Our professor defines for an $n-1$ dimensional surface $M$ in $\mathbb R^n$, given $\hat n$ a unit normal vector field on $M$, one can define the surface form to be the $n-1$ form $dS(v_1,\cdots , v_{n-1})=\det \begin{bmatrix} \hat n &v_1 &\cdots&v_{n-1}\end{bmatrix}$
My question and confusion arise from a specific problem, to find the surface integral of the lower hemisphere $M=\{(x,y,z)|x^2+y^2+z^2=r^2, z\leq 0\}$ oriented outwards (downwards). I started by using parametrization $\varphi(x,y)=(x,y,-\sqrt{r^2 -x^2-y^2})$. Then I found that normal vector given by $\vec v=D\varphi (e_1) \times D\varphi (e_2)$ is orientation reversing. So $$\int_M dS= \int_{\bar{D}_1(0)} \det\begin{bmatrix} -\frac{\vec v}{||{\vec v}||} & D\varphi(e_1) & D\varphi(e_2)\end{bmatrix}$$. But from this approach the answer I found was negative, which differs by a sign with the model answer. What did I do wrong?
Expanding on the question, if I have a surface parametrized as followed: $\varphi (x_1,...,x_n)=(x_1,...,x_n,f(x_1,...x_n))\in \mathbb R^{n+1}$, that matches with a desired orientation. Then consider $\psi(x_2,x_1,...,x_n)=\varphi(x_1,...,x_n)$, clearly $\psi$ reverses the orientation, so the surface integral using $psi$ should be the negative of the former. But in both integrals, the integrand is $\sqrt{||\nabla f||^2+1}$. Where does the negative sign cancel out?
You're only getting a negative sign because you're not paying attention to the $2$-form when you do the integral. You're just doing the double integral of a scalar function. Recall that the convention is that $$\int_\Omega f(x)dx_1\wedge\dots\wedge dx_n = \int_\Omega f\,dV$$ when we give $\Omega$ the standard orientation as a region in $\Bbb R^n$. As you pointed out, the parametrization of the lower hemisphere by the disk in the $xy$-plane is orientation-reversing, so you must pull back the area $2$-form $dS$ on the hemisphere and integrate it as a $2$-form $f\, dy\wedge dx=-f\,dx\wedge dy$ and then do the double integration. (It's always a point of confusion that you can do iterated integrals in any order by Fubini, but you must get the sign right at the beginning by having the form agree with the endowed orientation on the submanifold.)
To make this explicit, think about what happens when you analogously calculate the arclength of the upper and lower semicircles in $\Bbb R^2$. There the orientation issue is quite clear.