I am asked to evaluate $$\iint_{S} [ \nabla \phi \times \nabla \psi] \bullet n dS$$ where $\phi=(x+y+z)^2$ and $\psi=x^2-y^2+z^2$ where S is the curved surface of the hemisphere $x^2+y^2+z^2=1$ , $z \ge 0$.
My attempts:
I tried using that if$ \bar A= \phi \nabla \psi$ then $\nabla \times \bar A = \nabla \phi \times \nabla \psi$
So I calculated $$\phi \nabla \psi=(2x(x+y+z)^2,-2y(x+y+z)^2,2z(x+y+z)^2)$$
So essentially according to above this is what I need to integrate over S.
For my normal I thought this would be obtained by writing $$G(x,y,z)=1-x^2-y^2-z^2$$ and solving for the gradient and dividing by its norm. Doing this i get that $$\hat n=\frac{(x)i+(y)j+(z)k}{x^2+y^2+z^2}$$
For the dS, I thought if it is of the form $z=f(x,y)$ then we can obtain the dS as $\sqrt{1+f_{x}^2+f_{y}^2}$ ie $dS=\frac{1}{\sqrt{x^2+y^2}}$
But I have no clue if I am on the right path. I feel like I am not confident on it. Can anyone help? Any advice would be appreciated, I have been trying for several hours with no progress
Using Stokes' Theorem along with the product rule $\nabla \times (\phi \nabla \psi)=\nabla \phi \times \nabla \psi+\phi \nabla \times (\nabla \psi)$ and the identity $\nabla \times (\nabla \psi)=0$, we have
$$ \int_S (\nabla \phi \times \nabla \psi) \cdot \hat n\,dS=\oint_C \phi \nabla \psi \cdot d\vec \ell $$
where $C$ is the closed curve, traversed according to the right-hand rule, that bounds the $S$ oriented with unit normal $\hat n$.
Transforming to cylindrical coordinates $(\rho,\phi,z)$, and recognizing that on $C$, $\rho =1$ and $z=0$, we have $\phi(x,y,z) =(\sin(\phi)+\cos(\phi))^2$, $\nabla \psi(x,y,z) =2(\hat x\cos(\phi)-\hat y\sin(\phi))$, and $d\vec \ell=-\hat x\sin(\phi)+\hat y\cos(\phi)$. Then, the contour integral of interest becommes
$$\begin{align} \oint_C \phi \nabla \psi \cdot d\vec \ell&=-4\int_0^{2\pi}(\sin(\phi)+\cos(\phi))^2\,\sin(\phi)\cos(\phi)\,d\phi\\\\ &=-8\int_0^{2\pi}\sin^2(\phi)\cos^2(\phi)\,d\phi\\\\ &=-2\pi \end{align}$$
And we are done!