Calculate the integral $\int_{\gamma}\omega$ on the curve Calculate the integral $\int_{\gamma}\omega$ on the curve $\gamma(t)=(t\cos(2\pi t), t\sin(2\pi t), t^2)$, $t\in[0,1]$ that is contained in the paraboloid S such that:
$S=$ {$(x,y,z)$ | $z=x^2 + y^2$}
$\omega = ydx+xdy+(x^2 +y^2)dz$
The solution is:
$$t=0(start), t=1 (end) => \gamma(0)=(0,0,0), \gamma(1)=(1,0,1)$$
$$\int_{\gamma}\omega = f(end)-f(start)$$
$f(x,y,z)$ such that:$$\omega = f_1dx_1+f_2dx_2+f_3dx_3$$ $$=> f=xy+\frac{z^2}{2}$$ $$=> \int_\gamma \omega = \frac{1}{2}$$
 A: I think instead of going into notions of $1$-forms and stuff associated with differential geometry, it is better to define the line integral for the op using vector calculus and multivariable calculus because I think the op is confused by the "$dx,dy,dz$" .(For reference see Murray R Spiegel Vector Analysis of the Schaums Outline series). Here's a link.
Given a vector valued function
$\vec{f}:\mathbb{R}^{3}\to\mathbb{R}^{3}$ such that
$\vec{f}(x,y,z)=f_{1}\hat{i}+f_{2}\hat{j}+f_{3}\hat{k}$
$f_{1},f_{2},f_{3}$ are scalar functions of $(x,y,z)$
The line integral over a curve $\gamma$ of the function is defined as:-
$$\int_{\gamma}\vec{f}\cdot d\vec{r}=\int_{\gamma}(f_{1}\,dx+f_{2}\,dy+f_{3}\,dz)$$
Define:-
$\omega=y\hat{i}+x\hat{j}+(x^{2}+y^{2})\hat{k}$
Now on the curve $\gamma$ given, $x=t\cos(2\pi t)$ so $dx=\cos(2\pi t)-2\pi t\sin(2\pi t)$
Do the same for $dy$ and $dz$. Essentially since the curve is parametrized by $t$ we are just converting the integral into an integral over the parameter $t$.
Then
$$\int_{\gamma}\omega\cdot \vec{dr} = \int_{0}^{1} \Big(t\sin(2\pi t) d(t\cos(2\pi t))+t\cos(2\pi t)d(t\sin(2\pi t))+t^2d(t^{2}) \Big)$$
This is the general method for evaluating Line integrals. (Also note that the curve lies "on" the paraboloid and not "in"), i.e.  it lies on the surface. In terms of Physics, think of $\omega$ as a Force and you are calculating the work done by the force in moving a particle of unit mass along the curve $\gamma$ .
