Calculating the flux of the vector field + Gauss 
Calculate the flux $\int_BF · ν dS$ of the vector field $F : \mathbb{R}^3 \to \mathbb{R}^3, F(x, y, z) := (yz, xz, x^2)^T$ through the northern hemisphere B of radius 1 with 0 as its center, i.e. the surface
$B := \{(x, y, z) \in \mathbb{R}^3: x^2 + y^2 + z^2 = 1, z > 0\}$

This is what I got so far:
$z=\sqrt{1-x^2-y^2}$
$vdS=(\frac{x}{\sqrt{1-x^2-y^2}},\frac{y}{\sqrt{1-x^2-y^2}},1)$
$\int_{-1}^1 \int_{-\sqrt{1-x^2}}^\sqrt{1-x^2}(yz, xz, x^2)*vdS\;dydx$
$\int_{-1}^1 \int_{-\sqrt{1-x^2}}^\sqrt{1-x^2} y\sqrt{1-x^2-y^2},x\sqrt{1-x^2-y^2}, x^2 (\frac{x}{\sqrt{1-x^2-y^2}},\frac{y}{\sqrt{1-x^2-y^2}},1) \ dy\ dx\\
\int_{-1}^1 \int_{-\sqrt{1-x^2}}^\sqrt{1-x^2} x^2+2xy\; dy dx\\
\int_{-1}^1  2x^2\sqrt{1-x^2} dx\\
\frac\pi 2*\frac1 4(\frac\pi 2 * \frac1 4) = \frac \pi 4$
EDIT: I replaced the last two lines, but for some reason the answer doesn't match the one given in this post, which I am sure is correct, so if anyone sees my mistake and can kindly point it out, it would be great! found
Is this correct? And I also want to know how can I calculate using the following steps:
(i) Draw B and find an embedding of B.
(ii) Identify the outer normal vector ν on B.
(iii) Finally, calculate $\int_BF · ν dS.$
What do the first two steps add if I can calculate the flux directly?
I thank anyone who can help!
EDIT.NR2 : If I use Gauss theorem to find the flux is it possible for it to be different?
 A: Using spherical coordinates
$ \hat{n} dS = ( \sin \theta \cos \phi, \sin \theta \sin \phi, \cos \theta) \sin \theta \ d\theta \ d \phi $
$ \vec{F} = ( \sin \theta \cos \theta \sin \phi, \sin \theta \cos \theta \cos \phi, \sin^2 \theta \cos^2 \phi ) $
So
$ \vec{F} \cdot \hat{n} dS = (\sin^2 \theta \cos \theta \cos \phi \sin \phi + \sin^2 \theta \cos \theta \cos \phi \sin \phi + \sin^2 \theta \cos \theta \cos^2 \phi ) \sin \theta \ d\theta \ d\phi$
And this simplifies to
$ \vec{F} \cdot \hat{n} dS = \sin^3 \theta \cos \theta (\sin(2 \phi) + \cos^2(\phi) ) \ d\theta \ d\phi$
Integrating with respect to $\phi$ first, from $\phi = 0 $ to $\phi = 2 \pi$, we get the term $\sin(2 \phi) $ cancelled out, and the term $\cos^2 \phi$ contributing a fator of $ \pi $
i.e.
$\displaystyle \int_B \vec{F} \cdot \hat{n} dS = \pi \int_{\theta = 0}^{\dfrac{\pi}{2}} \sin^3 \theta \cos \theta \ d\theta = \dfrac{\pi}{4}$
A: Part (i) is probably to familiarize yourself with visualizing surfaces. In the figure, $B$ is the transparent hemisphere (orange) and its embedding in the $(x,y)$ plane is the ruled disk shown in the plane $z=0$ (light blue).

Part (ii) is a necessary step in computing the surface integral because the surface element $d\vec S$ depends on the normal vector. But when it comes to computing the normal to $B$, you can choose either $\vec s_u\times \vec s_v$ or $\vec s_v\times\vec s_u$, where $\vec s(u,v)$ is a vector-valued function that parameterizes $B$. Your choice affects the sign of the surface element and integral. The figure from part (i) can help decide which is the correct choice.

And yes, you can confirm your result with Gauss' theorem.
Note that the theorem applies to closed surfaces, and $B$ is not closed. So suppose we close it up by connecting to $B$ a simple region, say, the disk $D:=\{(x,y)\in\Bbb R^2\mid x^2+y^2\le1\}$. Now apply the theorem to $B\cup D$:
$$\iint_{B\cup D} \vec F\cdot \vec\nu\,d\vec S = \iiint_{\operatorname{int}(B\cup D)} \underbrace{\nabla\cdot \vec F}_{=0} \, dV = 0$$
which means $\iint_B=-\iint_D$. In the region $D$, we have $\vec F=\langle0,0,x^2\rangle$ and the outer normal is $\vec\nu=\langle0,0,-1\rangle$, so the flux across $D$ is
$$\iint_D \vec F\cdot\vec\nu \, d\vec S = \int_{-1}^1 \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \langle0,0,x^2\rangle\cdot\langle0,0,-1\rangle\,dy\,dx$$
and multiplying by $-1$ makes this agree with your integral.
