Given the vector field $F=((y-1)^2+z^2,(z^2-x^2),(y-1)^2)$ , Let $S$ be the surface of the sphere $x^2+y^2+z^2-2y=0$ which is above $xy$ meaning above the plane $z=0$ ' let $n$ be the unit vector normal outward the sphere ( pointing up) , show that the flux in the given direction is $0.25\pi$ $\iint_SF \cdot n dS=0.25\pi$ ( we need to solve using gauss and divergence theorem)
My try:
First I organized what is given:
our sphere and plane looks like this 
our field is $F=((y-1)^2+z^2,(z^2-x^2),(y-1)^2)$ and we have two equations
\begin{cases}
x^2+(y-1)^2+z^2=1\\
z=0
\end{cases}
since we need to solve using gauss and divergence theorem we have – $\iint_SF \cdot n dS=\iiint_V div(F) \cdot n dxdydz$
It is obvious that $div(F)=0$ , and we can also see that $S$ does not enclose the volume $V$ so we need to close it and on this we can apply the theorem.
we can close it with $D$ the projection on $Z=0$ and we get $\iint_S F \cdot ndS +\iint_D F \cdot ndD = \iiint_V div(F) \cdot n dV $ since the $div(F)=0$ we get that $\iint_S F \cdot ndS =- \iint_D F \cdot ndD$
Here I got stuck , I tried to calculate $\iint_D F \cdot ndD$ but I keep getting the wrong answer , assuming $D$ is the projection and according to the functions we have I got that it is the circle $x^2+(y-1)^2=1$ I tried using polar coordiantes where $0 \leq \theta \leq \pi$ and $0 \leq r \leq 1$ and I kept getting a different answer than $0.25 \pi$.
Can anyone give me any tips and hints? I want to understand this topic more it is interesting but I am having a hard time with it.
Thank you and sorry for the English mistakes hope The translations are understandable.
Best Answer
The spherical surface is $S1: x^2 + (y-1)^2 + z^2 = 1, z \geq 0$
Now applying divergence theorem, you correctly concluded that the net flux through the closed surface is zero. The disk that we use at $z= 0$ to close the surface is $S2: x^2 + (y-1)^2 \leq 1$.
We can parametrize $S2$ as $ \ x = r \cos\theta, y = 1 + r\sin\theta, 0 \leq r \leq 1, 0 \leq \theta \leq 2\pi$.
The outward normal vector to the disk is $(0, 0, -1)$ as it is in plane $z = 0$.
So $ \ \vec F \cdot \hat n = - (y-1)^2 = - r^2 \sin^2\theta$
The integral to find flux through the disk is,
$ \displaystyle \int_0^{2\pi} \int_0^1 - r^3 \sin^2\theta \ dr \ d\theta = - \frac{\pi}{4}$
We subtract this from the net flux through the closed surface to get flux through the spherical surface $S1$. Hence the answer is $ \displaystyle \frac{\pi}{4}$.