Calculate $\int_\phi \omega$, where:
$\phi(t): [0,1] \to \mathbb{R}^3$
$\phi(t) = \Big( 2 + 16t^2(t-1)^2, 4t(1-t), (1-t) \big( 1 + 16t^2(t-1)(2t-1) \big) \Big)$
$\omega = \big( 2xz^2 e^{x^2 + y^2} + 2xz \big)dx + \big( 2yz^2 e^{x^2 + y^2} + 2yz \big)dy + \big( 2z e^{x^2 + y^2} + x^2 + y^2 \big)dz$
For: $F(x, y, z) = z^2 e^{x^2 + y^2} + z(x^2 + y^2)$ we have that: $\omega = dF = \frac{\partial F}{\partial x} dx + \frac{\partial F}{\partial y} dy + \frac{\partial F}{\partial z} dz$
Therefore, from Gradient theorem: $\int_\phi \omega = \int_\phi \frac{\partial F}{\partial x} dx + \frac{\partial F}{\partial y} dy + \frac{\partial F}{\partial z} dz = \int_0^1 \Big( \frac{\partial F}{\partial x} \frac{dx}{dt} + \frac{\partial F}{\partial y} \frac{dy}{dt} + \frac{\partial F}{\partial z} \frac{dz}{dt} \Big) dt = F(\phi(1)) - F(\phi(0))$
$\phi(0) = (2, 0, 1)$ and $\phi(1) = (2, 0, 0)$
From that: $$\int_\phi \omega = z^2 e^{x^2 + y^2} + z(x^2 + y^2) = 0 - (e^4 + 4) = -e^4 - 4$$
Is that correct?