integration - Showing that a line integral along a given curve is in independent of one the curve components

Let $\gamma(t)=(\gamma_1(t),\gamma_2(t),\gamma_3(t))$, $t \in [0, 2 \pi]$, be a smooth curve in $\Bbb R^3$ where
$$\gamma_1(t)=cos(t),\,\gamma_2(t)=sin(t),\,\gamma_3(t)>0$$
Let $F$ be the vector field $F(x,y,z)=(2y^2,\,x^2,\,3z^2)$.
Prove that the line integral $\int_{\gamma} F \cdot dl$ is independent of $\gamma_3(t)$.

Let me emphasize that according to this problem, $\gamma_3(t)$ is not necessarily a closed curve.

I've tried to go be the definition of $\int_{\gamma} F \cdot dl$ and got:

\begin{align}
\int_{\gamma} F \cdot dl & = \int_0^{2 \pi} (\,2cos^2(t), \, sin^2(t),\, 3\gamma_3^2(t) \,) \cdot (\,-sin(t),\,cos(t),\,\gamma_3^{'}(t)\,) \, dt \\
& =\int_0^{2 \pi} (-2cos^2(t)sin(t)+cos(t)sin^2(t)+3\gamma_3^2(t) \gamma_3^{'}(t)) \,dt \\
& =\int_0^{2 \pi}3\gamma_3^2(t) \gamma_3^{'}(t) \,dt \\ & =\gamma_3^3(t)|_0^{2 \pi}
\end{align}

I don't know if the author of this problem forgot to mention that $\gamma_3(t)$ is a closed curve or not, because if $\gamma_3(t)$ is indeed closed the argument above solves the problem.

Do you have an idea of how to solve this problem without this information?

Answer

I think that the fact that the integral is independent on $\gamma_3$ means, that if we have another curve through the same end points, i.e. if we have $\overline \gamma_3$ such that $\gamma_3(0)=\overline\gamma_3(0)$ and $\gamma_3(2\pi)=\overline\gamma_3(2\pi)$, then the integrals are equal if $\overline\gamma_3$ replaces $\gamma_3$. And your computation reveals that this is the case for your integral, since $$\int _\gamma F\cdot dl=\gamma_3^3(2\pi)-\gamma_3^3(0)
=\overline\gamma_3^3(2\pi)-\overline\gamma_3^3(0) =\int _{\overline \gamma} F\cdot dl$$
where $\overline \gamma=(\gamma_1,\gamma_2,\overline\gamma_3)$.