Compute the integral $\int_X d\omega,$ where $ \ \omega \ $ is a $1$-form on $X$ and $X$ is a $2$-segment in $\mathbb{R}^3$ defined as follows: $$ \omega=[e^{xz}+2y e^{x^2+y^2}] \, dx + [\ln (2+z)+3xe^{x^2+y^2}] \, dy+3x^2y \, dz$$ and $X:E \to \mathbb{R}^3$ be such that $X(r, \theta) = (r \cos\theta, r \sin\theta, 1-r^2)$ with $(r, \theta) \in [0,1] \times [0, 2\pi].$
Answer:
We can use the Stoke's theorem as given below: $$\int_X d\omega=\int_{\partial X} \omega,$$ where $\partial X$ is just the unit circle $x^2+y^2 = 1$ in the $xy$-plane ($z=0$).
Thus the above integral becomes $$ \int_{\partial X} \omega = \int_{\partial X} (1+2y) \, dx + (3x+\ln 2) \, dy,$$ since $z=0$ and $x^2+y^2=1.$
Am I right so far?
On $\partial X$ we have $\omega = (1+2ey)dx + (\ln 2 + 3ex)dy$. Note that if we set $D = \{(x,y,0): x^2+y^2\le 1\}$, oriented upwards (as, presumably, $X$ is oriented upwards, coming from the canonical orientation on $(0,1)\times (0,2\pi)$), then $\partial D = \partial X$ and $$\int_{\partial X}\omega = \int_{\partial D}\omega = \int_D d\omega = \int_D e\,dx\wedge dy = e\,\text{area}(D) = \pi e.$$
This same principle applies whenever we have oriented manifolds with boundary: If $\partial M = \partial N$, then $\int_M \omega = \int_N \omega$ for any exact form $\omega$.