Compute $\iint\limits_S\vec{F}\cdot\vec{n}dS$

Question

The question I've been given is:

Compute $\iint\limits_S\vec{F}\cdot\vec{n}dS$ where $\vec{F}(x,y,z)=(x,y,z)$ for the surface/parametrisation in 1a).

In question 1a) I've calculated the parameterisation of surface $x^2-y^2+z^2=0$ as

$\vec{r}(\alpha,\beta)=(\alpha cos\beta, \alpha, \alpha sin\beta)$ with $\alpha\in [0,1], \beta \in [0,2\Pi]$.

I don't know what exactly this question is asking me to compute.

Answer 1

Actually , the question is to evaluate surface integral over the surface , you've parameterized in $1(a)$. So,

The formula is $\iint\limits_S\vec{F}\cdot\vec{n}dS=\iint\limits_S\vec{F}\cdot\vec{n}$ $|\vec{r_{\alpha}}\times \vec{r_{\beta}}| \cdot d\alpha d\beta$ , where $S:\{\vec{r}(\alpha,\beta)=(\alpha \cos\beta, \alpha, \alpha \sin\beta):\alpha\in [0,1], \beta \in [0,2\pi] \}$

Here, $\vec{n}$ is the outward normal to the surface $S$ can be given by $\frac{\vec{r_{\alpha}}\times \vec{r_{\beta}}}{|\vec{r_{\alpha}}\times \vec{r_{\beta}}| }$ .

You have $\vec{F}=(x,y,z)$ and , here you can use $x=\alpha \cos\beta$ , $y=\alpha$ ,$z=\alpha \sin\beta$

Luckily , $\vec{F}\cdot\vec{n}=0$ (here)

Hence,

$\iint\limits_S\vec{F}\cdot\vec{n}dS=0$.

EDIT: $ \vec{r_{\alpha}}$=$\frac{\partial \vec{r}}{\partial \alpha}$