the surface F, the vector field v, the point P and the normal vector n are defined as follows:
$$ F = \vec{x}(s,t) = \begin{bmatrix}s \\t \\s^2+t^2 \end{bmatrix} \ s,t \in [0,1]$$ $$\vec{n} = \begin{bmatrix} 0 \\ 0 \\ -1 \end{bmatrix} in \ point \ P = \ (0,0,0)$$ $$\vec{v}(x,y,z) = \begin{bmatrix} xy^2 \\ x^2y \\ 1\end{bmatrix}$$
a) Calculate the flow $\int\int_F \vec{v}dF$ of the vector field $\vec{v}$ through $F$.
My solution attempt using the divergence theorem:
$\int_C<\vec{v},\vec{n}>ds = \int\int_\Sigma div\ \vec{v}\ dA$ with $div\ \vec{v} = y^2+x^2$ $\rightarrow \int_0^1\int_0^1x^2+y^2dydx=\frac{2}{3}$
c) Calculate the integral of the vector field $v$ along the boundary $\partial F$ of $F$.
My solution attempt using Stokes' theorem:
$\int_{\partial F}\vec{v} \ dx = \int\int_\Sigma<curl \ \vec{v}, \vec{n}>dA$ and since $curl \ \vec{v} = 0$ the integral is also equal to zero.
Parts of the solution where I am unsure: Is the surface F closed? How can one quickly check this? If it isn't closed, I cannot apply the divergence theorem and hence have to calculate the given integral. How would I do this? Why do they give me the parametrization of F, and why do they give me the normal vector only in the point P?
Here an approach for part a). We compute the flow $\phi$ through surface $F$. The flow $d \phi$ through a particular surface element $dS$ is $d \phi = \textbf{v} \cdot \textbf{n} dS$. We are given parameters for $F$:
$$ \textbf{x} = \begin{bmatrix}x \\y \\z \end{bmatrix} = \begin{bmatrix}s \\t \\s^2+t^2 \end{bmatrix} $$
as well as a flow field $\textbf{v}$:
$$ \textbf{v} = \begin{bmatrix}xy^2 \\x^2y \\1 \end{bmatrix} $$
which at the surface $F$ is equal to:
$$ \textbf{v} = \begin{bmatrix}st^2 \\s^2t \\1 \end{bmatrix} $$
From the parameters we can compute the differential surface element $dS$:
$$ dS = \lvert \textbf{r}_s \times \textbf{r}_t \rvert dsdt $$
where:
$$\textbf{r}_s = \frac{\partial x}{\partial s} \textbf{i} + \frac{\partial y}{\partial s} \textbf{j} + \frac{\partial z}{\partial s} \textbf{k}$$
$$= 1 \textbf{i} + 0 \textbf{j} + 2s \textbf{k}$$
and similarly for $\textbf{r}_t$ we get $0 \textbf{i} + 1 \textbf{j} + 2t \textbf{k}$. Therefore the cross product $\textbf{r}_s \times \textbf{r}_t$ is:
$$ \textbf{r}_s \times \textbf{r}_t = -2s \textbf{i} - 2t \textbf{j} + 1 \textbf{k} $$
Now the flow through $F$ is obtained from integrating the flow through each surface element:
$$ \phi = \int d \phi = \int \textbf{v} \cdot \textbf{n} \lvert \textbf{r}_s \times \textbf{r}_t \rvert dsdt $$
The normal vector $\textbf{n}$ is related to $\textbf{r}_s$ and $\textbf{r}_t$ as follows:
$$ \textbf{n} = -\frac{\textbf{r}_s \times \textbf{r}_t }{\lvert \textbf{r}_s \times \textbf{r}_t \rvert} $$
which coincides with the given normal vector at $s,t = 0$. And so:
$$ \phi = \int_S \textbf{v} \cdot (-1)\frac{\textbf{r}_s \times \textbf{r}_t }{\lvert \textbf{r}_s \times \textbf{r}_t \rvert} \lvert \textbf{r}_s \times \textbf{r}_t \rvert dsdt = -\int_S (\textbf{r}_s \times \textbf{r}_t) \cdot \textbf{v} dsdt $$
$$ = \int_S \begin{bmatrix}2s \\2t \\-1 \end{bmatrix} \cdot \begin{bmatrix}st^2 \\s^2 t \\1 \end{bmatrix} dsdt $$
Which means:
$$ \phi = \int_S 4s^2 t^2 - 1 ds dt $$
$$ = \int_0^1 \int_0^1 4s^2 t^2 - 1 ds dt = -\frac{5}{9}$$