{l}} \vec{F}\left (\vec{v}(t)\right ) \cdot \vec{v}'(t) dt $$
where you are taking the dot product of the field with the tangent direction, but also evaluating the field at the particular point of contact along the trajectory $\vec{v}(t)$. Computing this isn't hard: \begin{eqnarray*} \vec{v}(t) & = & \left (\cos t, \sin t, 1\right ) \\ \vec{v}'(t) & = & \left (-\sin t, \cos t, 0\right ) \\ \vec{F}\left (\vec{v}(t)\right ) & = & \left (\sin^2t, \cos t, 1 \right ). \end{eqnarray*}
Now computing the integral we get:
\begin{eqnarray*} \int_{\mathscr{l}} \vec{F}\left (\vec{v}(t)\right ) \cdot \vec{v}'(t) dt & = & \int_0^{2\pi} \cos^2(t) - \sin^3(t) dt \\ & = & \int_0^{2\pi} \cos^2(t)dt - \int_0^{2\pi}\sin^3(t)dt \\ & = & \frac{1}{2}\int_0^{2\pi} \left (1 + \cos(2t)\right )dt - \int_0^{2\pi} \sin (t)\left (1 - \cos^2(t)\right )dt \\ & = & \pi + \left . \left(\cos t - \frac{1}{3}\cos^3t\right ) \right |_0^{2\pi} \\ & = & \pi. \end{eqnarray*}