Given $A(1,0,0),B(0,1,0),D(1,1,1)$ and functions $$f,g,h:\Bbb{R}^3\to \Bbb{R}, \ f(x,y,z)=x+e^{(x+y-z)^2},g(x,y,z)=-e^{(x+y-z)^2},h(x,y,z)=xy$$
Calculate $$\int\limits_\gamma f(x,y,z)dx+g(x,y,z)dy+h(x,y,z)dz\\\iint\limits_Sf(x,y,z)dydz+g(x,y,z)dzdx+h(x,y,z)dxdy$$ where $\gamma$ is oriented segment from $A$ to $B$ and $S$ is triangle $ADB$ oriented so that it's normal forms an acute angle with the positive part of $Oz$-axis.
My attempt:
For the first integral I've used the parametrisation $x=1-t,y=t,z=0$ where $t\in[0,1]$ we have $x_t' = -1,y_t' = 1$ so $$\int\limits_\gamma f(x,y,z)dx+g(x,y,z)dy+h(x,y,z)dz=\int_0^1(1-t+e^1)x_t'+(-e^1)y_t'=\int_0^1t-1-2e=-1/2-2e$$ For the second integral I've used the parametrisation $x=u,y=v,z=u+v-1$ where $0\leq u \leq 1,0\leq v\leq 1-u$, the normal vector is $\vec n = (-z_x',-z_y',1)=(-1,-1,1)$ and $\frac{\vec n}{||\vec n||} = (-\frac{1}{\sqrt3},-\frac{1}{\sqrt3},\frac{1}{\sqrt3})$ so the angle between $Oz$ is the $\theta$ such that $\cos(\theta) = \frac{1}{\sqrt3}$ so $\theta<\frac\pi2$ so we have $$\iint\limits_Sf(x,y,z)dydz+g(x,y,z)dzdx+h(x,y,z)dxdy=\int_0^1\int_0^{1-u}-1(u+e^1)-(-e^1)+1(uv) \ dvdu=\int_0^1\frac {u^3}2-\frac u2 du = -\frac18$$
This was one problem from the last year exam and it felt too easy so I'm concerned I made a mistake.