I have this vector field :
$\overrightarrow{F}$ = $x$$\overrightarrow{i}$ + $y$$\overrightarrow{j}$ + $2(1-z)$$\overrightarrow{k}$
And I want to calculate the flux integral over the surface $S_1$ :
$z = 1 - x^2 - y^2$ for $z > 0$
However, I'm not certain of how to apply the divergence theorem here and I reach different results depending on the process. My reasoning is that since $z > 0$, then the theorem cannot be directly applied. Given this, I tried to make a new surface $S_2$:
$1 = x^2 + y^2$ such that $S_1$ $\cup$ $S_2$ is closed (a paraboloid).
Then, would this following application be correct?
$\iint\limits_{S_1}{{\vec F\cdot d\vec S}} + \iint\limits_{S_2}{{\vec F\cdot d\vec S}}= \iiint\limits_{E}{{{\mathop{\rm div}\nolimits} \vec F\,dV}}\\$
Since ${\rm div}\vec F = \vec \nabla \cdot \vec F$ = $1 + 1 -2 = 0$ then we have:
$\iint\limits_{S_1}{{\vec F\cdot d\vec S}} + \iint\limits_{S_2}{{\vec F\cdot d\vec S}}= 0$
$\iint\limits_{S_1}{{\vec F\cdot d\vec S}} = -\iint\limits_{S_2}{{\vec F\cdot d\vec S}}$
$S_2$ is the unitary circle at $z = 0$, so $S_2: x^2 + y^2 = 1$
Therefore,
$\iint\limits_{S_1}{{\vec F\cdot d\vec S}} = -\iint\limits_{S_2}{{<x,y,2(1-z)> \cdot <2x,2y,0> dA}}$
$=-\iint\limits_{S_2}{{<x,y,2> \cdot <2x,2y,0> dA}}$
$=-2\iint\limits_{S_2}{{x^2 + y^2 dA}}$
$=-2$$\int_{0}^{2\pi}\int_{0}^1{{r^3 drd\theta}}$
$=-1/2$$\int_{0}^{2\pi}1 = -\pi$
But when I try to directly integrate for $S_1$ I get this:
$\iint\limits_{S_1}{{\vec F\cdot d\vec S}} = \iint\limits_{S_1}{{<x,y,2(1-1+x^2 + y^2)> \cdot <2x,2y,1> dA}} $
$= 4\iint\limits_{S_2}{x^2 + y^2dA}$
$= 4$$\int_{0}^{2\pi}\int_{0}^1{{r^3 drd\theta}} = 2\pi \neq -\pi$
I'm not too sure where I went wrong. Any help would be appreciated. Thanks.
Regardless of whether the question states $ \ z \gt 0$ or $z \geq 0, \ z = 1 - x^2 - y^2$ represents only the paraboloid surface and does not include disk at $z = 0$.
Paraboloid surface $S_1: z = 1- x^2 - y^2, 0 \lt z \lt 1$
To apply divergence theorem we close the paraboloid surface with a disk $ \ S2: x^2+y^2 \leq 1$ at $z=0$. We later on will have to subtract flux through the disk from the total flux to get flux through the paraboloid surface.
$\vec F = (x, y, 2-2z)$
So $\nabla \cdot \vec F = 1 + 1 -2 = 0$ and hence flux through the entire surface ($S_1 \cup S_2$) is zero.
Flux through surface $S_2$:
Outward Normal vector to the surface $S_2$ is $(0, 0, -1)$.
So flux through $S_2$ is given by,
$\displaystyle \iint_{x^2+y^2 \leq 1} (x, y, 2) \cdot (0, 0, -1) \ dx \ dy = - 2\pi$
So flux through $S_1$ is $2\pi$.