I need to compute the integral: $$ \int_{\mathbb{S}^3}\alpha, $$ where $\alpha\in \Omega^3(\mathbb{R}^4)$ is given by $\alpha=-4t\;dx\wedge dy\wedge dz+4z\;dx\wedge dy\wedge dt-4y\;dx\wedge dz\wedge dt+4x\;dy\wedge dz\wedge dt$.
My problem is that I am getting two different results. If I apply the definition: $$ \begin{array}{rccl} \varphi^+:& \mathbb{S}^3\cap\{t \geq 0\}:=\mathbb{S}^{3+}&\longrightarrow&\mathbb{D}^3\\ &(x,y,z,t)&\longmapsto&(x,y,z) \end{array} $$ $$ \begin{array}{rccl} \varphi^-:& \mathbb{S}^{3-}\cap\{t \leq 0\}:=\mathbb{S}^{3-}&\longrightarrow&\mathbb{D}^3\\ &(x,y,z,t)&\longmapsto&(x,y,z) \end{array} $$ The inverses: $$ \begin{array}{rccl} (\varphi^+)^{-1}:& \mathbb{D}^3&\longrightarrow&\mathbb{S}^{3+}\\ &(u,v,w)&\longmapsto&(u,v,w,\sqrt{1-u^2-v^2-w^2}) \end{array} $$ $$ \begin{array}{rccl} (\varphi^-)^{-1}:& \mathbb{D}^3&\longrightarrow&\mathbb{S}^{3-}\\ &(u,v,w)&\longmapsto&(u,v,w,-\sqrt{1-u^2-v^2-w^2}) \end{array} $$ If $\beta=t\;dx\wedge dy\wedge dz$: $$ \int_{\mathbb{S}^3}\beta=\int_{\mathbb{S}^{3-}}\beta+\int_{\mathbb{S}^{3+}}\beta=\int_{\mathbb{D}^3}\left[(\varphi^-)^{-1}\right]^*(\beta)+\int_{\mathbb{D}^3}\left[(\varphi^+)^{-1}\right]^*(\beta) $$ $$=-\int_{\mathbb{D}^3}\sqrt{1-u^2-v^2-w^2}\;du\;dv\;dw+\int_{\mathbb{D}^3}\sqrt{1-u^2-v^2-w^2}\;du\;dv\;dw=0 $$ And also $0$ for the other terms.
But using Stokes theorem I have: $$ \int_{\mathbb{S}^3}\alpha=\int_{\partial(\mathbb{D}^4)}\alpha=\int_{\mathbb{D}^4}\partial\alpha=\int_{\mathbb{D}^4}16\;dx\wedge dy\wedge dz\wedge dt=16\text{Vol}(\mathbb{D}^4)\not=0 $$ Where is the mistake? Which one is correct?
The Stokes's Theorem application is correct. The mistake comes in half of the first part: With the boundary orientation on $\Bbb S^3$, the chart $\varphi^+$ is orientation-reversing. So you should get two positive terms for your integrals. (In your computation, in the $\varphi^-$ integral, $-4t\ge 0$, and in the $\varphi^+$ integral, you must change sign to compensate for the orientation reversal, and so $-(-4t)\ge 0$ once again.)