Problem while calculating the area of $S^2$ using differential forms I am trying to calculate the area of the semi-sphere $S^2_-$ using differential forms, I have the only local chart $(A,\phi,S^2_-)$ given by $A=\{(x,y)\in\mathbb{R}:x^2+y^2< 1\}$ and $\phi(x,y)={^t(-x, y, -\sqrt{1-x^2-y^2})}$.
Also I have the unitary normal vector field $N(x,y)={^t(-x, y, -\sqrt{1-x^2-y^2})}$, so that $\{N,\frac{\partial{\phi}}{\partial{x}},\frac{\partial{\phi}}{\partial{y}}\}$ form an oriented basis of $\mathbb{R}^3$.
Now if I calculate by hand $\phi^*(\iota(N)dx\wedge dy\wedge dz)$ I get
\begin{align}
&\phi^*(-xdy\wedge dz-y dx\wedge dz-\sqrt{1-x^2-y^2}dx\wedge dy)\\
&=-(-x)dy\wedge\left(\frac{x}{\sqrt{1-x^2-y^2}}dx+\frac{y}{\sqrt{1-x^2-y^2}}dy\right)\\
&\qquad-yd(-x)\wedge\left(\frac{x}{\sqrt{1-x^2-y^2}}dx+\frac{y}{\sqrt{1-x^2-y^2}}dy\right)\\
&\qquad-\sqrt{1-x^2-y^2}d(-x)\wedge dy\\
&=\left(\frac{-x^2}{\sqrt{1-x^2-y^2}}+\frac{y^2}{\sqrt{1-x^2-y^2}}+\sqrt{1-x^2-y^2}\right)dx\wedge dy
\end{align}
so $\int_{S^2_-}\iota(N)dx\wedge dy\wedge dz=\int_{A}\phi^*(\iota(N)dx\wedge dy\wedge dz)=\int_A \frac{1-2x^2}{\sqrt{1-x^2-y^2}}dxdy$.
At this point I know there is a mistake because last integral should be $\int_A \frac{1}{\sqrt{1-x^2-y^2}}dxdy$ but I really can't figure out where this mistake is.

There's an extra minus sign in the definition of the normal vector field: it should be $$\mathbf{N}=\left(x,y,-\sqrt{1-x^2-y^2}\right).$$ Indeed, consider the point $(1,0,0)$ of the sphere, whose normal vector should be $(1,0,0)$ and not $(-1,0,0)$ since the normal vector by definition points outward and not inward. After fixing that, letting $$\gamma=\iota(\mathbf{N})(dx\wedge dy\wedge dz)=x\,dy\wedge dz - y\,dx\wedge dz -\sqrt{1-x^2-y^2}\,dx\wedge dy$$ be the area form, by definition $$Area(S^2_{-}) = \int_{[S^2_{-}]} \gamma = \int_A \phi^{\star}(\gamma)$$ where \begin{align*} \phi^{\star} (\gamma) &= -x\, dy \wedge \left( \frac{x}{\sqrt{1-x^2-y^2}}dx + \frac{y}{\sqrt{1-x^2-y^2}}dy \right) \\ & \ \ \ \ - y\, d(-x) \wedge \left( \frac{x}{\sqrt{1-x^2-y^2}}dx + \frac{y}{\sqrt{1-x^2-y^2}}dy \right) \\ &\ \ \ \ -\sqrt{1-x^2-y^2}\, d(-x)\wedge dy \\ &= \left(\frac{x^2}{\sqrt{1-x^2-y^2}} +\frac{y^2}{\sqrt{1-x^2-y^2}} +\sqrt{1-x^2-y^2}\right)\,dx\wedge dy \\ &= \frac{1}{\sqrt{1-x^2-y^2}}\,dx\wedge dy\end{align*} and you can then proceed as you intended to.
