How many cubic metres of fluid cross the upper hemisphere $x^{2}+y^{2}+z^{2}=1$, $z\ge0$ per second if the velocity of the flow is $\mathbf{u} = \mathbf{i}+x\mathbf{j}+z\mathbf{k}$ metres per second.
I don't really know how to start here, I know I have to parametrise the hemisphere but i'm not really sure how to, then I must find the surface normal to this and so on but I don't really know how to parametrise this surface.
On
Hint: Spherical coordinates: $${\bf x}(u,v) = (\cos u \cos v, \cos u \sin v, \sin u), \quad (u,v) \in ]0, \pi/2[ \times ]0, 2\pi[.$$
On
Since you already asked a similiar question just before, I'll tell you how to generally solve these, rather than give you an exact answer.
There is no general way of parametrizing, but most of the time, these vector calculus "homeworks" will use either spherical or cylindrical surfaces.
The coordinate transforms are as follows:
$$ \psi_s:\mathbb{R}^3\rightarrow\mathbb{R}^3,\ \psi_s(r,\vartheta,\varphi)=(r\sin\vartheta\cos\varphi,r\sin\vartheta\sin\varphi,r\cos\vartheta)$$ and $$ \psi_c:\mathbb{R}^3\rightarrow\mathbb{R}^3,\ \psi_c(\rho,\varphi,z)=(\rho\cos\varphi,\rho\sin\varphi,z) $$ Where $\psi_s$ is the transition function from spherical to cartesian and $\psi_c$ is the transition function from cylindrical to cartesian. Also here $r,\rho\ge 0$, $0\le\vartheta\le\pi$ and $0\le\varphi\le2\pi$.
Now, if you have a cylindrical or spherical surface, then you can often parametrize them by taking one of the coordinates from the $(r,\vartheta,\varphi)$ and $(\rho,\varphi,z)$ tuples constant.
In this case, you need to use the spherical coordinate system, and keep $r$ constant, but since you only need the upper hemisphere, you also have a restriction on $\vartheta$: $0\le\vartheta\le\pi/2$.
The general calculation of the surface normal is as follows. If you have a $$ \psi:\mathbb{R}^2\rightarrow\mathbb{R}^3,\ (u,v)\mapsto\psi(u,v) $$ embedding, then the $$ \frac{\partial\psi}{\partial u},\ \frac{\partial\psi}{\partial v} $$ vector fields will be tangent basis vector fields to the surface. Then, the $$ \mathbf{n}=\frac{\partial\psi}{\partial u}\times\frac{\partial\psi}{\partial v} $$ vector field will be a normal vector field to the surface.
In this case, the vectorial area element will be $$ d\mathbf{A}=\frac{\partial\psi}{\partial u}\times\frac{\partial\psi}{\partial v}dudv, $$ and the scalar area element will be $$ dA=\left\|\frac{\partial\psi}{\partial u}\times\frac{\partial\psi}{\partial v}\right\|dudv $$.
But note, that if your surface has some obvious geometric quality, as Simon S said above, then you can often calculate the normal by geometric intuition, rather than the laborous calculation of cross products.
Hint: The normal is in the radial direction, so it's just $\hat n = {1 \over \sqrt{x^2 + y^2 + z^2}}\langle x,y,z\rangle$.
Write down the surface integral over the hemisphere $S$ and it's ugly:
$$I = \iint_S {1 \over \sqrt{x^2 + y^2 + z^2}}\langle x,y,z\rangle \cdot \langle 1,x,z\rangle \ dA = \iint_S (x + xy + z^2) \ dA$$
To me at least, this suggests spherical polars aren't going to save us a pesky calculation. Who doesn't like nice, clean easy calculations? Hmm, how about Gauss's theorem. Looks promising as the divergence of the flow is just $\nabla\cdot\langle 1,x,z\rangle = 1$.
If you do use Gauss's theorem, don't forget to consider the second boundary of the hemisphere, the disk $D = \{ (x,y,z) \in \mathbb R^3 : x^2 + y^2 \leq 1, z = 0 \}$. It's normal is $-\hat z$.