Integral of a differential 2-form in a parametrized surface I want to calculate the integral $\mathop{\iint}_{\phi}xdx\land dy+ydy\land dz+z dz\land dx$
considering $\left\{ \phi\left(u,v\right):=\left(u+v,uv,u^{2}-v^{2}\right):u,v\in\left[0,1\right]\right\} $
And the Stokes Theorem says that  integral of a differential form ω over the boundary of some orientable manifold Ω is equal to the integral of its exterior derivative dω over the whole of Ω.
I know that $d\left(Fdx+Gdy+Hdz\right)=\left(G_{x}-F_{y}\right)dx\land dy+\left(H_{y}-G_{z}\right)dy\land dz+\left(F_{z}-H_{x}\right)dz\land dx$
How should I do that? I only find examples of vector fields and I don't know if this should be converted to something similar to a vector field.
 A: The surface is parameterized as $(x, y, z) = (u+v, uv, u^2-v^2)$ where $u,v \in [0,1]$ giving
$$
dx = du + dv, \quad
dy = u \, dv + v \, du, \quad
dz = 2u \, du - 2v \, dv
$$
Let us take one term separately. 
For the first term the differential form becomes
$$
x \, dx \wedge dy 
= (u+v) \, (du+dv) \wedge (u \, dv + v \, du) 
= (u+v)(u-v) \, du \wedge dv
= (u^2-v^2) \, du \wedge dv
$$
resulting in the integral
$$
\iint_\phi x \, dx \wedge dy 
= \iint_{[0,1]^2} (u^2-v^2) \, du \wedge dv
= \int_0^1 \left( \int_0^1 (u^2-v^2) \, du \right) dv \\
= \int_0^1 \left[ \frac13 u^3-uv^2 \right]_{u=0}^1 dv 
= \int_0^1 \left( \frac13 - v^2 \right) dv 
= \left[ \frac13 v - \frac13 v^3 \right]_0^1 
= 0.
$$
For the second term the differential form becomes
$$
y \, dy \wedge dz
= uv \, (u \, dv + v \, du) \wedge (2u \, du - 2v \, dv)
= uv \, (2u^2 \, dv \wedge du - 2v^2 \, du \wedge dv) 
\\
= -2uv (u^2+v^2) \, du \wedge dv
= -2(u^3v + uv^3) \, du \wedge dv
$$
resulting in the integral
$$
\iint_\phi y \, dy \wedge dz
= \iint_{[0,1]^2} -2(u^3v + uv^3) \, du \wedge dv
= -2 \int_0^1 \left( \int_0^1 (u^3v + uv^3) \, du \right) \, dv \\
= -2 \int_0^1 \left[ \frac14 u^4v + \frac12 u^2v^3 \right]_0^1 \, dv
= -2 \int_0^1 \left( \frac14 v + \frac12 v^3 \right) \, dv
= -2 \cdot 0 = 0.
$$
For the third term the differential form becomes
$$
z \, dz \wedge dx
= (u^2-v^2) (2u \, du - 2v \, dv) \wedge (du+dv)
= (u^2-v^2) (2u \, du \wedge dv - 2v \, dv \wedge du)
= (u^2-v^2) 2(u+v) \, du \wedge dv
= 2(u^3+u^2v-uv^2-v^3) \, du \wedge dv
$$
resulting in the integral
$$
\iint_{\phi} z \, dz \wedge dx
= \iint_{[0,1]^2} 2(u^3+u^2v-uv^2-v^3) \, du \wedge dv
= \int_0^1 \left( \int_0^1 2(u^3+u^2v-uv^2-v^3) \, du \right) \, dv
= \int_0^1 \left[ 2(\frac14 u^4 + \frac13 u^3v - \frac12 u^2v^2 - uv^3) \right]_0^1 \, dv
= \int_0^1 2(\frac14 + \frac13 v - \frac12 v^2 - v^3) \, dv
= \left[ \frac14 v + \frac16 v^2 - \frac16 v^3 - \frac14 v^4 \right]_0^1 \\
= \frac14 + \frac16 - \frac16 - \frac14
= 0.
$$
Thus, the value of the integral is $0$.
