An Application of Stokes's Theorem Let $D^2=\{(x,y)\in \mathbf R^2: x^2+y^2\leq 1\}$ be the unit disc in $\mathbf R^2$, and $D^3=\{(x,y,z)\in \mathbf R^3: x^2+y^2+y^2\leq 1\}$ be the unit disc in $\mathbf R^3$.

Let $i_{\pm}:D^2\to D^3$ be two maps defined as
  $$
i_{\pm}(x, y)=\left(x, y, \pm\sqrt{(1-x^2-y^2)}\right)
$$
  Let $\omega$ be a closed $2$-form in a smooth manifold $M$, and $F:D^3\to M$ be a smooth map.
  Then I need to show that
  $$
\int_{D^2}(F\circ i_+)^*\omega = \int_{D^2} (F\circ i_-)^*\omega
$$
  where $D^2$ has the standard orientation governed by the form $dx\wedge dy$.

Attempt:
Since $\omega$ is a closed form, it seems that Stokes's Theorem might be useful.
Define $H_+=\{\left(x, y, \sqrt{1-x^2-y^2}\right): (x, y)\in D^2\}$ and $H_-=\{\left(x, y, -\sqrt{1-x^2-y^2}\right): (x,y)\in D^2\}$.
Now $H_+\cup D^2$ is the boundary of the "upper solid half unit disc" $D^3_+$ in $\mathbf R^3$ and $H_-\cup D^2$ is the boundary of the "lower solid half unit disc" $D^3_-$ in $\mathbf R^3$.
By Stokes we have
$$
\int_{H_+\cup D^2}i_+^*(F^* \omega)= \int_{D^3_+}d(F^* \omega))=\int_{D^3_+} F^* d\omega = 0
$$
Therefore
$$
\int_{D^2}i_+^*(F^*\omega) = -\int_{H_+} i_+^*(F^*\omega)
$$
Similarly
$$
\int_{D^2}i_-^*(F^*\omega) = -\int_{H_-} i_-^*(F^*\omega)
$$
How do I proceed from here?
Thanks.
 A: Comments:
Your method seems to be a bit too complicated. It's possible that part of the reason you seem to be confused is that you're writing $D^2$ for two different things: One, the disc of radius $1$ inside $\mathbb{R}^2$; this is the $D^2$ that appears in your statement of the problem and defintions etc. 
However, in the attempt you write up, you also use $D^2$ as shorthand for the set of points $\lbrace (x, y, 0) : x^2 + y^2 \leq 1 \rbrace$ in $D^3 \subseteq \mathbb{R}^3$. Considering this set at all is a little strange; it's more or less irrelevant to the problem.
I also wish to remark that the integral
$$
  \int_{H_+} i_+^*(F^* \omega)
$$
makes no sense. $i_+^*(F^* \omega)$ is a $2$-form on $D^2 \subseteq \mathbb{R}^2$, but $H_+ \subseteq D^3$. This might be related to the use of $D^2$ for two different things.
Proof outline:
The idea of the proof should be that the integral of $F^*(d\omega)$ over $D^3$ (which we know is zero) is equal to the integral over the boundary $\partial D^3$ (by Stokes); this is a sphere that we can decompose into the discs $i_+(D^2)$ and $i_-(D^2)$. The rest is just being careful about orientations to check that
\begin{align}
  \int_{\partial D^3} F^* \omega &= \int_{i_+(D^2)} F^*\omega + \int_{i_-(D^2)} F^* \omega \\ &= \int_{D^2} (F \circ i_+)^* \omega - \int_{D^2} (F \circ i_-)^* \omega.
\end{align}
(Assuming $\partial D^3$ is given the "outward facing" unit normal as its orientation, and $D^2 \subseteq \mathbb{R}^2$ the standard orientation, the map $i_+$ is orientation preserving and the map $i_-$ is orientation reversing.)
