Evaluate the flux integral:
$\int_S F \cdot \hat n dS$
where $F = <y,x,z>$, S is the surface $z = 1- x^2 - y^2$ for $0 \leq z \leq 1$ and $n$ is the upward unit normal to the surface.
Now I know what the shape looks like and so deduce that the unit normal to the surface is:
$\hat n = <2x, 2y, 1>$
And so the surface integral becomes:
$\int_S <y,x,1-x^2-y^2> \cdot <2x, 2y, 1> = \int_S (4xy + 1 - x^2 - y^2) $
However from this point I am unsure how to proceed. We havent been thought divergence theorem.
Is there a way to convert the dS to a dA? Also I am unsure how it can be dS (i.e. d(w.r.t a surface) when there is only one integral? (i.a.w my question is why is there a surface 'd' when there is only a single integral since surfaces usually require 2)?
Any help is much appreciated. Also we haven't been thought the divergence theorem. Is there another way to do this? Perhaps by changing of variables or something?
The only thing you need is to find bounds of integration over $x$ and $y$.
if $z=0\to x^2+y^2=1$ that is a disc of radius one. If $z=1\to x^2+y^2=0$ that implies $x=0=y$. So the bounds of integration over $x$ is
$$x\in [-\sqrt{1-y^2}, +\sqrt{1-y^2}]$$
After performing integration over $x$, then perform the integration over $y$ for the bounds $$y\in [-1,1]$$.